I'm checking if an byte array contains a another Byte Array with this Code:
private int IndexOf(int index, byte[] AllBytes, byte[] searchByteArray)
{
for (int i = index; i <= AllBytes.Length - 1 - searchByteArray.Length - 1; i++)
{
for (int j = 0; j <= searchByteArray.Length - 1; j++)
{
if (AllBytes[i + j] == searchByteArray[j])
{
if (j + 1 == searchByteArray.Length)
return i;
}
else
break;
}
}
return -1;
}
That works perfect and I get the index of the first byte back.
But the problem is I want to check very large Data.
My "big" Array contains around 900000000 Bytes and my searchArray about 10-20 Bytes. In that way my function is very very slow. Is there a way to make a better performance?
Thanks.
You could adapt the Boyer–Moore string-search algorithm to bytes.
First, you create an array int[256] where the array indexes correspond to all possible byte values. The array contains the positions of each byte value in the search pattern starting from the end, and the search pattern length for values not appearing in the search pattern.
Then you compare the last position of the pattern with a position in the input array. If the values do not match, you can advance the search position by the value in the table found at the index equal to the value in the input array. If they match, compare the pattern with the input.
Example
intput: 100 206 002 250 123 075 074 109 184 222
search pattern: 200 109 100 150 123
^
|
The table
...
[108] = 5
[109] = 3
[110] = 5
...
the value 123 does not match 109, therefore you look up the table at position 109 and get 3. I.e. you can increase the search position by 3, so that the two 109 values line up. If the value was 108, you could have moved by 5 positions (the full search pattern length).
The link above explains the algorithm in more detail.
I needed something similar. What I needed to do was determine if an array of bytes is present within a different array of bytes. I did not need to know how many times one array appeared within the other array or even the position.
My search for a quick solution which performs well didn't yield any acceptable results. I've seen the Boyer–Moore string-search algorithm as well as other supposedly fast search algorithms suggested as answers to virtually every question on the subject I found yet I have been unable to find a single implementation of one of those algorithms which is actually faster than unoptimized brute code.
So I decided to go with brute force since the code is vastly superior than the Boyer–Moore implementations. Being a little on the obsessive side I did make a few basic optimizations to the brute force approach. I present it here for others who happen upon this in the future.
In my testing my optimizations resulted in code which executes very close to 100% faster when tested against purely pseudo-random data. My code is not designed to allow you to find all the occurrences of one array within another but it can be easily modified to do so. I'll leave that to the reader.
My code:
/// <summary>
/// Determines whether a byte array contains the specified sequence of bytes.
/// </summary>
/// <param name="caller">The byte array to be searched.</param>
/// <param name="array">The byte to be found.</param>
/// <returns>The first location of the sequence within the array, -1 if the sequence is not found.</returns>
/// <exception cref="ArgumentNullException"></exception>
/// <exception cref="ArgumentException"></exception>
public static int Contains(this byte[] caller, byte[] array)
{
byte startValue, endValue;
int result, arrayLength, searchBoundary, j, startLocation, endOffset;
if (caller == null)
throw new ArgumentNullException($"{nameof(caller)}");
if (array == null)
throw new ArgumentNullException($"{nameof(array)}");
if (caller.Length == 0 || array.Length == 0)
throw new ArgumentException($"Argument {(caller.Length == 0 ? nameof(caller) : nameof(array))} is empty.");
if (array.Length > caller.Length)
return -1;
startValue = array[0];
arrayLength = array.Length;
if (arrayLength > 1)
{
result = -1;
endValue = array[^1];
endOffset = arrayLength - 1;
searchBoundary = caller.Length - arrayLength;
startLocation = -1;
while ((startLocation = Array.IndexOf(caller, startValue, startLocation + 1)) >= 0)
{
if (startLocation > searchBoundary)
break;
if (caller[startLocation + endOffset] == endValue)
{
for (j = 1; j < endOffset && caller[startLocation + j] == array[j]; j++) { }
if (j == endOffset)
{
result = startLocation;
break;
}
}
}
}
else
{
result = Array.IndexOf(caller, startValue);
}
return result;
}
Also the version of the original posters code I used for test purposes:
public static int IndexOf(this byte[] AllBytes, int index, byte[] searchByteArray)
{
for (int i = index; i <= AllBytes.Length - 1 - searchByteArray.Length - 1; i++)
{
for (int j = 0; j <= searchByteArray.Length - 1; j++)
{
if (AllBytes[i + j] == searchByteArray[j])
{
if (j + 1 == searchByteArray.Length)
return i;
}
else
break;
}
}
return -1;
}
And finally my code which tests the functions:
private static void TestContains()
{
bool copyContained;
int result, offset;
byte[] containedValue, containingValue;
Random rng;
Stopwatch watch1, watch2;
List<Tuple<byte[], byte[], bool, int>> testList;
testList = new();
watch1 = new();
watch2 = new();
rng = new();
for (int j = 0; j < 120; j++)
{
Console.SetCursorPosition(0, 0);
Console.WriteLine($"Test number {j:#,0} in progress...");
for (int i = 32; i < 40000; i++)
{
offset = 0;
copyContained = rng.Next(0, 1000) != 0;
containingValue = new byte[i];
rng.NextBytes(containingValue);
containedValue = new byte[rng.Next(8, i)];
if (copyContained)
{
offset = rng.Next(0, i - containedValue.Length);
rng.NextBytes(containedValue);
Array.Copy(containedValue, 0, containingValue, offset, containedValue.Length);
}
testList.Add(new Tuple<byte[], byte[], bool, int>(containingValue, containedValue, copyContained, offset));
}
foreach (var testCase in testList)
{
watch1.Start();
result = testCase.Item1.Contains(testCase.Item2);
watch1.Stop();
watch2.Start();
testCase.Item1.IndexOf(0, testCase.Item2);
watch2.Stop();
if ((testCase.Item1.Length == 0 && result != -1) ||
(testCase.Item2.Length == 0 && result != -1) ||
(testCase.Item3 && result != testCase.Item4) ||
(!testCase.Item3 && result != -1))
Debugger.Break();
}
testList.Clear();
//CollectGarbage(0);
}
Console.WriteLine(
"\r\n\r\n" + $#"Comparison duration (using Contains): {watch1.Elapsed:hh\:mm\:ss\.ff}" + "\r\n" +
"\r\n" + $#"Comparison duration (using Contains2): {watch2.Elapsed:hh\:mm\:ss\.ff}" + "\r\n" +
"\r\n" + $#"Speed differential: {(watch1.Elapsed > watch2.Elapsed ? $"Contains2 is {((double)watch1.ElapsedTicks / watch2.ElapsedTicks - 1) * 100:#.00}%" : $"Contains is {((double)watch2.ElapsedTicks / watch1.ElapsedTicks - 1) * 100:#.00}%")} faster." +
"\r\n\r\n\r\n Press enter to continue.");
Console.ReadLine();
}
While a 100% improvement isn't going to make the difference in some cases, it's an improvement nonetheless and a pretty significant one. On the old desktop I am testing this on, a Core i5-4570, I can find the position of the array I am searching for in a 1 GB array in less than a second. I didn't specifically test large arrays as you can see because my use case is for smaller arrays, less than 65 KB.
