Python - list concatenation

Question

I have a question about list concatenation in Python, I have this piece of code:

def lista():
    word = sys.argv[1]
    l = []
    m = []
    for file_name in sys.argv[2:]:
        with open(file_name, "r") as f:
            for line in f:
                l + [len(re.findall(word, line))] #doesn't work
                m.append(len(re.findall(word, line))) #works
    print l
    print m
    return l

when I run this function I always get empty list l, but there are elements in m, why l+[elem] doesen't work for me?

note: l is a bad variable name: python.org/dev/peps/pep-0008 — jamylak
– jamylak, Commented Jul 13, 2012 at 9:14

Ivaylo Strandjev · Accepted Answer · 2012-04-20 09:31:35Z

4

You are never assigning a new value to l. You should use assignment. Try calling l = l + [len(re.findall(word,line))]

EDIT: another option is to use the += opertor: l+=[len(re.findall(word,line))]

answered Apr 20, 2012 at 9:31

Ivaylo Strandjev

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Abhijit · Accepted Answer · 2012-04-20 09:39:18Z

3

Because l+[len(re.findall(word,line))] is just adding the two list but throwing away the result. You probably want

l+=[len(re.findall(word,line))]

General observations:

Your variable names are not according to PEP-8
Appending a list is costly and generally not pythonic. You can use a generator for this purpose.
You should use a with statement with files, to ensure it is closed even if an exception occurs.

Here is the Program with suggested edits.

def lista():
    word=sys.argv[1]
    def search():
        for file_name in sys.argv[2:]:
            with open(file_name,"r") as fin:
                for line in fin:
                    yield len(re.findall(word,line))
    return [l for l in search()]

edited Apr 20, 2012 at 9:39

answered Apr 20, 2012 at 9:32

Abhijit

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Gareth Latty · Accepted Answer · 2012-04-20 09:37:35Z

2

Here you want to just use l.append() - what you are doing has the same effect, except it's less readable and slower.

python -m timeit -s "l=[]" "for i in range(1000):" "  l.append(i)"
10000 loops, best of 3: 106 usec per loop

python -m timeit -s "l=[]" "for i in range(1000):" "  l += [i]"
10000 loops, best of 3: 124 usec per loop

answered Apr 20, 2012 at 9:37

Gareth Latty

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Comments

bpgergo · Accepted Answer · 2012-04-20 09:34:31Z

1

l = l+[len(re.findall(word,line))] #works

answered Apr 20, 2012 at 9:34

bpgergo

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