I have two lists with me as per the following example.
List<string> words = new List<string>() {"V","H","M" };
List<int> numbers = new List<int>() {10,20,30 };
I need to pair the values of these two lists so that my output needs to be exactly like the following text.
Desired output : V10 H20 M30
You could use Zip method for that.
You can try the following:
String.Join(" ", words.Zip(numbers, (first, second) => first + second))
Try using Zip:
var result = words
.Zip(numbers, (w, n) => $"{w}{n}");
Console.Write(string.Join(" ", result));
I'm a bit late to the party but here's a very simple way of doing it without Zip: (x = item, y = index)
var mergedList = words.Select((x, y) => $"{x}{numbers.ElementAt(y)}");
Related
I have two lists of strings. My manager's requirement is to concatenate both lists and output as a string using LINQ.
List<string> upper = new List<string> { "A", "B", "C" };
List<string> lower = new List<string> { "a", "b", "c" };
The output should be something as below:
//string output = "A:a,B:b,C:c"
And to achieve this, I need to use LINQ Aggregate method or other extension methods available. Please help
You are looking for the method Zip and maybe combined with string.Join
Console.WriteLine(string.Join(",", upper.Zip(lower, (u, l) => u + ":" + l)));
This outputs: A:a,B:b,C:c
You need Zip method not Aggregate:
If you want the result as a List of String:
List<string> result = upper.Zip(lower, (first, second) => first + ":" + second).ToList();
If you want the result as a string you can either use String.Join method like this:
string output = string.Join(",", result);
Or Aggregate method as you are looking for like this:
string output = result.Aggregate((f, s) => f + "," + s);
Use Enumerable.Zip:
var results = upper.Zip(lower, (up, low) => $"{up}:{low}");
Now to convert it into a single string:
string output = string.Join(",", results);
I have a bunch of SKUs (stock keeping units) that represent a series of strings that I'd like to create a single Regex to match for.
So, for example, if I have SKUs:
var skus = new[] { "BATPAG003", "BATTWLP03", "BATTWLP04", "BATTWSP04", "SPIFATB01" };
...I'd like to automatically generate the Regex to recognize any one of the SKUs.
I know that I could do simply do "BATPAG003|BATTWLP03|BATTWLP04|BATTWSP04|SPIFATB01", but list of SKUs can be quite lengthy and I'd like to compress the resulting Regex to look like "BAT(PAG003|TW(LP0(3|4)|SP04))|SPIFATB01"
So this is a combinatorics exercise. I want to generate the all of the possible Regex to match any of my input strings, with the view that the shortest is probably the best.
I could, for example, produce any of these:
BATPAG003|BATTWLP03|BATTWLP04|BATTWSP04|SPIFATB01
BAT(PAG003|TW(LP0(3|4)|SP04))|SPIFATB01
BAT(PAG003|TW(LP(03|04)|SP04))|SPIFATB01
B(ATPAG003|ATTW(LP0(3|4)|ATSP04))|SPIFATB01
Any of those would work, but the shortest is probably the one I'd choose.