Given a point collection defined by x and y coordinates.
In this collection I get the start point, the end point and all the other n-2 points.
I have to find the shortest way between the start point and end point by going through all the other points. The shortest way is defined by its value and if possible the crossing point order.
At a first look this seems to be a graph problem, but i am not so sure about that right now, any way i am trying to find this shortest way by using only geometric relations since currently all the information that i have is only the x and y coordinates of the points, and which point is the start point and which is the end point.
My question is, can this way be found by using only geometric relations?
I am trying to implement this in C#, so if some helpful packages are available please let me know.
The simplest heuristic with reasonable performance is 2-opt. Put the points in an array, with the start point first and the end point last, and repeatedly attempt to improve the solution as follows. Choose a starting index i and an ending index j and reverse the subarray from i to j. If the total cost is less, then keep this change, otherwise undo it. Note that the total cost will be less if and only if d(p[i - 1], p[i]) + d(p[j], p[j + 1]) > d(p[i - 1], p[j]) + d(p[i], p[j + 1]), so you can avoid performing the swap unless it's an improvement.
There are a possible number of improvements to this method. 3-opt and k-opt consider more possible moves, resulting in better solution quality. Data structures for geometric search, kd-trees for example, decrease the time to find improving moves. As far as I know, the state of the art in local search algorithms for TSP is Keld Helsgaun's LKH.
Another family of algorithms is branch and bound. These return optimal solutions. Concorde (as far as I know) is the state of the art here.
Here's a Java implementation of the O(n^2 2^n) DP that Niklas described. There are many possible improvements, e.g., cache the distances between points, switch to floats (maybe), reorganize the iteration so that subsets are enumerating in increasing order of size (to allow only the most recent layer of minTable to be retained, resulting in a significant space saving).
class Point {
private final double x, y;
Point(double x, double y) {
this.x = x;
this.y = y;
}
double distanceTo(Point that) {
return Math.hypot(x - that.x, y - that.y);
}
public String toString() {
return x + " " + y;
}
}
public class TSP {
public static int[] minLengthPath(Point[] points) {
if (points.length < 2) {
throw new IllegalArgumentException();
}
int n = points.length - 2;
if ((1 << n) <= 0) {
throw new IllegalArgumentException();
}
byte[][] argMinTable = new byte[1 << n][n];
double[][] minTable = new double[1 << n][n];
for (int s = 0; s < (1 << n); s++) {
for (int i = 0; i < n; i++) {
int sMinusI = s & ~(1 << i);
if (sMinusI == s) {
continue;
}
int argMin = -1;
double min = points[0].distanceTo(points[1 + i]);
for (int j = 0; j < n; j++) {
if ((sMinusI & (1 << j)) == 0) {
continue;
}
double cost =
minTable[sMinusI][j] +
points[1 + j].distanceTo(points[1 + i]);
if (argMin < 0 || cost < min) {
argMin = j;
min = cost;
}
}
argMinTable[s][i] = (byte)argMin;
minTable[s][i] = min;
}
}
int s = (1 << n) - 1;
int argMin = -1;
double min = points[0].distanceTo(points[1 + n]);
for (int i = 0; i < n; i++) {
double cost =
minTable[s][i] +
points[1 + i].distanceTo(points[1 + n]);
if (argMin < 0 || cost < min) {
argMin = i;
min = cost;
}
}
int[] path = new int[1 + n + 1];
path[1 + n] = 1 + n;
int k = n;
while (argMin >= 0) {
path[k] = 1 + argMin;
k--;
int temp = s;
s &= ~(1 << argMin);
argMin = argMinTable[temp][argMin];
}
path[0] = 0;
return path;
}
public static void main(String[] args) {
Point[] points = new Point[20];
for (int i = 0; i < points.length; i++) {
points[i] = new Point(Math.random(), Math.random());
}
int[] path = minLengthPath(points);
for (int i = 0; i < points.length; i++) {
System.out.println(points[path[i]]);
System.err.println(points[i]);
}
}
}
The Euclidean travelling salesman problem can be reduced to this and it's NP-hard. So unless your point set is small or you have a very particular structure, you should probably look out for an approximation. Note that the Wikipedia article mentions the existence of a PTAS for the problem, which could turn out to be quite effective in practice.
