create specific permutation of a list in python

Question

I'm trying to write a function that would input a list (of pre-determined length) and output the permutation of that list according to my choice of re-ordering.

For example, suppose I have the list

    [1,'a',12,'b','poppy']

and I have the permutation

    (3,4,5,2,1)

Then the function output would be a list with the third element first, the fourth element second, etc

    [12,'b','poppy','a',1]

Padraic Cunningham · Accepted Answer · 2015-06-16T20:54:49.403

5

l = [1,'a',12,'b','poppy']

def p(l,t):
    return [l[i-1] for i in t]

print(p(l,(3,4,5,2,1)))
[12, 'b', 'poppy', 'a', 1]

indexing is 0 based so if you actually wanted to use the indexes for a 5 element list it would be (2,3,4,1,0) and [l[i] for i in t]

edited Jun 16 '15 at 20:54

answered Jun 16 '15 at 20:53

Padraic Cunningham

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1

This is basically [this](http://stackoverflow.com/questions/6098250/in-place-way-to-apply-a-permutation-to-a-list-inverse-of-sorting-by-key) with a -1. Should I dup-close or not? – DSM Jun 16 '15 at 20:54
1

This question is much clearer written than that similar question. – joeforker Jun 16 '15 at 20:55
@DSM, probably 50/50, the OP may have actually also meant to use zero based indexing so that would be an exact dupe then. – Padraic Cunningham Jun 16 '15 at 20:56
Thank you, that's great! – cnrk Jun 16 '15 at 20:58
I'm surprised no one proposed an itemgetter-based approach (although that'd need an extra list call). – DSM Jun 16 '15 at 20:59
@cnrk, no worries, just make sure if you use zero based indexing t remove the -1 – Padraic Cunningham Jun 16 '15 at 21:01
@DSM Ask and ye shall receive :) – chepner Jun 16 '15 at 21:13
@DSM, I suppose you could `itemgetter(*(i-1 for i in t))(l)` without needed a list call – Padraic Cunningham Jun 16 '15 at 21:26
1

@PadraicCunningham: yeah, but then the output is a tuple, not a list. – DSM Jun 16 '15 at 22:37
@DSM, On that note I am going to turn off my computer ;) – Padraic Cunningham Jun 16 '15 at 22:41

score 2 · Answer 2 · answered Jun 16 '15 at 21:12

Use operator.itemgetter. Since your permutations are for a 1-based list, you'll need to add a dummy element to the front of your list first. (You could subtract one from each element of p using a variety of techniques, but it's simpler to just "shift" the list before passing it to the function returned by itemgetter.)

>>> from operator import itemgetter
>>> p = (3,4,5,2,1)
>>> lst = [1,'a',12,'b','poppy']
>>> itemgetter(*p)([None] + lst)
(12, 'b', 'poppy', 'a', 1)

I was going to `itemgetter(*map(lambda x: x-1, t))(l)` but this is better. — Padraic Cunningham, Jun 16 '15 at 21:21

score 0 · Answer 3 · answered Jun 16 '15 at 20:57

0

The first element of the list is 0:

l = [1,'a',12,'b','poppy']
p = (2,3,4,1,0)
result = []
for i in p:
    result.append(l[i])

answered Jun 16 '15 at 20:57

joeforker

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create specific permutation of a list in python

3 Answers3