I have number list like below and I should check a condition to get most suitable match.
List<int> numbers= new List<int>();
numbers.Add(1000);
numbers.Add(3000);
numbers.Add(5500);
numbers.Add(7000);
If I send a value to check it should check like below examples
Scenario 1:
If I send a value less than or equal 1000 to check, it should return 1000
Scenario 2:
If I send a value between 1001 - 3000 to check, it should return 3000
Scenario 3:
If I send a value between 3001 - 5500 to check, it should return 5500
Scenario 4:
If I send a value between 5501 - 7000 to check, it should return 7000
Scenario 5:
If I send a value above 7000 to check, it should return nothing.
Can I do this with Linq? or what is the most efficient way to do this?
Update: the numbers in maxCheckPoint is dynamic and it can be any values. So we cannot hard coded and check
You can do this with LINQ:
int input = 1000;
int? result = numbers
.OrderBy(n => n) // get the numbers in ascending order
.SkipWhile(n => n < input) // skip until the remaining set >= input
.Cast<int?>() // cast to nullable int
.FirstOrDefault(); // take the first or default entry (if no items remain, it will be null)
Well, it's too late and my solution is far away from perfect, but:
int number = 400; //imput number
int closest = numbers.Aggregate((x, y) => Math.Abs(x - number) < Math.Abs(y - number) ? x : y); // searching the closer one
int compare;
try{
if (numbers.IndexOf(closest) != 0)
{
compare = Math.Max(closest, numbers[numbers.IndexOf(closest) + 1]); // if it's not 0th and last
}
else
{ compare = closest; // if closest with index 0}
}// check which closest number is bigger
catch{
compare = closest; // if closest is last
}
Console.WriteLine(compare);
I am trying to filter a list of int based on multiples of a specific number, but I am not sure how to do this. I have searched this forum and found nothing related, but apologies in advance if I'm wrong.
Here is my code:
int val = 28;
List<int> divisible = new List<int>();
for (int i = 1; i <= val; i++) {
divisible.Add(i);
}
foreach(var d in divisible)
{
if(val % d == 0)
{
// do something
}
else
{
// get rid of all multiples of the number that "val" is not divisible by
}
}
Basically, this code should create a divisible list from 1 to 28. If val is divisible by one of the numbers in the list, thats fine, but if it falls into else, I want to be able to filter out all multiples of that number out of the current list we are looping through.
The next number that wouldn't be divisible would be 3 in this example, so in the else get rid of 6, 9, 12, ... etc.
Your code is fine, but you're just missing the actual code to remove the item. But there is a caveat: You cannot modify a list when you are looping through it using foreach. There are a couple ways to handle that:
Depending on your requirements, maybe just don't add them in the first place. Move your val % d == 0 condition into the for loop that adds the values, and just don't add the values that are divisible by d.
Make a new list (List<int> toRemove) where you keep track of all the values you need to remove. After you're done the foreach loop, loop through your toRemove list and use divisible.Remove(value) to remove those.
Change your foreach to a for loop, which will allow you to use divisible.RemoveAt(i). But you will have to make sure you don't skip a value on the next iteration of the for loop (since removing a value changes the size of the list).
I agree with Gabriel. You cannot alter a underlying enumeration while traversing it with with foreach. The easiest thing to do would be to convert it to a for loop.
Also in the initial population of your list try using the newer way
var divisible = Enumerable.Range(1, val).ToList();
then do
for(int 0 = 1; i < val; i++)
{
if(val % d == 0)
{
// do something
}
else
{
divisible.RemoveAt(i);
}
}
I have a function that takes in a list of numbers and will return how many even numbers and odd numbers there are in the list. However, I passed in a list of numbers but I'm getting 0 results.
Here is my function -
public static string HowManyEvenAndOdds(List<int> numbers)
{
int numOfOdds = 0;
int numOfEvens = 0;
int numOfBoth = 0;
foreach (int i in numbers) {
bool isEven = i % 2 == 0;
bool isOdd = i % 3 == 0;
numOfBoth = isEven && isOdd ? numOfBoth++ : numOfBoth;
numOfEvens = isEven ? numOfEvens++ : numOfEvens;
numOfOdds = isOdd ? numOfOdds++ : numOfOdds;
}
return string.Format("This list has {0} odd numbers,\n{1} even numbers,\nand {2} numbers that are even and odd.", numOfOdds, numOfEvens, numOfBoth);
}
Any ideas on what I'm doing wrong here? I debugged through it but none of the lists are incrementing.
