Python: list grouping

Question

I've got a list with (for example) 100 entries of the sort ['A0', 'B0', 'A1', 'B1', 'A2', 'B2', ... 'A99', 'B99'].

I'd now like to make this into a list of 50 entries with each entry a tuple (Ai, Bi) such that they are grouped together. So the result should be

[('A0','B0'),('A1','B1'),('A2','B2'),...,('A99','B99')]. Is there a shortcut to achieve this or do I have to use a loop like

for i in numpy.arange(0,100,2):
    newlist.add((oldlist[i], oldlist[i+1]))

I'm trying to do quick and advanced programming in python so I'd prefer using shortcuts, list comprehension, ... and not simple for loops where possible

score 3 · Accepted Answer · answered Mar 29 '20 at 17:18

3

This is the most pythonic way I can think of:

list(zip(oldlist[::2], oldlist[1::2]))

answered Mar 29 '20 at 17:18

Djib2011

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score 1 · Answer 2 · answered Mar 29 '20 at 17:17

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The code below should simply it:

list_1 = ['A0', 'B0', 'A1', 'B1', 'A2', 'B2']
list_of_groups = [x for x in zip(*(iter(list_1),) * 2)]

answered Mar 29 '20 at 17:17

jammin0921

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score 1 · Answer 3 · answered Mar 29 '20 at 17:20

1

If you really want, you can do this:

a = ['A0', 'B0', 'A1', 'B1', 'A2', 'B2']
list(zip(*[iter(a)]*2))
# [('A0', 'B0'), ('A1', 'B1'), ('A2', 'B2')]

answered Mar 29 '20 at 17:20

Chrispresso

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ansev · Answer 4 · 2020-03-29T17:39:34.890

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Another way wiith numpy

list(map(tuple, np.array(l).reshape(-1, 2)))
#[('A0', 'B0'), ('A1', 'B1'), ('A2', 'B2')]

edited Mar 29 '20 at 17:39

answered Mar 29 '20 at 17:22

ansev

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You don't really need the division — `reshape(-1, 2)` works fine. – Mark Mar 29 '20 at 17:25
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thanks for remembering, I forgot something basic :) @ Mark Meyer – ansev Mar 29 '20 at 17:40

Python: list grouping

4 Answers4