UPDATE: Since your instances seem to have only few nodes, you can use a simple exponential-time dynamic programming approach. Let f(S, p) be the minimum cost to connect all the points in the set S, ending at the points p. We have f({start}, start) = 0 and we are looking for f(P, end), where P is the set of all points. To compute f(S, p), we can check all potential predecessors of p in the tour, so we have
f(S, p) = MIN(q in S \ {p}, f(S \ {p}, q) + distance(p, q))
You can represent S as a bitvector to save space (just use an single-word integer for maximum simplicity). Also use memoization to avoid recomputing subproblem results.
The runtime will be O(2^n * n^2) and the algorithm can be implemented with a rather low constant factor, so I predict it to be able to solve instance with n = 25 within seconds a reasonable amount of time.
This can be solved using an evolutionary algorithm.
Look at this: http://johnnewcombe.net/blog/post/23
You might want to look at TPL (Task Parallel Library) to speed up the application.
EDIT
I found this Link which has a Traveling Salesman algorithm:
http://msdn.microsoft.com/en-us/magazine/gg983491.aspx
The Source Code is said to be at:
http://archive.msdn.microsoft.com/mag201104BeeColony
I'm trying to get my head around a problem of identifying the best match of English words from a dictionary file to a given string.
For example ("lines" being a List of dictionary words):
string testStr = "cakeday";
for (int x= 0; x<= testStr.Length; x++)
{
string test = testStr.Substring(x);
if (test.Length > 0)
{
string test2 = testStr.Remove(counter);
int count = (from w in lines where w.Equals(test) || w.Equals(test2) select w).Count();
Console.WriteLine("Test: {0} / {1} : {2}", test, test2, count);
}
}
Gives the output:
Test: cakeday / : 0
Test: akeday / c : 1
Test: keday / ca : 0
Test: eday / cak : 0
Test: day / cake : 2
Test: ay / caked : 1
Test: y / cakeda : 1
Obviously "day / cake" is the best fit for the string however if I were to introduce a 3rd word into the string e.g "cakedaynow" it doesnt work so well.
I know the example is primitive, its more a proof of concept and was wondering if anyone had any experience with this type of string analysis?
Thanks!
You'll want to research the class of algorithms appropriate to what you're trying to do. Start with Approximate string matching on Wikipedia.
Also, here's a Levenshtein Edit Distance implementation in C# to get you started:
using System;
namespace StringMatching
{
/// <summary>
/// A class to extend the string type with a method to get Levenshtein Edit Distance.
/// </summary>
public static class LevenshteinDistanceStringExtension
{
/// <summary>
/// Get the Levenshtein Edit Distance.
/// </summary>
/// <param name="strA">The current string.</param>
/// <param name="strB">The string to determine the distance from.</param>
/// <returns>The Levenshtein Edit Distance.</returns>
public static int GetLevenshteinDistance(this string strA, string strB)
{
if (string.IsNullOrEmpty(strA) && string.IsNullOrEmpty(strB))
return 0;
if (string.IsNullOrEmpty(strA))
return strB.Length;
if (string.IsNullOrEmpty(strB))
return strA.Length;
int[,] deltas; // matrix
int lengthA;
int lengthB;
int indexA;
int indexB;
char charA;
char charB;
int cost; // cost
// Step 1
lengthA = strA.Length;
lengthB = strB.Length;
deltas = new int[lengthA + 1, lengthB + 1];
// Step 2
for (indexA = 0; indexA <= lengthA; indexA++)
{
deltas[indexA, 0] = indexA;
}
for (indexB = 0; indexB <= lengthB; indexB++)
{
deltas[0, indexB] = indexB;
}
// Step 3
for (indexA = 1; indexA <= lengthA; indexA++)
{
charA = strA[indexA - 1];
// Step 4
for (indexB = 1; indexB <= lengthB; indexB++)
{
charB = strB[indexB - 1];
// Step 5
if (charA == charB)
{
cost = 0;
}
else
{
cost = 1;
}
// Step 6
deltas[indexA, indexB] = Math.Min(deltas[indexA - 1, indexB] + 1, Math.Min(deltas[indexA, indexB - 1] + 1, deltas[indexA - 1, indexB - 1] + cost));
}
}
// Step 7
return deltas[lengthA, lengthB];
}
}
}
Why not:
Check all the strings inside the search word extracting from current search position to all possible lengths of the string and extract all discovered words. E.g.:
var list = new List<string>{"the", "me", "cat", "at", "theme"};
const string testStr = "themecat";
var words = new List<string>();
var len = testStr.Length;
for (int x = 0; x < len; x++)
{
for(int i = (len - 1); i > x; i--)
{
string test = testStr.Substring(x, i - x + 1);
if (list.Contains(test) && !words.Contains(test))
{
words.Add(test);
}
}
}
words.ForEach(n=> Console.WriteLine("{0}, ",n));//spit out current values
Output:
theme, the, me, cat, at
Edit
Live Scenario 1:
For instance let's say you want to always choose the longest word in a jumbled sentence, you could read from front forward, reducing the amount of text read till you are through. Using a dictionary makes it much easier, by storing the indexes of the discovered words, we can quickly check to see if we have stored a word containing another word we are evaluating before.
