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0

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Process a list with a loop, taking 100 elements each time and automatically less than 100 at the end of the list

Is there a way to use a loop that takes the first 100 items in a big list, does something with them, then the next 100 etc but when it is nearing the end it automatically shortens the "100" step to the items remaining.

Currently I have to use two if loops:

for (int i = 0; i < listLength; i = i + 100)
{
    if (i + 100 < listLength)
    {
        //Does its thing with a bigList.GetRange(i, 100)
    }
    else
    {
        //Does the same thing with bigList.GetRange(i, listLength - i)
    }
}

Is there a better way of doing this? If not I will at least make the "thing" a function so the code does not have to be copied twice.

over 4 years ago · Santiago Trujillo
3 answers
Answer question

0

You can make use of LINQ Skip and Take and your code will be cleaner.

for (int i = 0; i < listLength; i=i+100)
{
    var items = bigList.Skip(i).Take(100); 
    // Do something with 100 or remaining items
}

Note: If the items are less than 100 Take would give you the remaining ones.

over 4 years ago · Santiago Trujillo Report

0

I didn't like any of the answers listed, so I made my own extension:

public static class IEnumerableExtensions
{
    public static IEnumerable<IEnumerable<T>> MakeGroupsOf<T>(this IEnumerable<T> source, int count)
    {
        var grouping = new List<T>();
        foreach (var item in source)
        {
            grouping.Add(item);
            if(grouping.Count == count)
            {
                yield return grouping;
                grouping = new List<T>();
            }
        }

        if (grouping.Count != 0)
        {
            yield return grouping;
        }
    }
}

Then you can use it:

foreach(var group in allItems.MakeGroupsOf(100))
{
    // Do something
}
over 4 years ago · Santiago Trujillo Report

0

You can keep an explicit variable for the end point:

for (int i = 0, j; i < listLength; i = j)
{
    j = Math.min(listLength, i + 100);
    // do your thing with bigList.GetRange(i, j)
}
over 4 years ago · Santiago Trujillo Report
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