Thanks
your are not calculating odd in the correct way
i%3 does not catch 5 which is also an odd number, try this instead
bool isEven = i % 2 == 0;
bool isOdd =!isEven;
I agree with Schachaf Gortler's answer as well as p.s.w.g's comment. Just do:
foreach (var number in numbers)
{
// A number is even if, and only if, it's evenly divisible by 2
if (number % 2 == 0)
numEvens++;
// A number is odd if, and only if, it's NOT evenly divisible by 2
// Alternatively, a number is odd if it isn't even and vice versa
else
numOdds++;
}
As p.s.w.g. mentioned, there's no such thing as a number that's both even and odd, so eliminate that completely.
Incidentally, numOfEvens++ retrieves the value and then increments it, which is why your code didn't work.
I think you should have a look at your test for isOdd
Use the Linq Count extension.
int numOfOdds = numbers.Count(x => x % 2 != 0);
int numOfEvens = numbers.Count(x => x % 2 == 0);
Of course you don't need to evaluate both expressions, as per the comment below.
Well,
int a = 20;
int b = 30;
int c = 40;
int d = 50;
if (a > b,c,d)
how would i approach this, i have no idea i fail at every turn, its been hours
If there is a short quantity of numbers, you can simply use the boolean logic:
if (a > b && a > c && a > d)
{
}
If you don't know in advance the quantity of numbers, what about creating a collection and compare the first number to the numbers from the collection through a loop?
var numbers = { 30, 40, 50 };
if (!numbers.Any(c => 20 <= c))
{
}
You can put them in an array:
int a = 20;
int[] others = { 30, 40, 50 };
if(others.All(o => a > o))
{
// do something
}
Put them all in a list and do this:
if(list.All(x=> a > x))
Or in one line:
if(new List<int>{a, b, c, d}.All(x=> a > x))
EDIT
I changed the Max() to All(x => a > x) because the a > x will not return a true when a == x whereas Max() will do that.
Non-LINQ example:
if (Math.Max(a, Math.Max(b, Math.Max(c, d))) == a)
{
}
If all you want to know is if the number x is greater than the other numbers, you could either compare them explicitly like if(x>b & b>c) or use something like if(list.All(x=> a > x))
as mentioned above. If you have many numbers and all you want is the higher number, you could sort the list using a quick sort that could be efficient and get the first item.
It's a bit different if you need to compare them and get different comparissons then probably the easiest thing is to loop through the list.
I want to generate a shuffled merged list that will keep the internal order of the lists.
For example:
list A: 11 22 33
list B: 6 7 8
valid result: 11 22 6 33 7 8
invalid result: 22 11 7 6 33 8
Just randomly select a list (e.g. generate a random number between 0 and 1, if < 0.5 list A, otherwise list B) and then take the element from that list and add it to you new list. Repeat until you have no elements left in each list.
Generate A.Length random integers in the interval [0, B.Length). Sort the random numbers, then iterate i from 0..A.Length adding A[i] to into position r[i]+i in B. The +i is because you're shifting the original values in B to the right as you insert values from A.
This will be as random as your RNG.
None of the answers provided in this page work if you need the outputs to be uniformly distributed.
To illustrate my examples, assume we are merging two lists A=[1,2,3], B=[a,b,c]
In the approach mentioned in most answers (i.e. merging two lists a la mergesort, but choosing a list head randomly each time), the output [1 a 2 b 3 c] is far less likely than [1 2 3 a b c]. Intuitively, this happens because when you run out of elements in a list, then the elements on the other list are appended at the end. Because of that, the probability for the first case is 0.5*0.5*0.5 = 0.5^3 = 0.125, but in the second case, there are more random random events, since a random head has to be picked 5 times instead of just 3, leaving us with a probability of 0.5^5 = 0.03125. An empirical evaluation also easily validates these results.
The answer suggested by #marcog is almost correct. However, there is an issue where the distribution of r is not uniform after sorting it. This happens because original lists [0,1,2], [2,1,0], [2,1,0] all get sorted into [0,1,2], making this sorted r more likely than, for example, [0,0,0] for which there is only one possibility.
There is a clever way of generating the list r in such a way that it is uniformly distributed, as seen in this Math StackExchange question: https://math.stackexchange.com/questions/3218854/randomly-generate-a-sorted-set-with-uniform-distribution
To summarize the answer to that question, you must sample |B| elements (uniformly at random, and without repetition) from the set {0,1,..|A|+|B|-1}, sort the result and then subtract its index to each element in this new list. The result is the list r that can be used in replacement at #marcog's answer.