To start with I've tried to implement this function:
Func<IEnumerable<string>, IEnumerable<string>> regexify = null;
regexify = xs =>
from n in Enumerable.Range(1, 10)
let g = xs.ToArray().Where(s => !String.IsNullOrWhiteSpace(s)).GroupBy(x => new String(x.Take(n).ToArray()), x => new String(x.Skip(n).ToArray()))
let parts = g.SelectMany(x => x.Count() > 1 ? regexify(x).Select(y => x.Key + "(" + String.Join("|", y) + ")") : new [] { x.Key + String.Join("", x) })
let regex = String.Join("|", parts)
orderby regex.Length
select regex;
This takes my source skus and results in a list of possible Regex's, but it's not working. This is the current output:
BATPAG003|BATTWLP03|BATTWLP04|BATTWSP04|SPIFATB01
BATPAG003|BATTWLP03|BATTWLP04|BATTWSP04|SPIFATB01
BATPAG003|BATTWLP0(3|4)|BATTWLP0(3|4)|BATTWLP0(3|4)|BATTWLP0(3|4)|BATTWLP0(3|4)|BATTWLP0(3|4)|BATTWLP0(3|4)|BATTWLP0(3|4)|BATTWLP0(3|4)|BATTWLP0(3|4)|BATTWSP04|SPIFATB01
BATPAG003|BATTWLP(03|04)|BATTWLP(03|04)|BATTWLP(03|04)|BATTWLP(03|04)|BATTWLP(03|04)|BATTWLP(03|04)|BATTWLP(03|04)|BATTWLP(03|04)|BATTWLP(03|04)|BATTWLP(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4))|BATTWSP04|SPIFATB01
BATPAG003|BATTWL(P03|P04)|BATTWL(P03|P04)|BATTWL(P03|P04)|BATTWL(P03|P04)|BATTWL(P03|P04)|BATTWL(P03|P04)|BATTWL(P03|P04)|BATTWL(P03|P04)|BATTWL(P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4))|BATTWL(P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)))|BATTWSP04|SPIFATB01
BATPAG003|BATTW(LP03|LP04|SP04)|BATTW(LP03|LP04|SP04)|BATTW(LP03|LP04|SP04)|BATTW(LP03|LP04|SP04)|BATTW(LP03|LP04|SP04)|BATTW(LP03|LP04|SP04)|BATTW(LP03|LP04|SP04)|BATTW(LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|SP04)|BATTW(LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4))|SP04)|BATTW(L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4))|L(P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)))|SP04)|SPIFATB01
BATPAG003|BATT(WLP03|WLP04|WSP04)|BATT(WLP03|WLP04|WSP04)|BATT(WLP03|WLP04|WSP04)|BATT(WLP03|WLP04|WSP04)|BATT(WLP03|WLP04|WSP04)|BATT(WLP03|WLP04|WSP04)|BATT(WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WSP04)|BATT(WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4))|WSP04)|BATT(WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4))|WL(P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)))|WSP04)|BATT(W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|SP04)|W(LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4))|SP04)|W(L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4))|L(P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)))|SP04))|SPIFATB01
BAT(PAG003|TWLP03|TWLP04|TWSP04)|BAT(PAG003|TWLP03|TWLP04|TWSP04)|BAT(PAG003|TWLP03|TWLP04|TWSP04)|BAT(PAG003|TWLP03|TWLP04|TWSP04)|BAT(PAG003|TWLP03|TWLP04|TWSP04)|BAT(PAG003|TWLP0(3|4)|TWLP0(3|4)|TWLP0(3|4)|TWLP0(3|4)|TWLP0(3|4)|TWLP0(3|4)|TWLP0(3|4)|TWLP0(3|4)|TWLP0(3|4)|TWLP0(3|4)|TWSP04)|BAT(PAG003|TWLP(03|04)|TWLP(03|04)|TWLP(03|04)|TWLP(03|04)|TWLP(03|04)|TWLP(03|04)|TWLP(03|04)|TWLP(03|04)|TWLP(03|04)|TWLP(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4))|TWSP04)|BAT(PAG003|TWL(P03|P04)|TWL(P03|P04)|TWL(P03|P04)|TWL(P03|P04)|TWL(P03|P04)|TWL(P03|P04)|TWL(P03|P04)|TWL(P03|P04)|TWL(P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4))|TWL(P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)))|TWSP04)|BAT(PAG003|TW(LP03|LP04|SP04)|TW(LP03|LP04|SP04)|TW(LP03|LP04|SP04)|TW(LP03|LP04|SP04)|TW(LP03|LP04|SP04)|TW(LP03|LP04|SP04)|TW(LP03|LP04|SP04)|TW(LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|SP04)|TW(LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4))|SP04)|TW(L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