Example:
var list = new List<string>{"the", "me", "cat", "at", "theme", "crying", "them"};
const string testStr = "themecatcryingthem";
var words = new Dictionary<int, string>();
var len = testStr.Length;
for (int x = 0; x < len; x++)
{
int n = len > 28 ? 28 : len;//assuming 28 is the maximum length of an english word
for(int i = (n - 1); i > x; i--)
{
string test = testStr.Substring(x, i - x + 1);
if (list.Contains(test))
{
if (!words.ContainsValue(test))
{
bool found = false;//to check if there's a shorter item starting from same index
var key = testStr.IndexOf(test, x, len - x);
foreach (var w in words)
{
if (w.Value.Contains(test) && w.Key != key && key == (w.Key + w.Value.Length - test.Length))
{
found = true;
}
}
if (!found && !words.ContainsKey(key)) words.Add(key, test);
}
}
}
}
words.Values.ToList().ForEach(n=> Console.WriteLine("{0}, ",n));//spit out current values
Output:
theme, cat, crying, them
I have the following homework problem:
There are many ways one can implement the strcmp() function.
Note that strcmp(str1,str2) returns a negative number if str1 is alphabetically above str2, 0 if both are equal and postiveve if str2
is alphabetically above str1.
In it can be implemented in C as follows:
int mystrcmp(const char *s1, const char *s2)
{
while (*s1==*s2)
{
if(*s1=='\0')
return(0);
s1++;
s2++;
}
return(*s1-*s2);
}
So now I want to implement it in C# without using any of the built in methods of .NET. How can I accomplish this?
To avoid using any of the methods available in .NET or the BCL, you'd have to avoid the Length property of string (as properties are implemented by one or two methods). And you'd have to avoid the [] indexer property as well, for the same reason.
So you're pretty stuffed.
You make the assumption that the numeric character code indicates a human-significant sort ordering. It doesn't - the character codes in C# are Unicode, which has a lot of alphabets in it, some of which use a mixture of the Western alphabet (low values) with their own additional characters (high values).
So either you can reproduce a vast amount of character set information in your own code, so you know how to order two characters from Unicode, or you need to call a method in the BCL.
Get a copy of .NET Reflector and inspect how the Compare()/CompareTo() methods of System.String and System.Globalization.CompareInfo are implemented in mscorlib.
one way could be like this. Code is edited based on comments...
public static int mystrcmp(string st1, string st2)
{
int iST1 = 0, iST2=0;
for (int i = 0; i < (st1.Length > st2.Length ? st1.Length : st2.Length); i++)
{
iST1 += (i >= st1.Length ? 0 : st1[i]) - (i >= st2.Length ? 0 : st2[i]);
if (iST2 < 0)
{
if (iST1 < 0)
iST2 += iST1;
if (iST1 > 0)
iST2 += -iST1;
}
else
{
iST2 += iST1;
}
}
return iST2;
}
Don't use char* if you want to do this. Char* is unicode, you need ascii.
Your best bet would be using byte*. Then you can use the algorithm you currently have.
Calculate the Levenshtein distance between the two strings. and return that...
Here is a .net implementation of the Levenshtein distance from dot net Pearls:
using System;
/// <summary>
/// Contains approximate string matching
/// </summary>
static class LevenshteinDistance
{
/// <summary>
/// Compute the distance between two strings.
/// </summary>
/// <param name=s>The first of the two strings.</param>
/// <param name=t>The second of the two strings.</param>
/// <returns>The Levenshtein cost.</returns>
public static int Compute(string s, string t)
{
int n = s.Length;
int m = t.Length;
int[,] d = new int[n + 1, m + 1];
// Step 1
if (n == 0)
{
return m;
}
if (m == 0)
{
return n;
}
// Step 2
for (int i = 0; i <= n; d[i, 0] = i++)
{
}
for (int j = 0; j <= m; d[0, j] = j++)
{
}
// Step 3
for (int i = 1; i <= n; i++)
{
//Step 4
for (int j = 1; j <= m; j++)
{
// Step 5
int cost = (t[j - 1] == s[i - 1]) ? 0 : 1;
// Step 6
d[i, j] = Math.Min(
Math.Min(d[i - 1, j] + 1, d[i, j - 1] + 1),
d[i - 1, j - 1] + cost);
}
}
// Step 7
return d[n, m];
}
}
class Program
{
static void Main()
{
Console.WriteLine(LevenshteinDistance.Compute("aunt", "ant"));
Console.WriteLine(LevenshteinDistance.Compute("Sam", "Samantha"));
Console.WriteLine(LevenshteinDistance.Compute("flomax", "volmax"));
}
}
Write it as you would in C, but using array subscript notation, instead of pointer notation.
Increment the index.
How do you convert a numerical number to an Excel column name in C# without using automation getting the value directly from Excel.
Excel 2007 has a possible range of 1 to 16384, which is the number of columns that it supports. The resulting values should be in the form of excel column names, e.g. A, AA, AAA etc.
Here's how I do it:
private string GetExcelColumnName(int columnNumber)
{
string columnName = "";
while (columnNumber > 0)
{
int modulo = (columnNumber - 1) % 26;
columnName = Convert.ToChar('A' + modulo) + columnName;
columnNumber = (columnNumber - modulo) / 26;
}
return columnName;
}
If anyone needs to do this in Excel without VBA, here is a way:
=SUBSTITUTE(ADDRESS(1;colNum;4);"1";"")
where colNum is the column number
And in VBA:
Function GetColumnName(colNum As Integer) As String
Dim d As Integer
Dim m As Integer
Dim name As String
d = colNum
name = ""
Do While (d > 0)
m = (d - 1) Mod 26
name = Chr(65 + m) + name
d = Int((d - m) / 26)
Loop
GetColumnName = name
End Function
You might need conversion both ways, e.g from Excel column adress like AAZ to integer and from any integer to Excel. The two methods below will do just that. Assumes 1 based indexing, first element in your "arrays" are element number 1.
