CODEWITHSUNDEEP
pythonarrayslistnumpy
yatu
yatu

Reputation: 88236

get a chunk of items every n items in list

Say I have the following list:

l = [4,3,1,5,3,5,8,11,10,4,12,2,1]

What is the most pythonic way to slice l in order to get chunks of length n leaving k items between chunks?

For example, if n=2 and k=3 the result should be:

[4,3,5,8,12,2]

Upvotes: 2

Views: 205

Answers (4)

Jakob
Jakob

Reputation: 82

My take would be list comprehensions, even though you need to use two because otherwise it would just produce a list of lists.

l = [4,3,1,5,3,5,8,11,10,4,12,2,1]
n = 2
k = 3

chunk_l = [y for x in [l[i:i+n] for i in range(0,len(l), n+k)] for y in x]
print(chunk_l)
#[4, 3, 5, 8, 12, 2]

Upvotes: 1

Sheldore
Sheldore

Reputation: 39052

One alternate solution using slicing could be following

l = np.array([4,3,1,5,3,5,8,11,10,4,12,2,1])
A = l[:][::5]
B = l[1:][::5] 
final = np.insert(B, np.arange(len(A)), A)
# array([ 4,  3,  5,  8, 12,  2])

l[:][::5] gives you every 5th element starting from the first element and l[1:][::5] gives you every 5th element starting from the second element. You then merge the two together

Upvotes: 0

subjord
subjord

Reputation: 109

A possible solution is

l = [4,3,1,5,3,5,8,11,10,4,12]
k,n=3,2
res=[]
while l:
    res+=l[:n]
    l=l[n+k:]
print(res)

This can be probably be heavily optimized in way of pythonicizity.

Upvotes: 0

akuiper
akuiper

Reputation: 214957

Use a list comprehension:

[e for i in range(0, len(l), n+k) for e in l[i:i+n]]
# [4, 3, 5, 8, 12, 2]

A numpy solution would be:

import numpy as np

idx = (np.arange(0, len(l), n+k)[:,None] + np.arange(n)).ravel()
np.array(l)[idx]
# array([ 4,  3,  5,  8, 12,  2])

Upvotes: 5

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