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

Question

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.

Answer 1

mmm also need for SQL is :

l = ["foo" , "baar" , 6]
where_clause = "..... IN ("+(','.join([ f"'{x}'" for x in l]))+")"
>> "..... IN ('foo','baar','6')"

enjoit

Answer 2

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)

Answer 3

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)) 

Answer 4

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)

Answer 5

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.

Answer 6

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).

Answer 7

>>> 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'.

Answer 8

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

Answer 9

",".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'

Answer 10

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

print str(l)[1:-1]

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

Answer 11

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

Answer 12

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.

Answer 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' 

Answer 14

@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))

Answer 15

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.)