No limits on size here, so you can use adresses like ERROR and that would be column number 2613824 ...
public static string ColumnAdress(int col)
{
if (col <= 26) {
return Convert.ToChar(col + 64).ToString();
}
int div = col / 26;
int mod = col % 26;
if (mod == 0) {mod = 26;div--;}
return ColumnAdress(div) + ColumnAdress(mod);
}
public static int ColumnNumber(string colAdress)
{
int[] digits = new int[colAdress.Length];
for (int i = 0; i < colAdress.Length; ++i)
{
digits[i] = Convert.ToInt32(colAdress[i]) - 64;
}
int mul=1;int res=0;
for (int pos = digits.Length - 1; pos >= 0; --pos)
{
res += digits[pos] * mul;
mul *= 26;
}
return res;
}
Sorry, this is Python instead of C#, but at least the results are correct:
def ColIdxToXlName(idx):
if idx < 1:
raise ValueError("Index is too small")
result = ""
while True:
if idx > 26:
idx, r = divmod(idx - 1, 26)
result = chr(r + ord('A')) + result
else:
return chr(idx + ord('A') - 1) + result
for i in xrange(1, 1024):
print "%4d : %s" % (i, ColIdxToXlName(i))
I discovered an error in my first post, so I decided to sit down and do the the math. What I found is that the number system used to identify Excel columns is not a base 26 system, as another person posted. Consider the following in base 10. You can also do this with the letters of the alphabet.
Space:.........................S1, S2, S3 : S1, S2, S3
....................................0, 00, 000 :.. A, AA, AAA
....................................1, 01, 001 :.. B, AB, AAB
.................................... …, …, … :.. …, …, …
....................................9, 99, 999 :.. Z, ZZ, ZZZ
Total states in space: 10, 100, 1000 : 26, 676, 17576
Total States:...............1110................18278
Excel numbers columns in the individual alphabetical spaces using base 26. You can see that in general, the state space progression is a, a^2, a^3, … for some base a, and the total number of states is a + a^2 + a^3 + … .
Suppose you want to find the total number of states A in the first N spaces. The formula for doing so is A = (a)(a^N - 1 )/(a-1). This is important because we need to find the space N that corresponds to our index K. If I want to find out where K lies in the number system I need to replace A with K and solve for N. The solution is N = log{base a} (A (a-1)/a +1). If I use the example of a = 10 and K = 192, I know that N = 2.23804… . This tells me that K lies at the beginning of the third space since it is a little greater than two.
The next step is to find exactly how far in the current space we are. To find this, subtract from K the A generated using the floor of N. In this example, the floor of N is two. So, A = (10)(10^2 – 1)/(10-1) = 110, as is expected when you combine the states of the first two spaces. This needs to be subtracted from K because these first 110 states would have already been accounted for in the first two spaces. This leaves us with 82 states. So, in this number system, the representation of 192 in base 10 is 082.
The C# code using a base index of zero is
private string ExcelColumnIndexToName(int Index)
{
string range = string.Empty;
if (Index < 0 ) return range;
int a = 26;
int x = (int)Math.Floor(Math.Log((Index) * (a - 1) / a + 1, a));
Index -= (int)(Math.Pow(a, x) - 1) * a / (a - 1);
for (int i = x+1; Index + i > 0; i--)
{
range = ((char)(65 + Index % a)).ToString() + range;
Index /= a;
}
return range;
}
//Old Post
A zero-based solution in C#.
private string ExcelColumnIndexToName(int Index)
{
string range = "";
if (Index < 0 ) return range;
for(int i=1;Index + i > 0;i=0)
{
range = ((char)(65 + Index % 26)).ToString() + range;
Index /= 26;
}
if (range.Length > 1) range = ((char)((int)range[0] - 1)).ToString() + range.Substring(1);
return range;
}
This answer is in javaScript:
function getCharFromNumber(columnNumber){
var dividend = columnNumber;
var columnName = "";
var modulo;
while (dividend > 0)
{
modulo = (dividend - 1) % 26;
columnName = String.fromCharCode(65 + modulo).toString() + columnName;
dividend = parseInt((dividend - modulo) / 26);
}
return columnName;
}
Easy with recursion.
public static string GetStandardExcelColumnName(int columnNumberOneBased)
{
int baseValue = Convert.ToInt32('A');
int columnNumberZeroBased = columnNumberOneBased - 1;
string ret = "";
if (columnNumberOneBased > 26)
{
ret = GetStandardExcelColumnName(columnNumberZeroBased / 26) ;
}
return ret + Convert.ToChar(baseValue + (columnNumberZeroBased % 26) );
}
I'm surprised all of the solutions so far contain either iteration or recursion.
Here's my solution that runs in constant time (no loops). This solution works for all possible Excel columns and checks that the input can be turned into an Excel column. Possible columns are in the range [A, XFD] or [1, 16384]. (This is dependent on your version of Excel)
private static string Turn(uint col)
{
if (col < 1 || col > 16384) //Excel columns are one-based (one = 'A')
throw new ArgumentException("col must be >= 1 and <= 16384");
if (col <= 26) //one character
return ((char)(col + 'A' - 1)).ToString();
else if (col <= 702) //two characters
{
char firstChar = (char)((int)((col - 1) / 26) + 'A' - 1);
char secondChar = (char)(col % 26 + 'A' - 1);
if (secondChar == '#') //Excel is one-based, but modulo operations are zero-based
secondChar = 'Z'; //convert one-based to zero-based
return string.Format("{0}{1}", firstChar, secondChar);
}
else //three characters
{
char firstChar = (char)((int)((col - 1) / 702) + 'A' - 1);
char secondChar = (char)((col - 1) / 26 % 26 + 'A' - 1);
char thirdChar = (char)(col % 26 + 'A' - 1);
if (thirdChar == '#') //Excel is one-based, but modulo operations are zero-based
thirdChar = 'Z'; //convert one-based to zero-based
return string.Format("{0}{1}{2}", firstChar, secondChar, thirdChar);
}
}
Same implementation in Java
public String getExcelColumnName (int columnNumber)
{
int dividend = columnNumber;
int i;
String columnName = "";
int modulo;
while (dividend > 0)
{
modulo = (dividend - 1) % 26;
i = 65 + modulo;
columnName = new Character((char)i).toString() + columnName;
dividend = (int)((dividend - modulo) / 26);
}
return columnName;
}
int nCol = 127;
string sChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol >= 26)
{
int nChar = nCol % 26;
nCol = (nCol - nChar) / 26;
// You could do some trick with using nChar as offset from 'A', but I am lazy to do it right now.