Original Answer:
static IEnumerable<T> MergeShuffle<T>(IEnumerable<T> lista, IEnumerable<T> listb)
{
var first = lista.GetEnumerator();
var second = listb.GetEnumerator();
var rand = new Random();
bool exhaustedA = false;
bool exhaustedB = false;
while (!(exhaustedA && exhaustedB))
{
bool found = false;
if (!exhaustedB && (exhaustedA || rand.Next(0, 2) == 0))
{
exhaustedB = !(found = second.MoveNext());
if (found)
yield return second.Current;
}
if (!found && !exhaustedA)
{
exhaustedA = !(found = first.MoveNext());
if (found)
yield return first.Current;
}
}
}
Second answer based on marcog's answer
static IEnumerable<T> MergeShuffle<T>(IEnumerable<T> lista, IEnumerable<T> listb)
{
int total = lista.Count() + listb.Count();
var random = new Random();
var indexes = Enumerable.Range(0, total-1)
.OrderBy(_=>random.NextDouble())
.Take(lista.Count())
.OrderBy(x=>x)
.ToList();
var first = lista.GetEnumerator();
var second = listb.GetEnumerator();
for (int i = 0; i < total; i++)
if (indexes.Contains(i))
{
first.MoveNext();
yield return first.Current;
}
else
{
second.MoveNext();
yield return second.Current;
}
}
Rather than generating a list of indices, this can be done by adjusting the probabilities based on the number of elements left in each list. On each iteration, A will have A_size elements remaining, and B will have B_size elements remaining. Choose a random number R from 1..(A_size + B_size). If R <= A_size, then use an element from A as the next element in the output. Otherwise use an element from B.
int A[] = {11, 22, 33}, A_pos = 0, A_remaining = 3;
int B[] = {6, 7, 8}, B_pos = 0, B_remaining = 3;
while (A_remaining || B_remaining) {
int r = rand() % (A_remaining + B_remaining);
if (r < A_remaining) {
printf("%d ", A[A_pos++]);
A_remaining--;
} else {
printf("%d ", B[B_pos++]);
B_remaining--;
}
}
printf("\n");
As a list gets smaller, the probability an element gets chosen from it will decrease.
This can be scaled to multiple lists. For example, given lists A, B, and C with sizes A_size, B_size, and C_size, choose R in 1..(A_size+B_size+C_size). If R <= A_size, use an element from A. Otherwise, if R <= A_size+B_size use an element from B. Otherwise C.
Here is a solution that ensures a uniformly distributed output, and is easy to reason why. The idea is first to generate a list of tokens, where each token represent an element of a specific list, but not a specific element. For example for two lists having 3 elements each, we generate this list of tokens: 0, 0, 0, 1, 1, 1. Then we shuffle the tokens. Finally we yield an element for each token, selecting the next element from the corresponding original list.
public static IEnumerable<T> MergeShufflePreservingOrder<T>(
params IEnumerable<T>[] sources)
{
var random = new Random();
var queues = sources
.Select(source => new Queue<T>(source))
.ToArray();
var tokens = queues
.SelectMany((queue, i) => Enumerable.Repeat(i, queue.Count))
.ToArray();
Shuffle(tokens);
return tokens.Select(token => queues[token].Dequeue());
void Shuffle(int[] array)
{
for (int i = 0; i < array.Length; i++)
{
int j = random.Next(i, array.Length);
if (i == j) continue;
if (array[i] == array[j]) continue;
var temp = array[i];
array[i] = array[j];
array[j] = temp;
}
}
}
Usage example:
var list1 = "ABCDEFGHIJKL".ToCharArray();
var list2 = "abcd".ToCharArray();
var list3 = "#".ToCharArray();
var merged = MergeShufflePreservingOrder(list1, list2, list3);
Console.WriteLine(String.Join("", merged));
Output:
ABCDaEFGHIb#cJKLd
This might be easier, assuming you have a list of three values in order that match 3 values in another table.
You can also sequence with the identity using identity (1,2)
Create TABLE #tmp1 (ID int identity(1,1),firstvalue char(2),secondvalue char(2))
Create TABLE #tmp2 (ID int identity(1,1),firstvalue char(2),secondvalue char(2))
Insert into #tmp1(firstvalue,secondvalue) Select firstvalue,null secondvalue from firsttable
Insert into #tmp2(firstvalue,secondvalue) Select null firstvalue,secondvalue from secondtable
Select a.firstvalue,b.secondvalue from #tmp1 a join #tmp2 b on a.id=b.id
DROP TABLE #tmp1
DROP TABLE #tmp2