4))|L(P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)))|SP04))|BAT(PAG003|T(WLP03|WLP04|WSP04)|T(WLP03|WLP04|WSP04)|T(WLP03|WLP04|WSP04)|T(WLP03|WLP04|WSP04)|T(WLP03|WLP04|WSP04)|T(WLP03|WLP04|WSP04)|T(WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WSP04)|T(WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4))|WSP04)|T(WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4))|WL(P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)))|WSP04)|T(W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|SP04)|W(LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4))|SP04)|W(L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4))|L(P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)))|SP04)))|SPIFATB01
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WL(P03|P04)|TWL(P03|P04)|TWL(P03|P04)|TWL(P03|P04)|TWL(P03|P04)|TWL(P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4))|TWL(P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)))|TWSP04)|T(PAG003|TW(LP03|LP04|SP04)|TW(LP03|LP04|SP04)|TW(LP03|LP04|SP04)|TW(LP03|LP04|SP04)|TW(LP03|LP04|SP04)|TW(LP03|LP04|SP04)|TW(LP03|LP04|SP04)|TW(LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|SP04)|TW(LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4))|SP04)|TW(L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4))|L(P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)))|SP04))|T(PAG003|T(WLP03|WLP04|WSP04)|T(WLP03|WLP04|WSP04)|T(WLP03|WLP04|WSP04)|T(WLP03|WLP04|WSP04)|T(WLP03|WLP04|WSP04)|T(WLP03|WLP04|WSP04)|T(WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WLP0(3|4)|WSP04)|T(WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(03|04)|WLP(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4))|WSP04)|T(WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P03|P04)|WL(P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4))|WL(P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)))|WSP04)|T(W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP03|LP04|SP04)|W(LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|LP0(3|4)|SP04)|W(LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(03|04)|LP(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4))|SP04)|W(L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P03|P04)|L(P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4)|P0(3|4))|L(P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(03|04)|P(0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)|0(3|4)))|SP04)))))|SPIFATB01
It feels very close, and I think I'm slightly doing something wrong.
This works if each SKU id have the same length.
// ...
string regexStr = Calculate(skus);
// ...
public static string Calculate(IEnumerable<string> rest) {
if (rest.First().Length > 0) {
string[] groups = rest.GroupBy(r => r[0])
.Select(g => g.Key + Calculate(g.Select(e => e.Substring(1))))
.ToArray();
return groups.Length > 1 ? "(" + string.Join("|", groups) + ")" : groups[0];
} else {
return string.Empty;
}
}
This is what I finally worked out:
var skus = new[] { "BATPAG003", "BATTWLP03", "BATTWLP04", "BATTWSP04", "SPIFATB01" };
Func<IEnumerable<IGrouping<string, string>>, IEnumerable<string>> regexify = null;
Func<IEnumerable<string>, IEnumerable<string>> generate =
xs =>
from n in Enumerable.Range(2, 20)
let g = xs.GroupBy(x => new String(x.Take(n).ToArray()), x => new String(x.Skip(n).ToArray()))
where g.Count() != xs.Count()
from r in regexify(g)
select r;
regexify = gxs =>
{
if (!gxs.Any())
{
return new [] { "" };
}
else
{
var rs = regexify(gxs.Skip(1)).ToArray();
return
from f in gxs.Take(1)
from z in new [] { String.Join("|", f) }.Concat(f.Count() > 1 ? generate(f) : Enumerable.Empty<string>())