sCol = sChars[nChar] + sCol;
}
sCol = sChars[nCol] + sCol;
Update: Peter's comment is right. That's what I get for writing code in the browser. :-) My solution was not compiling, it was missing the left-most letter and it was building the string in reverse order - all now fixed.
Bugs aside, the algorithm is basically converting a number from base 10 to base 26.
Update 2: Joel Coehoorn is right - the code above will return AB for 27. If it was real base 26 number, AA would be equal to A and the next number after Z would be BA.
int nCol = 127;
string sChars = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol > 26)
{
int nChar = nCol % 26;
if (nChar == 0)
nChar = 26;
nCol = (nCol - nChar) / 26;
sCol = sChars[nChar] + sCol;
}
if (nCol != 0)
sCol = sChars[nCol] + sCol;
..And converted to php:
function GetExcelColumnName($columnNumber) {
$columnName = '';
while ($columnNumber > 0) {
$modulo = ($columnNumber - 1) % 26;
$columnName = chr(65 + $modulo) . $columnName;
$columnNumber = (int)(($columnNumber - $modulo) / 26);
}
return $columnName;
}
Just throwing in a simple two-line C# implementation using recursion, because all the answers here seem far more complicated than necessary.
/// <summary>
/// Gets the column letter(s) corresponding to the given column number.
/// </summary>
/// <param name="column">The one-based column index. Must be greater than zero.</param>
/// <returns>The desired column letter, or an empty string if the column number was invalid.</returns>
public static string GetColumnLetter(int column) {
if (column < 1) return String.Empty;
return GetColumnLetter((column - 1) / 26) + (char)('A' + (column - 1) % 26);
}
Although there are already a bunch of valid answers1, none get into the theory behind it.
Excel column names are bijective base-26 representations of their number. This is quite different than an ordinary base 26 (there is no leading zero), and I really recommend reading the Wikipedia entry to grasp the differences. For example, the decimal value 702 (decomposed in 26*26 + 26) is represented in "ordinary" base 26 by 110 (i.e. 1x26^2 + 1x26^1 + 0x26^0) and in bijective base-26 by ZZ (i.e. 26x26^1 + 26x26^0).
Differences aside, bijective numeration is a positional notation, and as such we can perform conversions using an iterative (or recursive) algorithm which on each iteration finds the digit of the next position (similarly to an ordinary base conversion algorithm).
The general formula to get the digit at the last position (the one indexed 0) of the bijective base-k representation of a decimal number m is (f being the ceiling function minus 1):
m - (f(m / k) * k)
The digit at the next position (i.e. the one indexed 1) is found by applying the same formula to the result of f(m / k). We know that for the last digit (i.e. the one with the highest index) f(m / k) is 0.
This forms the basis for an iteration that finds each successive digit in bijective base-k of a decimal number. In pseudo-code it would look like this (digit() maps a decimal integer to its representation in the bijective base -- e.g. digit(1) would return A in bijective base-26):
fun conv(m)
q = f(m / k)
a = m - (q * k)
if (q == 0)
return digit(a)
else
return conv(q) + digit(a);
So we can translate this to C#2 to get a generic3 "conversion to bijective base-k" ToBijective() routine:
class BijectiveNumeration {
private int baseK;
private Func<int, char> getDigit;
public BijectiveNumeration(int baseK, Func<int, char> getDigit) {
this.baseK = baseK;
this.getDigit = getDigit;
}
public string ToBijective(double decimalValue) {
double q = f(decimalValue / baseK);
double a = decimalValue - (q * baseK);
return ((q > 0) ? ToBijective(q) : "") + getDigit((int)a);
}
private static double f(double i) {
return (Math.Ceiling(i) - 1);
}
}
Now for conversion to bijective base-26 (our "Excel column name" use case):
static void Main(string[] args)
{
BijectiveNumeration bijBase26 = new BijectiveNumeration(
26,
(value) => Convert.ToChar('A' + (value - 1))
);
Console.WriteLine(bijBase26.ToBijective(1)); // prints "A"
Console.WriteLine(bijBase26.ToBijective(26)); // prints "Z"
Console.WriteLine(bijBase26.ToBijective(27)); // prints "AA"
Console.WriteLine(bijBase26.ToBijective(702)); // prints "ZZ"
Console.WriteLine(bijBase26.ToBijective(16384)); // prints "XFD"
}
Excel's maximum column index is 16384 / XFD, but this code will convert any positive number.
