get all elements from list of lists in rows of 5

Question

so say I have a list of lists

x = [['#', '#', '#', '#', '#'], ['#', '0', ' ', ' ', '#'], ['#', '#', '#', ' ', '#']]

Say I need to split this into 3 rows of strings, how do i do this?

Here is how I can do it BUT it is not scalable, say i had tons more lists, then I would have to write so many print statments. I thought about a for statment

print "".join(mlist[0])
print "".join(mlist[1])
print "".join(mlist[2])

I was thinking about something like this but it didn't work

zert = ""
total = 0
for a in mlist:
     for b in a:
        if total < 6:
            print zert
            total = 0
            zert = ''
        zert += b

        total += 1

^ the problem above is that i would need to save a first, then iterate over it BUT just checking if there is not an inbuilt function? I tried ''.join(mlist) but that does work since its lists in a list?

Is there a simpler way to do this?

You want each inner list on a line, rather than specifically 3? — doctorlove, Aug 15 '13 at 11:25

score 4 · Accepted Answer · answered Aug 15 '13 at 11:26

4

You can use a list comprehension to join lists:

print '\n'.join([''.join(inner) for inner in mlist])

The list comprehension creates a string of each nested list, then we join that new list of rows into a larger string with newlines.

Demo:

>>> mlist = [['#', '#', '#', '#', '#'], ['#', '0', ' ', ' ', '#'], ['#', '#', '#', ' ', '#']]
>>> print '\n'.join([''.join(inner) for inner in mlist])
#####
#0  #
### #

You could also have used a for loop:

for inner in mlist:
    print ''.join(inner)

answered Aug 15 '13 at 11:26

Martijn Pieters

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296
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I think you can use a generator expression here, can't you? `'\n'.join(''.join(inner) for inner in mlist)` – Benjamin Hodgson Aug 15 '13 at 11:51
@poorsod: Yes, but that is slower than using a list comprehension. See [list comprehension without \[ \], Python](http://stackoverflow.com/q/9060653) – Martijn Pieters Aug 15 '13 at 11:52
@MartijnPieters Thanks for the link! Is `join` the only case where the list comp is faster than the generator exp? – Benjamin Hodgson Aug 15 '13 at 11:55
1

@poorsod: `str.join()` is a specific case where using a list comp is better, because `str.join()` **has** to use a list internally (it iterates over the strings twice, once to calculate the output length, and a second time to build the string). – Martijn Pieters Aug 15 '13 at 11:57
@poorsod: In most cases, avoiding having to build a list is better. – Martijn Pieters Aug 15 '13 at 11:57

score 1 · Answer 2 · answered Aug 15 '13 at 11:27

1

>>> orig_list = [['#', '#', '#', '#', '#'], ['#', '0', ' ', ' ', '#'], ['#', '#', '#', ' ', '#']]
>>> new_list = [ "".join(x) for x in orig_list ]
>>> new_list
['#####', '#0  #', '### #']
>>> print("\n".join(new_list))
#####
#0  #
### #

answered Aug 15 '13 at 11:27

DhruvPathak

42,059
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score 1 · Answer 3 · answered Aug 15 '13 at 11:28

1

Working with what you have so far, to just directly print it.

for a in mlist:
  print "".join(a)

answered Aug 15 '13 at 11:28

doctorlove

18,872
2
46
62

score 1 · Answer 4 · answered Aug 15 '13 at 11:29

1

or most simply

for a in x:
    print "".join(a)

answered Aug 15 '13 at 11:29

brainovergrow

458
4
13

get all elements from list of lists in rows of 5

4 Answers4