How would you make a comma-separated string from a list of strings?

593

What would be your preferred way to concatenate strings from a sequence such that between every two consecutive pairs a comma is added. That is, how do you map, for instance, ['a', 'b', 'c'] to 'a,b,c'? (The cases ['s'] and [] should be mapped to 's' and '', respectively.)

I usually end up using something like ''.join(map(lambda x: x+',',l))[:-1], but also feeling somewhat unsatisfied.

This question is tagged with python string list

~ Asked on 2008-09-04 21:04:04

14 Answers


1099

my_list = ['a', 'b', 'c', 'd']
my_string = ','.join(my_list)
'a,b,c,d'

This won't work if the list contains integers


And if the list contains non-string types (such as integers, floats, bools, None) then do:

my_string = ','.join(map(str, my_list)) 

~ Answered on 2008-09-04 21:06:12


83

Why the map/lambda magic? Doesn't this work?

>>> foo = ['a', 'b', 'c']
>>> print(','.join(foo))
a,b,c
>>> print(','.join([]))

>>> print(','.join(['a']))
a

In case if there are numbers in the list, you could use list comprehension:

>>> ','.join([str(x) for x in foo])

or a generator expression:

>>> ','.join(str(x) for x in foo)

~ Answered on 2008-09-04 21:08:29


21

",".join(l) will not work for all cases. I'd suggest using the csv module with StringIO

import StringIO
import csv

l = ['list','of','["""crazy"quotes"and\'',123,'other things']

line = StringIO.StringIO()
writer = csv.writer(line)
writer.writerow(l)
csvcontent = line.getvalue()
# 'list,of,"[""""""crazy""quotes""and\'",123,other things\r\n'

~ Answered on 2016-02-10 15:43:06


14

Here is a alternative solution in Python 3.0 which allows non-string list items:

>>> alist = ['a', 1, (2, 'b')]
  • a standard way

    >>> ", ".join(map(str, alist))
    "a, 1, (2, 'b')"
    
  • the alternative solution

    >>> import io
    >>> s = io.StringIO()
    >>> print(*alist, file=s, sep=', ', end='')
    >>> s.getvalue()
    "a, 1, (2, 'b')"
    

NOTE: The space after comma is intentional.

~ Answered on 2008-10-01 09:23:00


13

@Peter Hoffmann

Using generator expressions has the benefit of also producing an iterator but saves importing itertools. Furthermore, list comprehensions are generally preferred to map, thus, I'd expect generator expressions to be preferred to imap.

>>> l = [1, "foo", 4 ,"bar"]
>>> ",".join(str(bit) for bit in l)
'1,foo,4,bar' 

~ Answered on 2008-09-05 16:29:25


13

Don't you just want:

",".join(l)

Obviously it gets more complicated if you need to quote/escape commas etc in the values. In that case I would suggest looking at the csv module in the standard library:

https://docs.python.org/library/csv.html

~ Answered on 2008-09-04 21:09:33


8

>>> my_list = ['A', '', '', 'D', 'E',]
>>> ",".join([str(i) for i in my_list if i])
'A,D,E'

my_list may contain any type of variables. This avoid the result 'A,,,D,E'.

~ Answered on 2017-05-20 11:31:01


5

l=['a', 1, 'b', 2]

print str(l)[1:-1]

Output: "'a', 1, 'b', 2"

~ Answered on 2008-09-15 18:08:41


4

@jmanning2k using a list comprehension has the downside of creating a new temporary list. The better solution would be using itertools.imap which returns an iterator

from itertools import imap
l = [1, "foo", 4 ,"bar"]
",".join(imap(str, l))

~ Answered on 2008-09-04 21:57:51


3

If you want to do the shortcut way :) :

','.join([str(word) for word in wordList])

But if you want to show off with logic :) :

wordList = ['USD', 'EUR', 'JPY', 'NZD', 'CHF', 'CAD']
stringText = ''

for word in wordList:
    stringText += word + ','

stringText = stringText[:-2]   # get rid of last comma
print(stringText)

~ Answered on 2020-06-26 16:53:06


3

Here is an example with list

>>> myList = [['Apple'],['Orange']]
>>> myList = ','.join(map(str, [i[0] for i in myList])) 
>>> print "Output:", myList
Output: Apple,Orange

More Accurate:-

>>> myList = [['Apple'],['Orange']]
>>> myList = ','.join(map(str, [type(i) == list and i[0] for i in myList])) 
>>> print "Output:", myList
Output: Apple,Orange

Example 2:-

myList = ['Apple','Orange']
myList = ','.join(map(str, myList)) 
print "Output:", myList
Output: Apple,Orange

~ Answered on 2017-05-15 11:25:37


1

I would say the csv library is the only sensible option here, as it was built to cope with all csv use cases such as commas in a string, etc.

To output a list l to a .csv file:

import csv
with open('some.csv', 'w', newline='') as f:
    writer = csv.writer(f)
    writer.writerow(l)  # this will output l as a single row.  

It is also possible to use writer.writerows(iterable) to output multiple rows to csv.

This example is compatible with Python 3, as the other answer here used StringIO which is Python 2.

~ Answered on 2018-06-18 15:08:24


0

Unless I'm missing something, ','.join(foo) should do what you're asking for.

>>> ','.join([''])
''
>>> ','.join(['s'])
's'
>>> ','.join(['a','b','c'])
'a,b,c'

(edit: and as jmanning2k points out,

','.join([str(x) for x in foo])

is safer and quite Pythonic, though the resulting string will be difficult to parse if the elements can contain commas -- at that point, you need the full power of the csv module, as Douglas points out in his answer.)

~ Answered on 2008-09-04 21:10:25


-1

My two cents. I like simpler an one-line code in python:

>>> from itertools import imap, ifilter
>>> l = ['a', '', 'b', 1, None]
>>> ','.join(imap(str, ifilter(lambda x: x, l)))
a,b,1
>>> m = ['a', '', None]
>>> ','.join(imap(str, ifilter(lambda x: x, m)))
'a'

It's pythonic, works for strings, numbers, None and empty string. It's short and satisfies the requirements. If the list is not going to contain numbers, we can use this simpler variation:

>>> ','.join(ifilter(lambda x: x, l))

Also this solution doesn't create a new list, but uses an iterator, like @Peter Hoffmann pointed (thanks).

~ Answered on 2018-01-07 13:56:43


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