As an added bonus, we can now easily convert to any bijective base. For example for bijective base-10:
static void Main(string[] args)
{
BijectiveNumeration bijBase10 = new BijectiveNumeration(
10,
(value) => value < 10 ? Convert.ToChar('0'+value) : 'A'
);
Console.WriteLine(bijBase10.ToBijective(1)); // prints "1"
Console.WriteLine(bijBase10.ToBijective(10)); // prints "A"
Console.WriteLine(bijBase10.ToBijective(123)); // prints "123"
Console.WriteLine(bijBase10.ToBijective(20)); // prints "1A"
Console.WriteLine(bijBase10.ToBijective(100)); // prints "9A"
Console.WriteLine(bijBase10.ToBijective(101)); // prints "A1"
Console.WriteLine(bijBase10.ToBijective(2010)); // prints "19AA"
}
1 This generic answer can eventually be reduced to the other, correct, specific answers, but I find it hard to fully grasp the logic of the solutions without the formal theory behind bijective numeration in general. It also proves its correctness nicely. Additionally, several similar questions link back to this one, some being language-agnostic or more generic. That's why I thought the addition of this answer was warranted, and that this question was a good place to put it.
2 C# disclaimer: I implemented an example in C# because this is what is asked here, but I have never learned nor used the language. I have verified it does compile and run, but please adapt it to fit the language best practices / general conventions, if necessary.
3 This example only aims to be correct and understandable ; it could and should be optimized would performance matter (e.g. with tail-recursion -- but that seems to require trampolining in C#), and made safer (e.g. by validating parameters).
I wanted to throw in my static class I use, for interoping between col index and col Label. I use a modified accepted answer for my ColumnLabel Method
public static class Extensions
{
public static string ColumnLabel(this int col)
{
var dividend = col;
var columnLabel = string.Empty;
int modulo;
while (dividend > 0)
{
modulo = (dividend - 1) % 26;
columnLabel = Convert.ToChar(65 + modulo).ToString() + columnLabel;
dividend = (int)((dividend - modulo) / 26);
}
return columnLabel;
}
public static int ColumnIndex(this string colLabel)
{
// "AD" (1 * 26^1) + (4 * 26^0) ...
var colIndex = 0;
for(int ind = 0, pow = colLabel.Count()-1; ind < colLabel.Count(); ++ind, --pow)
{
var cVal = Convert.ToInt32(colLabel[ind]) - 64; //col A is index 1
colIndex += cVal * ((int)Math.Pow(26, pow));
}
return colIndex;
}
}
Use this like...
30.ColumnLabel(); // "AD"
"AD".ColumnIndex(); // 30
private String getColumn(int c) {
String s = "";
do {
s = (char)('A' + (c % 26)) + s;
c /= 26;
} while (c-- > 0);
return s;
}
Its not exactly base 26, there is no 0 in the system. If there was, 'Z' would be followed by 'BA' not by 'AA'.
if you just want it for a cell formula without code, here's a formula for it:
IF(COLUMN()>=26,CHAR(ROUND(COLUMN()/26,1)+64)&CHAR(MOD(COLUMN(),26)+64),CHAR(COLUMN()+64))
In Delphi (Pascal):
function GetExcelColumnName(columnNumber: integer): string;
var
dividend, modulo: integer;
begin
Result := '';
dividend := columnNumber;
while dividend > 0 do begin
modulo := (dividend - 1) mod 26;
Result := Chr(65 + modulo) + Result;
dividend := (dividend - modulo) div 26;
end;
end;
A little late to the game, but here's the code I use (in C#):
private static readonly string _Alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
public static int ColumnNameParse(string value)
{
// assumes value.Length is [1,3]
// assumes value is uppercase
var digits = value.PadLeft(3).Select(x => _Alphabet.IndexOf(x));
return digits.Aggregate(0, (current, index) => (current * 26) + (index + 1));
}
In perl, for an input of 1 (A), 27 (AA), etc.
sub excel_colname {
my ($idx) = #_; # one-based column number
--$idx; # zero-based column index
my $name = "";
while ($idx >= 0) {
$name .= chr(ord("A") + ($idx % 26));
$idx = int($idx / 26) - 1;
}
return scalar reverse $name;
}
Though I am late to the game, Graham's answer is far from being optimal. Particularly, you don't have to use the modulo, call ToString() and apply (int) cast. Considering that in most cases in C# world you would start numbering from 0, here is my revision:
public static string GetColumnName(int index) // zero-based
{
const byte BASE = 'Z' - 'A' + 1;
string name = String.Empty;
do
{
name = Convert.ToChar('A' + index % BASE) + name;
index = index / BASE - 1;
}
while (index >= 0);
return name;
}
More than 30 solutions already, but here's my one-line C# solution...
public string IntToExcelColumn(int i)
{
return ((i<16926? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + (i<2730? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + (i<26? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + ((char)((i%26)+65)));
}
After looking at all the supplied Versions here, I decided to do one myself, using recursion.
