Divide a list into 3 parts in python

Question

I am looking to divide a list into 3 parts in python. If the list doesn't divide by 3, I would like it to 'fill' the first column then the second, then the third.

e.g:

1|2|3

1|3|
2|


1|3|
2|4|

1|3|5
2|4|

1|3|5
2|4|6

And so on.

I am currently doing something like this, which doesn't give the desired results. Have also played around with using % to no avail

l = [1,2,3,4,5]
a = [:l/3]
b = [l/3:(l/3)*2 ]
c = [(l/3)*2 :]

All help appreciated, thanks

Note The answer to Splitting a list of into N parts of approximately equal length gives the opposite result, so please don't mark this question as already answered

I didn't get you. Say the list is `[1,2,3,4,5]`. What o/p do you want? — nitish712, Feb 03 '14 at 13:12
What does "fill" mean? What does "doesn't give the desired results" mean? What does `1|2|3` mean? Is that a list? Does it relate with `[1,2,3,4,5]` of the second example in anyway? — Bakuriu, Feb 03 '14 at 13:22

score 2 · Answer 1 · answered Feb 03 '14 at 13:21

There is a recipe in itertools that will do this:

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

You could use this as-is or modify for your needs.

score 0 · Answer 2 · answered Feb 03 '14 at 13:33

0

Here is some code I just drew up.

import math
startArray = [1,2,3,4,5,6,7,8,9]
secLength = math.ceil(len(startArray) / 3)
newArray = []
newArray.append(startArray[:secLength])
newArray.append(startArray[secLength : 2*secLength])
newArray.append(startArray[2*secLength:]);
print(newArray)

For this example, the return is [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

If your array is [1,2,3,4,5,6,7], it returns [[1, 2, 3], [4, 5, 6], [7]]

answered Feb 03 '14 at 13:33

erdekhayser

6,537
2
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When I ran this (after adding an integer cast to the slices) my results were [1, 2], [3, 4], [5, 6, 7]] – holmeswatson Feb 03 '14 at 14:04
Where are you integer casting? Where I use math.ceil? – erdekhayser Feb 03 '14 at 15:34
math.ceil() returns an integer, therefore the integer cast does not matter – erdekhayser Feb 03 '14 at 16:11

score 0 · Answer 3 · answered Feb 03 '14 at 14:05

0

Continued to try work this out, got this:

import math
l = [1,2,3,4,5,6,7]

col = 3
height = int(math.ceil(len(l)/col))

print height
a = l[:height+1]
b = l[height+1:(height+1)*2]
c = l[(height+1)*2:]

print l
print a
print b
print c

answered Feb 03 '14 at 14:05

holmeswatson

969
3
14
39

That is practically the same exact thing that I already posted... – erdekhayser Feb 03 '14 at 16:11
And it doesn't work? This printed this for me:`[1, 2, 3, 4] [5, 6, 7] []` – erdekhayser Feb 03 '14 at 16:15
I figured out what the confusion between our codes are. Yours works 100% in Python 2, mine works 100% in Python 3. Must be different behaviors somewhere – erdekhayser Feb 03 '14 at 16:19
in python 2 ceil seems to return a float. but even when i cast it, I don't get the same result as you, strange – holmeswatson Feb 04 '14 at 11:18

Divide a list into 3 parts in python

3 Answers3