带有2d数组的Python .join（）

Question

Say I have this array:

foo = [["a", "b"], ["c", "d"]]

If I want to print the elements of the inner arrays, I would do something like this:

for array_ in foo:
    for element in array_:
        print element

After doing this, I would get this result:

a
b
c
d

How can I print each inner array vertically, then join it with the other array, like this?

ac
bd

Answer 1

One way is to use zip() :

>>> foo = [["a", "b"], ["c", "d"]]
>>> [''.join(a) for a in zip(*foo)]
['ac', 'bd']

zip(*foo) returns list of tuples, where the i-th tuple contains the i-th element from each of the input lists. And join() concatenates the content of the lists.

If the lists in the list of lists variable foo are large, you could use an efficient version of zip() from the itertools module:

>>> from itertools import izip
>>> foo = [["a", "b"], ["c", "d"]]
>>> [''.join(a) for (a) in izip(*foo)]
['ac', 'bd']

Answer 2

A shorter version using map() :

import operator as op
map(op.add ,*foo)
=> ['ac', 'bd']

Answer 3

map will work also:

["".join(x) for x in map(None,*foo)]

Just join again to get just the strings:

"\n".join(["".join(x) for x in map(None,*foo)])

In this case map behaves like zip, it transposes the elements:

In [39]: foo = [["a", "b"], ["c", "d"]]

In [40]: map(None,*foo)
Out[40]: [('a', 'c'), ('b', 'd')]

Answer 4

In [1]: foo = [["a", "b"], ["c", "d"]]

In [2]: print "\n".join("".join(f) for f in zip(*foo))
ac
bd

The inner call to zip will transpose the array:

In [3]: zip(*foo)
Out[3]: [('a', 'c'), ('b', 'd')]

The first join will concatenate each tuple together:

In [4]: ["".join(element) for element in zip(*foo)]
Out[4]: ['ac', 'bd']

Finally each element in the list is joined by the outer call to join