Split string by comma, ignoring comma inside string. Am trying CSV

Question

I have a string like this:

s = '1,2,"hello, there"'

And I want to turn it into a list:

[1,2,"hello, there"]

Normally I'd use split:

my_list = s.split(",")

However, that doesn't work if there's a comma in a string.

So, I've read that I need to use cvs, but I don't really see how. I've tried:

from csv import reader
s = '1,2,"hello, there"'
ll = reader(s)
print ll 
for row in ll:
    print row

Which writes:

<_csv.reader object at 0x020EBC70>

['1']
['', '']
['2']
['', '']
['hello, there']

I've also tried with

ll = reader(s, delimiter=',')

where did you get that you need to use a csv to split string? — Padraic Cunningham
– Padraic Cunningham, Commented Apr 17, 2015 at 21:55
@user984003, please accept one or many of the answers that solve your problem. — gabe
– gabe, Commented Apr 18, 2015 at 22:54

dkroy · Accepted Answer · 2015-04-17 21:55:41Z

2

It is that way because you provide the csv reader input as a string. If you do not want to use a file or a StringIO object just wrap your string in a list as shown below.

>>> import csv
>>> s = ['1,2,"hello, there"']
>>> ll = csv.reader(s, delimiter=',', quotechar='"', quoting=csv.QUOTE_MINIMAL)
>>> list(ll)
[['1', '2', 'hello, there']]

answered Apr 17, 2015 at 21:55

dkroy

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1 Comment

Ivan Anishchuk Over a year ago

You mention StringIO and file but actually use a list in your example. You can use use any iterable there: list or tuple will work just as fine.

gabe · Accepted Answer · 2015-04-17 21:50:56Z

1

It sounds like you probably want to use the csv module. To use the reader on a string, you want a StringIO object.

As an example:

>> import csv, StringIO
>> print list(csv.reader(StringIO.StringIO(s)))
[['1', '2', 'hello, there']]

To clarify, csv.reader expects a buffer object, not a string. So StringIO does the trick. However, if you're reading this csv from a file object, (a typical use case) you can just as easily give the file object to the reader and it'll work the same way.

answered Apr 17, 2015 at 21:50

gabe

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2 Comments

user984003 Over a year ago

That seems to work, thanks! Anyone want to tell me why it was downvoted? It does turn the int into strings, but I'm ok with that.

Ivan Anishchuk Over a year ago

You can use use any iterable, actually: list or tuple will work just as fine.

Ivan Anishchuk · Accepted Answer · 2015-04-17 22:19:08Z

1

It's usually easier to re-use than to invent a bicycle... You just to use csv library properly. If you can't for some reason, you can always check the source code out and learn how's the parsing done there.

Example for parsing a single string into a list. Notice that the string in wrapped in list.

>>> import csv
>>> s = '1,2,"hello, there"'
>>> list(csv.reader([s]))[0]
['1', '2', 'hello, there']

edited Apr 17, 2015 at 22:19

answered Apr 17, 2015 at 21:55

Ivan Anishchuk

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1 Comment

Ivan Anishchuk Over a year ago

Thanks for pointing it out. Apparently I wasn't the first with this answer. It's a little weird for me to use SO mobile app.

meiamsome · Accepted Answer · 2015-04-17 21:56:32Z

0

You can split first by the string delimiters, then by the commas for every even index (The ones not in the string)

import itertools

new_data = s.split('"')
for i in range(len(new_data)):
    if i % 2 == 1: # Skip odd indices, making them arrays
       new_data[i] = [new_data[i]]
    else:
        new_data[i] = new_data[i].split(",")
data = itertools.chain(*new_data)

Which goes something like:

'1,2,"hello, there"'
['1,2,', 'hello, there']
[['1', '2'], ['hello, there']]
['1', '2', 'hello, there']

But it's probably better to use the csv library if that's what you're working with.

answered Apr 17, 2015 at 21:56

meiamsome

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Comments

Ben · Accepted Answer · 2015-04-17 22:04:46Z

0

You could also use ast.literal_eval if you want to preserve the integers:

>>> from ast import literal_eval
>>> literal_eval('[{}]'.format('1,2,"hello, there"'))
[1, 2, 'hello, there']

answered Apr 17, 2015 at 22:04

Ben

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