Iterating over a 2 dimensional python list

Question

I have created a 2 dimension array like:

rows =3
columns= 2
mylist = [[0 for x in range(columns)] for x in range(rows)]
for i in range(rows):
    for j in range(columns):
        mylist[i][j] = '%s,%s'%(i,j)
print mylist

Printing this list gives an output:

[  ['0,0', '0,1'], ['1,0', '1,1'], ['2,0', '2,1']   ]

where each list item is a string of the format 'row,column'

Now given this list, i want to iterate through it in the order:

'0,0'
'1,0'
'2,0'
'0,1'
'1,1'
'2,1'

that is iterate through 1st column then 2nd column and so on. How do i do it with a loop ?

This Question pertains to pure python list while the question which is marked as same pertains to numpy arrays. They are clearly different

@NominSim that refers to a numpy array, which doesn't necessarily translate to his problem — Ivo Flipse, May 14 '13 at 16:58

score 137 · Answer 1 · answered May 14 '13 at 16:58

137

same way you did the fill in, but reverse the indexes:

>>> for j in range(columns):
...     for i in range(rows):
...        print mylist[i][j],
... 
0,0 1,0 2,0 0,1 1,1 2,1
>>>

answered May 14 '13 at 16:58

Iliyan Bobev

3,070
2
20
24

24

This should be the accepted answer, because it is simple and easy to understand. One may not be familiar with the clever functions used in the other answers, but with only a quick look at the code, it's obvious how this solution works. Because of that, this answer is the most Pythonic. – Mr. Lance E Sloan Nov 05 '13 at 12:28
2

Hmm.. What if each of the children lists doesn't have the save length (rows here) ? – sonlexqt Dec 13 '16 at 17:45
@sonlexqt iterate indexes of the longest row and use `try ... except IndexError` around the print operand. (On my cell so can't provide exact example) – Iliyan Bobev Dec 14 '16 at 03:44
But this doesn't work if the number of rows is more than the number of columns. – TamingofTheShru Dec 19 '20 at 19:11

score 102 · Answer 2 · edited Feb 03 '16 at 00:50

102

This is the correct way.

>>> x = [ ['0,0', '0,1'], ['1,0', '1,1'], ['2,0', '2,1'] ]
>>> for i in range(len(x)):
        for j in range(len(x[i])):
                print(x[i][j])


0,0
0,1
1,0
1,1
2,0
2,1
>>>

edited Feb 03 '16 at 00:50

Kenneth Reitz

8,559
4
31
34

answered May 14 '13 at 16:57

mrKelley

3,365
2
21
28

5

This isn't very pythonic. – nelsontruran Feb 23 '20 at 19:11

Alexey Kachayev · Accepted Answer · 2013-05-14T17:02:16.577

38

Use zip and itertools.chain. Something like:

>>> from itertools import chain
>>> l = chain.from_iterable(zip(*l))
<itertools.chain object at 0x104612610>
>>> list(l)
['0,0', '1,0', '2,0', '0,1', '1,1', '2,1']

edited May 14 '13 at 17:02

answered May 14 '13 at 16:53

Alexey Kachayev

6,106
27
24

@AndyHayden Actually, it's ok for active interactive shell user. Updated answer with direct variable name. – Alexey Kachayev May 14 '13 at 17:03
2

name 'l' is not defined – Sérgio Mar 02 '18 at 17:53
1

@Sérgio l is a list... – Sep 03 '18 at 17:23

score 7 · Answer 4 · answered May 14 '13 at 16:55

>>> mylist = [["%s,%s"%(i,j) for j in range(columns)] for i in range(rows)]
>>> mylist
[['0,0', '0,1', '0,2'], ['1,0', '1,1', '1,2'], ['2,0', '2,1', '2,2']]
>>> zip(*mylist)
[('0,0', '1,0', '2,0'), ('0,1', '1,1', '2,1'), ('0,2', '1,2', '2,2')]
>>> sum(zip(*mylist),())
('0,0', '1,0', '2,0', '0,1', '1,1', '2,1', '0,2', '1,2', '2,2')

score 4 · Answer 5 · answered May 14 '13 at 16:52

4

zip will transpose the list, after that you can concatenate the outputs.

In [3]: zip(*[ ['0,0', '0,1'], ['1,0', '1,1'], ['2,0', '2,1'] ])
Out[3]: [('0,0', '1,0', '2,0'), ('0,1', '1,1', '2,1')]

answered May 14 '13 at 16:52

Adrian Ratnapala

5,485
2
29
39

2

But @Alexey's answer is better. – Adrian Ratnapala May 14 '13 at 16:54
Why is Alexey's answer better? This uses built-in functions only, and doesn't seem substantially more resource intensive. – Z4-tier Apr 25 '20 at 21:27
@Z4-tier This response returns a list with 2 tuples. Alexey's answer returns a single list with all values – Vincent LE GARREC Nov 28 '20 at 08:11

score 3 · Answer 6 · answered May 14 '13 at 17:06

3

Ref: zip built-in function

zip() in conjunction with the * operator can be used to unzip a list

unzip_lst = zip(*mylist)
for i in unzip_lst:
    for j in i:
        print j

answered May 14 '13 at 17:06

Tanky Woo

4,906
9
44
75

score 1 · Answer 7 · answered May 14 '13 at 16:55

1

>>> [el[0] if i < len(mylist) else el[1] for i,el in enumerate(mylist + mylist)]
['0,0', '1,0', '2,0', '0,1', '1,1', '2,1']

answered May 14 '13 at 16:55

dansalmo

11,506
5
58
53

I like zip method better, but I thought I would throw out a wacky alternative. – dansalmo May 14 '13 at 16:56
does this evaluate `len(myList)` only once, or does it recalculate it every iteration? – Z4-tier Apr 25 '20 at 21:29
@Z4-tier https://stackoverflow.com/questions/1115313/cost-of-len-function – dansalmo Apr 26 '20 at 22:04

Iterating over a 2 dimensional python list

7 Answers7

Linked