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? — Jayanth Koushik
– Jayanth Koushik, Commented Feb 19, 2014 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
– mgilson, Commented Feb 19, 2014 at 14:27
Also, if nLIdx.__len__()>1 can be written as if len(nLIdx)>1 or just if nLIdx — Noel Evans
– Noel Evans, Commented Feb 19, 2014 at 14:28
@NoelEvans if len(nLIdx) > 1 (greater than 1, meaning at least 2 elements) is not equivalent to if nLIdx. — sebastian
– sebastian, Commented Feb 19, 2014 at 14:31

thefourtheye · Accepted Answer · 2014-02-19 14:59:53Z

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, 2014 at 14:59

answered Feb 19, 2014 at 14:33

thefourtheye

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2 Comments

loopbackbee Over a year ago

I think you can further avoid the if and nLTest[idx] += str(items.setdefault(item, 0) or "")

thefourtheye Over a year ago

@goncalopp Awesome :) Thanks man. Updated my answer with that.

jonrsharpe · Accepted Answer · 2014-02-19 14:32:57Z

1

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.

answered Feb 19, 2014 at 14:32

jonrsharpe

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Comments

loopbackbee · Accepted Answer · 2014-02-19 14:43:19Z

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, 2014 at 14:43

loopbackbee

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2 Comments

thefourtheye Over a year ago

You can simply do defaultdict(int)

loopbackbee Over a year ago

@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

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