Split a List into Smaller Lists of N Size

Split a List into smaller lists of N size

public static List<List<float[]>> SplitList(List<float[]> locations, int nSize=30)  
{        
    var list = new List<List<float[]>>(); 

    for (int i = 0; i < locations.Count; i += nSize) 
    { 
        list.Add(locations.GetRange(i, Math.Min(nSize, locations.Count - i))); 
    } 

    return list; 
} 

Generic version:

public static IEnumerable<List<T>> SplitList<T>(List<T> locations, int nSize=30)  
{        
    for (int i = 0; i < locations.Count; i += nSize) 
    { 
        yield return locations.GetRange(i, Math.Min(nSize, locations.Count - i)); 
    }  
} 

How to split an n-size list into smaller lists of maximum 2 items?

I believe this is what you're looking for:

n = [1, 2, 3, 4, 5, 6, 7]

[[i, j] for i, j in zip(n[1:], n[2:])]

Output:

[[2, 3], [3, 4], [4, 5], [5, 6], [6, 7]]

Divide a list into smaller lists

So, I hardly tried to make it in one line. Here is what I ended up with

import math

nb_classes = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
N = 3

lists = [nb_classes[math.ceil(i): math.ceil(i + len(nb_classes) / N)] for i in (len(nb_classes) / N * j for j in range(N))]

print(lists)  # [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11]]

But this may be a little bit complicated, you just need to ceil your indexes

Split a python list into other sublists i.e smaller lists

I'd say

chunks = [data[x:x+100] for x in range(0, len(data), 100)]

If you are using python 2.x instead of 3.x, you can be more memory-efficient by using xrange(), changing the above code to:

chunks = [data[x:x+100] for x in xrange(0, len(data), 100)]

How do I split a list into equally-sized chunks?

Here's a generator that yields evenly-sized chunks:

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in range(0, len(lst), n):
        yield lst[i:i + n]

import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
 [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]

For Python 2, using xrange instead of range:

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in xrange(0, len(lst), n):
        yield lst[i:i + n]

---

Below is a list comprehension one-liner. The method above is preferable, though, since using named functions makes code easier to understand. For Python 3:

[lst[i:i + n] for i in range(0, len(lst), n)]

For Python 2:

[lst[i:i + n] for i in xrange(0, len(lst), n)]

Efficient way to divide a list into lists of n size

You'll want to do something that makes use of List.subList(int, int) views rather than copying each sublist. To do this really easily, use Guava's Lists.partition(List, int) method:

List<Foo> foos = ...
for (List<Foo> partition : Lists.partition(foos, n)) {
  // do something with partition
}

Note that this, like many things, isn't very efficient with a List that isn't RandomAccess (such as a LinkedList).

Splitting a list into N parts of approximately equal length

This code is broken due to rounding errors. Do not use it!!!

assert len(chunkIt([1,2,3], 10)) == 10  # fails

---

Here's one that could work:

def chunkIt(seq, num):
    avg = len(seq) / float(num)
    out = []
    last = 0.0

    while last < len(seq):
        out.append(seq[int(last):int(last + avg)])
        last += avg

    return out

Testing:

>>> chunkIt(range(10), 3)
[[0, 1, 2], [3, 4, 5], [6, 7, 8, 9]]
>>> chunkIt(range(11), 3)
[[0, 1, 2], [3, 4, 5, 6], [7, 8, 9, 10]]
>>> chunkIt(range(12), 3)
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11]]

Split list into N sublist with balanced size C#

Your are rounding off, instead of up:

var playersXGroup = (int)Math.Round(players.Count / numberGroupsRound);

should be

var playersXGroup = (int)Math.Ceiling(players.Count / numberGroupsRound);

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