Python, slicing lists?

Question

i really need your help with slicing a list.

lets say I've got a list of lists like this:

field=[[1, 2, 4, 4], [4, 1, 4, 2], [2, 1, 4, 3], [2, 4, 2, 3], [1, 2, 3, 4]]

and I have to write a function which will return a string that will look like this:

thus fat i managed to do this:

def out(field):
    b=''
    for list in field:
        list1=list[::-1]
        b+=''.join((str(sth) for sth in list1))+'\n'
    return b

which returns this:

Can you please explain the reason why you need the output you stated? Doesn't seem logical.. — Heskja, Dec 20 '14 at 12:21
@Heskja: transposition of the matrix, presumably, but it needs a rotation too. — Martijn Pieters, Dec 20 '14 at 12:24
we are programming this: https://www.youtube.com/watch?v=xKF6KNIh_mw (without graphics tho) — Rok Dolinar, Dec 20 '14 at 15:25

Martijn Pieters · Accepted Answer · 2014-12-20T12:33:30.507

3

You need to transpose your list of rows to a list of columns, as well as reverse the results of the transposition to get a proper rotation.

zip(*field) transposes the rows to columns, reversed() then reverses the results. Combined with a list comprehension you can do this all in one expression:

def out(field):
    return '\n'.join([''.join(map(str, c)) for c in reversed(list(zip(*field)))])

or spelled out into an explicit loop:

def out(field):
    b = []
    for columns in reversed(list(zip(*field))):
        b.append(''.join(map(str, columns)))
    return '\n'.join(b)

The list() call around the zip() call allows us to reverse the iterator result in Python 3; it can be dropped in Python 2.

Demo:

>>> field=[[1, 2, 4, 4], [4, 1, 4, 2], [2, 1, 4, 3], [2, 4, 2, 3], [1, 2, 3, 4]]
>>> [''.join(map(str, c)) for c in reversed(list(zip(*field)))]
['42334', '44423', '21142', '14221']
>>> '\n'.join([''.join(map(str, c)) for c in reversed(list(zip(*field)))])
'42334\n44423\n21142\n14221'
>>> print('\n'.join([''.join(map(str, c)) for c in reversed(list(zip(*field)))]))
42334
44423
21142
14221

edited Dec 20 '14 at 12:33

answered Dec 20 '14 at 12:21

Martijn Pieters

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`print("\n".join(["".join(map(str,x)) for x in reversed(zip(*field))]))`, `c[::-1]` gives wrong output – Padraic Cunningham Dec 20 '14 at 12:27
2 problems: #1 TypeError: argument to reversed() must be a sequence #2: output is wrong, i need 42334 in first row, not 43324 – Rok Dolinar Dec 20 '14 at 12:28
@RokDolinar: Are you using Python 3? – Martijn Pieters Dec 20 '14 at 12:29
@RokDolinar, `reversed(list(zip(*field))` – Padraic Cunningham Dec 20 '14 at 12:29
Yeah, the newest version. Working in pycharm 3.4.1 – Rok Dolinar Dec 20 '14 at 12:30
Working as intended! Thank you very much! – Rok Dolinar Dec 20 '14 at 12:34
@MartijnPieters Knowing you there probably is; is there a benefit to the `[` `]` brackets in the first one-liner? I'd have thought it would be better to omit them and pass join a generator rather a list (obviously pretty trivial in this example, but wondered more generally) – GP89 Dec 20 '14 at 12:37
@GP89: yes, since `str.join()` has to scan the input *twice* it'll convert an iterator to a list first *anyway*. Using a list comprehension is then faster. – Martijn Pieters Dec 20 '14 at 12:39
@GP89: see [list comprehension without \[ \], Python](http://stackoverflow.com/a/9061024) – Martijn Pieters Dec 20 '14 at 12:39
ahh I thought you'd be right :) thanks- wasn't aware it did two loops! – GP89 Dec 20 '14 at 12:43

score 0 · Answer 2 · answered Dec 20 '14 at 13:11

def out(field):
    b=''
    a=[]
    while len(field[0])>0:
        for list in field:
            a.append(list.pop())
        a.append("\n")
    a="".join(str(x) for x in a)
    return a 
x=[[1, 2, 4, 4], [4, 1, 4, 2], [2, 1, 4, 3], [2, 4, 2, 3], [1, 2, 3, 4]]
print out(x)

Python, slicing lists?

2 Answers2