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.
Georgy
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asked Sep 4, 2008 at 21:04
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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 Sep 4, 2008 at 21:06
Mark BiekMark Biek
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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]
Georgy
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answered Sep 4, 2008 at 21:08
jmanning2kjmanning2k
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",".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 Feb 10, 2016 at 15:43
Ricky SahuRicky Sahu
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@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 Sep 5, 2008 at 16:29
Aaron MaenpaaAaron Maenpaa
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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 Oct 1, 2008 at 9:23
jfsjfs
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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:
//docs.python.org/library/csv.html
twasbrillig
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answered Sep 4, 2008 at 21:09
Douglas LeederDouglas Leeder
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>>> 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 May 20, 2017 at 11:31
ShameemShameem
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l=['a', 1, 'b', 2]
print str[l][1:-1]
Output: "'a', 1, 'b', 2"
AlvaroAV
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answered Sep 15, 2008 at 18:08
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@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 Sep 4, 2008 at 21:57
Peter HoffmannPeter Hoffmann
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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 May 15, 2017 at 11:25
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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 Jun 26, 2020 at 16:53
faiz-efaiz-e
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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 Sep 4, 2008 at 21:10
David SingerDavid Singer
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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 Jun 18, 2018 at 15:08
Ron KalianRon Kalian
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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
answered Nov 26, 2021 at 14:44
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 Jan 7, 2018 at 13:56
RobertoRoberto
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