Iterating over a 2 dimensional python list [duplicate]

Answer 1

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

Answer 2

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
>>> 

Answer 3

>>> 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')

Answer 4

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']

Answer 5

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')]

Answer 6

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
>>>