vectorizing a for loop in python

Question

I'm a newbie to python and had a question to ask about vectorizing a code

def makeNames2(nList):
  for nLi in nList:
    nLIdx=[i for i,j in enumerate(nList) if j==nLi]
    if nLIdx.__len__()>1:
        for i,j in enumerate(nLIdx):
            if i>0: nList[j]=nList[j]+str(i)
  return nList

which does the following:

>>> nLTest=['asda','asda','test','ada','test','yuil','test']
>>> print(makenames2(nLTest)
['asda', 'asda1', 'test', 'ada', 'test1', 'yuil', 'test2']

The code works fine, but I was wondering if there is a way to vectorize the for loops?

EDIT

Thanks everyone for all the three answers. This is exactly what I was interested in and would have liked to selected all answers. I can't select more than one, but all of them work.

possible duplicate of [how do I parallelize a simple python loop?](http://stackoverflow.com/questions/9786102/how-do-i-parallelize-a-simple-python-loop) — Jayanth Koushik, Feb 19 '14 at 14:27
If you mean vectorize as in SSE3 stuff -- not with python out of the box. For some problems, you might be able to do it using 3rd party packages, but even then, it's hard to say when things are actually being vectorized vs. just pushed into a different language (e.g. C) in the implementation. You *can* parallelize it using `multiprocessing` (or sometimes `threading` depending on the problem and python implementation) — mgilson, Feb 19 '14 at 14:27
Also, if nLIdx.__len__()>1 can be written as if len(nLIdx)>1 or just if nLIdx — Noel Evans, Feb 19 '14 at 14:28
vectorize means several things, which is why it'd be helpful if the OP clarifies — Useless, Feb 19 '14 at 14:28
@NoelEvans `if len(nLIdx) > 1` (greater than 1, meaning at least 2 elements) is not equivalent to `if nLIdx`. — sebastian, Feb 19 '14 at 14:31
in Matlab, usually I can do stuff like `nList[nLIdx] = ...`, but in python it gives me an error that I can't use a list to subset (per se) a list or replace to a subset of a list — uday, Feb 19 '14 at 14:33
your algorithm doesn't lend itself well to vectorization anyways, because the value of element `n` depends on the values of all values `0:n-1` - a parallel solution will end up doing lots of duplicate calculations that are avoided in the below answers. but numpy will let you do that sort of thing, for some sorts of lists (not for string lists, i don't think). — Corley Brigman, Feb 19 '14 at 14:49
i agree and parallel code would be an overkill. I wanted to reduce two loops to one loop at least, and I am glad I asked this question - I got to learn how to implement the equivalent of Matlab's `find` function in python (i.e. using `dict`). — uday, Feb 19 '14 at 14:52

thefourtheye · Accepted Answer · 2014-02-19T14:59:53.627

3

nLTest, items = ['asda','asda','test','ada','test','yuil','test'], {}
for idx, item in enumerate(nLTest):
    nLTest[idx] += str(items.setdefault(item, 0) or "")
    items[item] += 1
print nLTest

Output

['asda', 'asda1', 'test', 'ada', 'test1', 'yuil', 'test2']

edited Feb 19 '14 at 14:59

answered Feb 19 '14 at 14:33

thefourtheye

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I think you can further avoid the if and `nLTest[idx] += str(items.setdefault(item, 0) or "")` – loopbackbee Feb 19 '14 at 14:57
@goncalopp Awesome :) Thanks man. Updated my answer with that. – thefourtheye Feb 19 '14 at 15:00

score 1 · Answer 2 · answered Feb 19 '14 at 14:32

You could simplify it a bit:

def makenames(lst):
    seen = {}
    for index, name in enumerate(lst):
        if name in seen:
            seen[name] += 1
            lst[index] = "{0}{1}".format(name, seen[name])
        else:
            seen[name] = 0
    return lst

This removes one of the for loops, operating in O(n) (dictionary access is O(1)).

Note that this modifies the list in-place; you may wish to have a new output list to append to instead. You could also simplify this slightly using defaultdict or Counter from the collections module.

score 1 · Answer 3 · answered Feb 19 '14 at 14:43

1

This is arguably more readable, avoids O(n^2). It's also not in-place.

from collections import defaultdict
def makeNames3(nList):
    counter= defaultdict(lambda:0)
    def posfix(x):
        n= counter[x]
        counter[x]+=1
        return str(n) if n>0 else ""
    return [x+posfix(x) for x in nList]

answered Feb 19 '14 at 14:43

loopbackbee

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You can simply do `defaultdict(int)` – thefourtheye Feb 19 '14 at 14:44
@thefourtheye You're right, I didn't know `int` had a default value. I'm not sure if this a case of *explicit is better than implicit*, though – loopbackbee Feb 19 '14 at 14:50

vectorizing a for loop in python

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