from r in rs
select f.Key + (f.Count() == 1 ? z : $"({z})") + (r != "" ? "|" + r : "");
}
};
Then using this query:
generate(skus).OrderBy(x => x).OrderBy(x => x.Length);
...I got this result:
BAT(PAG003|TW(LP0(3|4)|SP04))|SPIFATB01
BAT(PAG003|TWLP0(3|4)|TWSP04)|SPIFATB01
BA(TPAG003|TTW(LP0(3|4)|SP04))|SPIFATB01
BAT(PAG003|TW(LP(03|04)|SP04))|SPIFATB01
BAT(PAG003|TW(LP03|LP04|SP04))|SPIFATB01
BAT(PAG003|TWLP(03|04)|TWSP04)|SPIFATB01
BATPAG003|BATTW(LP0(3|4)|SP04)|SPIFATB01
BA(TPAG003|TT(WLP0(3|4)|WSP04))|SPIFATB01
BA(TPAG003|TTW(LP(03|04)|SP04))|SPIFATB01
BA(TPAG003|TTW(LP03|LP04|SP04))|SPIFATB01
BA(TPAG003|TTWLP0(3|4)|TTWSP04)|SPIFATB01
BAT(PAG003|TWL(P0(3|4))|TWSP04)|SPIFATB01
BAT(PAG003|TWL(P03|P04)|TWSP04)|SPIFATB01
BATPAG003|BATT(WLP0(3|4)|WSP04)|SPIFATB01
BATPAG003|BATTW(LP(03|04)|SP04)|SPIFATB01
BATPAG003|BATTW(LP03|LP04|SP04)|SPIFATB01
BA(TPAG003|TT(WLP(03|04)|WSP04))|SPIFATB01
BA(TPAG003|TTWLP(03|04)|TTWSP04)|SPIFATB01
BAT(PAG003|TWLP03|TWLP04|TWSP04)|SPIFATB01
BATPAG003|BATT(WLP(03|04)|WSP04)|SPIFATB01
BA(TPAG003|TT(WL(P0(3|4))|WSP04))|SPIFATB01
BA(TPAG003|TT(WL(P03|P04)|WSP04))|SPIFATB01
BA(TPAG003|TT(WLP03|WLP04|WSP04))|SPIFATB01
BA(TPAG003|TTWL(P0(3|4))|TTWSP04)|SPIFATB01
BA(TPAG003|TTWL(P03|P04)|TTWSP04)|SPIFATB01
BATPAG003|BATT(WL(P0(3|4))|WSP04)|SPIFATB01
BATPAG003|BATT(WL(P03|P04)|WSP04)|SPIFATB01
BATPAG003|BATT(WLP03|WLP04|WSP04)|SPIFATB01
BATPAG003|BATTWLP0(3|4)|BATTWSP04|SPIFATB01
BATPAG003|BATTWLP(03|04)|BATTWSP04|SPIFATB01
BA(TPAG003|TTWLP03|TTWLP04|TTWSP04)|SPIFATB01
BATPAG003|BATTWL(P0(3|4))|BATTWSP04|SPIFATB01
BATPAG003|BATTWL(P03|P04)|BATTWSP04|SPIFATB01
The only problem with my approach was computation time. Some of my source lists have nearly 100 SKUs. Some of the runs were taking longer than I care to wait for and had to break it down into smaller chunks and then manually concatenate.
Take the entire list of all of your sku's and make a single ternary tree regex.
When you add or delete sku's, regenerate the regex. Maybe your database
generates on a weekly basis.
This utility makes a regex of 10,000 strings in less than half a second
and size is not important, it could be 300,000 strings.
For example, here is regex of 175,000 word dictionary.
I'm having a string like "a.b.c.d.e".
If I want to get an array like "a.b.c.d.e", "b.c.d.e", "c.d.e", "d.e", "e" in C#. What's the simplest approach?
Something like this will do:
var stringParts = input.Split('.');
var result = Enumerable.Range(0, stringParts.Length)
.Select(i => string.Join(".", stringParts.Skip(i)));
But like I said in my comment, please show the code you came up with and why you want to make it a one-liner, which usually doesn't serve any benefit. This isn't codegolf.
If you really do it with one statement, you can try this:
var str = "a.b.c.d.e";
var parts = str.Split('.')
.Select((x,idx) => new { idx })
.Select(p => string.Join(".",
str.Split('.').Skip(p.idx))).ToList();
This could be more efficient if you use Split first:
var parts = str.Split('.');
var result = parts
.Select((x,idx) => new { idx })
.Select(p => string.Join(".",
parts.Skip(p.idx))).ToList();
You can also do it without creating anonymous type(s), just create an int variable:
int i = 0;
var result = parts
.Select(p => string.Join(".", parts.Skip(i++)))
.ToList();
This is fairly neat:
var text = "a.b.c.d.e";
var results =
text
.Split('.')
.Reverse()
.Scan("", (a, x) => x + "." + a)
.Select(x => x.TrimEnd('.'))
.Reverse();
You do need to add the Microsoft Reactive Extensions Team's "Interactive Extensions" to get the Scan operator. Use NuGet and look for "Ix-Main".