Here is my vb.net Version:
Function CL(ByVal x As Integer) As String
If x >= 1 And x <= 26 Then
CL = Chr(x + 64)
Else
CL = CL((x - x Mod 26) / 26) & Chr((x Mod 26) + 1 + 64)
End If
End Function
Refining the original solution (in C#):
public static class ExcelHelper
{
private static Dictionary<UInt16, String> l_DictionaryOfColumns;
public static ExcelHelper() {
l_DictionaryOfColumns = new Dictionary<ushort, string>(256);
}
public static String GetExcelColumnName(UInt16 l_Column)
{
UInt16 l_ColumnCopy = l_Column;
String l_Chars = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
String l_rVal = "";
UInt16 l_Char;
if (l_DictionaryOfColumns.ContainsKey(l_Column) == true)
{
l_rVal = l_DictionaryOfColumns[l_Column];
}
else
{
while (l_ColumnCopy > 26)
{
l_Char = l_ColumnCopy % 26;
if (l_Char == 0)
l_Char = 26;
l_ColumnCopy = (l_ColumnCopy - l_Char) / 26;
l_rVal = l_Chars[l_Char] + l_rVal;
}
if (l_ColumnCopy != 0)
l_rVal = l_Chars[l_ColumnCopy] + l_rVal;
l_DictionaryOfColumns.ContainsKey(l_Column) = l_rVal;
}
return l_rVal;
}
}
Here is an Actionscript version:
private var columnNumbers:Array = ['A', 'B', 'C', 'D', 'E', 'F' , 'G', 'H', 'I', 'J', 'K' ,'L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];
private function getExcelColumnName(columnNumber:int) : String{
var dividend:int = columnNumber;
var columnName:String = "";
var modulo:int;
while (dividend > 0)
{
modulo = (dividend - 1) % 26;
columnName = columnNumbers[modulo] + columnName;
dividend = int((dividend - modulo) / 26);
}
return columnName;
}
JavaScript Solution
/**
* Calculate the column letter abbreviation from a 1 based index
* #param {Number} value
* #returns {string}
*/
getColumnFromIndex = function (value) {
var base = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'.split('');
var remainder, result = "";
do {
remainder = value % 26;
result = base[(remainder || 26) - 1] + result;
value = Math.floor(value / 26);
} while (value > 0);
return result;
};
These my codes to convert specific number (index start from 1) to Excel Column.
public static string NumberToExcelColumn(uint number)
{
uint originalNumber = number;
uint numChars = 1;
while (Math.Pow(26, numChars) < number)
{
numChars++;
if (Math.Pow(26, numChars) + 26 >= number)
{
break;
}
}
string toRet = "";
uint lastValue = 0;
do
{
number -= lastValue;
double powerVal = Math.Pow(26, numChars - 1);
byte thisCharIdx = (byte)Math.Truncate((columnNumber - 1) / powerVal);
lastValue = (int)powerVal * thisCharIdx;
if (numChars - 2 >= 0)
{
double powerVal_next = Math.Pow(26, numChars - 2);
byte thisCharIdx_next = (byte)Math.Truncate((columnNumber - lastValue - 1) / powerVal_next);
int lastValue_next = (int)Math.Pow(26, numChars - 2) * thisCharIdx_next;
if (thisCharIdx_next == 0 && lastValue_next == 0 && powerVal_next == 26)
{
thisCharIdx--;
lastValue = (int)powerVal * thisCharIdx;
}
}
toRet += (char)((byte)'A' + thisCharIdx + ((numChars > 1) ? -1 : 0));
numChars--;
} while (numChars > 0);
return toRet;
}
My Unit Test:
[TestMethod]
public void Test()
{
Assert.AreEqual("A", NumberToExcelColumn(1));
Assert.AreEqual("Z", NumberToExcelColumn(26));
Assert.AreEqual("AA", NumberToExcelColumn(27));
Assert.AreEqual("AO", NumberToExcelColumn(41));
Assert.AreEqual("AZ", NumberToExcelColumn(52));
Assert.AreEqual("BA", NumberToExcelColumn(53));
Assert.AreEqual("ZZ", NumberToExcelColumn(702));
Assert.AreEqual("AAA", NumberToExcelColumn(703));
Assert.AreEqual("ABC", NumberToExcelColumn(731));
Assert.AreEqual("ACQ", NumberToExcelColumn(771));
Assert.AreEqual("AYZ", NumberToExcelColumn(1352));
Assert.AreEqual("AZA", NumberToExcelColumn(1353));
Assert.AreEqual("AZB", NumberToExcelColumn(1354));
Assert.AreEqual("BAA", NumberToExcelColumn(1379));
Assert.AreEqual("CNU", NumberToExcelColumn(2413));
Assert.AreEqual("GCM", NumberToExcelColumn(4823));
Assert.AreEqual("MSR", NumberToExcelColumn(9300));
Assert.AreEqual("OMB", NumberToExcelColumn(10480));
Assert.AreEqual("ULV", NumberToExcelColumn(14530));
Assert.AreEqual("XFD", NumberToExcelColumn(16384));
}
Sorry, this is Python instead of C#, but at least the results are correct:
def excel_column_number_to_name(column_number):
output = ""
index = column_number-1
while index >= 0:
character = chr((index%26)+ord('A'))
output = output + character
index = index/26 - 1
return output[::-1]
for i in xrange(1, 1024):
print "%4d : %s" % (i, excel_column_number_to_name(i))
Passed these test cases:
Column Number: 494286 => ABCDZ
Column Number: 27 => AA
Column Number: 52 => AZ
For what it is worth, here is Graham's code in Powershell:
function ConvertTo-ExcelColumnID {
param (
[parameter(Position = 0,
HelpMessage = "A 1-based index to convert to an excel column ID. e.g. 2 => 'B', 29 => 'AC'",
Mandatory = $true)]
[int]$index
);
[string]$result = '';
if ($index -le 0 ) {
return $result;
}
while ($index -gt 0) {
[int]$modulo = ($index - 1) % 26;
$character = [char]($modulo + [int][char]'A');
$result = $character + $result;
[int]$index = ($index - $modulo) / 26;
}
return $result;
}
Another VBA way
Public Function GetColumnName(TargetCell As Range) As String
GetColumnName = Split(CStr(TargetCell.Cells(1, 1).Address), "$")(1)
End Function
Here's my super late implementation in PHP. This one's recursive. I wrote it just before I found this post. I wanted to see if others had solved this problem already...
public function GetColumn($intNumber, $strCol = null) {
if ($intNumber > 0) {
$intRem = ($intNumber - 1) % 26;
$strCol = $this->GetColumn(intval(($intNumber - $intRem) / 26), sprintf('%s%s', chr(65 + $intRem), $strCol));
}
return $strCol;
}