I actually kind of like this question, not necessarily production but a bit of brain-bendy fun:
"a.b.c.d.e".Split('.').Reverse()
.Aggregate(Enumerable.Empty<string>(), (acc, c) =>
acc.Concat(new [] { c+(acc.LastOrDefault()??"") })
).Reverse()
Dotnetfiddle
What this does is move through each character in the split array and build up a new array by prepending the last value in the array with the current character. It's a fairly common functional programming technique.
Well, this is how I might write it.. I know, not "one line", but if you're gonna use (and I do recommend) a method anyway..
IEnumerable<string> AllComponentPartsForward (string s) {
IEnumerable<string> p = s.Split('.');
while (p.Any()) {
yield return string.Join(".", p); // p.ToArray() for .NET 3.5
p = p.Skip(1);
}
}
(I suppose it could be "more efficient" with IndexOf/Substring, but that's also harder for me to write and reason about!)
I have the following string arrays:
var myArray1 = new string[] {"A", "B", "C"};
var myArray2 = new string[] {"t1", "t2"};
I would like to be able to produce a final string that looks like this:
var myFinalString = "t1.A, t2.A, t1.B, t2.B, t1.C, t2.C";
I know I can iterate through each element of the array and build the string manually. However, I'd like to know if there's a better way. I tried to figure out how to make String.Join() method work, but couldn't :-(.
I don't know that any direct method exists, but this one-liner
return
from a in myArray
from b in tTerms
select string.Format("{0}.{1}", b, a)
should do it.
This works:
var query = from x in new[]{"A", "B", "C"}
from y in new[]{"t1", "t2"}
select y + "." + x;
var result = string.Join(", ", query.ToArray());
The term for such sequence is "Cartesian Product".
Here is long blog by Eric Lippert on it Computing a Cartesian Product with LINQ
As it is already shown in other answers sequence of tuples can be obtained with following code and than aggregated (using Join in case of string, or Aggregate for other type of result) to produce your final string:
var product =
from first in s1
from second in s2
select new[] { first, second };
var result String.Join(", ",
product.Select(p => String.Format("{0}.{1}", p.first, p.second));
If the two sequences are of the same length, you could probably put Enumerable.Zip to work for you.
var myFinalString = myArray.Zip(mySecondArray, (f, s) => f + "." + s);
I have a string containing up to 9 unique numbers from 1 to 9 (myString) e.g. "12345"
I have a list of strings {"1"}, {"4"} (myList) .. and so on.
I would like to know how many instances in the string (myString) are contained within the list (myList), in the above example this would return 2.
so something like
count = myList.Count(myList.Contains(myString));
I could change myString to a list if required.
Thanks very much
Joe
I would try the following:
count = mylist.Count(s => myString.Contains(s));
It is not perfectly clear what you need, but these are some options that could help:
myList.Where(s => s == myString).Count()
or
myList.Where(s => s.Contains(myString)).Count()
the first would return the number of strings in the list that are the same as yours, the second would return the number of strings that contain yours. If neither works, please make your question more clear.
If myList is just List<string>, then this should work:
int count = myList.Count(x => myString.Contains(x));
If myList is List<List<string>>:
int count = myList.SelectMany(x => x).Count(s => myString.Contains(s));
Try
count = myList.Count(s => s==myString);
This is one approach, but it's limited to 1 character matches. For your described scenario of numbers from 1-9 this works fine. Notice the s[0] usage which refers to the list items as a character. For example, if you had "12" in your list, it wouldn't work correctly.
string input = "123456123";
var list = new List<string> { "1", "4" };
var query = list.Select(s => new
{
Value = s,
Count = input.Count(c => c == s[0])
});
foreach (var item in query)
{
Console.WriteLine("{0} occurred {1} time(s)", item.Value, item.Count);
}
For multiple character matches, which would correctly count the occurrences of "12", the Regex class comes in handy:
var query = list.Select(s => new
{
Value = s,
Count = Regex.Matches(input, s).Count
});
try
var count = myList.Count(x => myString.ToCharArray().Contains(x[0]));
this will only work if the item in myList is a single digit
Edit: as you probably noticed this will convert myString to a char array multiple times so it would be better to have
var myStringArray = myString.ToCharArray();
var count = myList.Count(x => myStringArray.Contains(x[0]));