I've got:
words = ['hello', 'world', 'you', 'look', 'nice']
I want to have:
'"hello", "world", "you", "look", "nice"'
What's the easiest way to do this with Python?
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9 Answers
Update 2021: With f strings in Python3
>>> words = ['hello', 'world', 'you', 'look', 'nice']
>>> ', '.join(f'"{w}"' for w in words)
'"hello", "world", "you", "look", "nice"'
Original Answer (Supports Python 2.6+)
>>> words = ['hello', 'world', 'you', 'look', 'nice']
>>> ', '.join('"{0}"'.format(w) for w in words)
'"hello", "world", "you", "look", "nice"'
3 Comments
jamylak
@Meow that uses
repr which is a lil hacky in this specific case as opposed to being clear with the quotes
Meow
@jamlak ok, repr just seemed safer to me incase you have quotes in your string.
jamylak
@Meow good point about the quotes, I guess that this one has the advantage of being explicit while the other approach does support the quotes thing
You can try this :
str(words)[1:-1]
2 Comments
Dejell
how does it add quotes?
Félix Caron
This adds single quotation marks instead of double, but quotes inside the words will be automatically escaped. +1 for cleverness.
you may also perform a single format call
>>> words = ['hello', 'world', 'you', 'look', 'nice']
>>> '"{0}"'.format('", "'.join(words))
'"hello", "world", "you", "look", "nice"'
Update: Some benchmarking (performed on a 2009 mbp):
>>> timeit.Timer("""words = ['hello', 'world', 'you', 'look', 'nice'] * 100; ', '.join('"{0}"'.format(w) for w in words)""").timeit(1000)
0.32559704780578613
>>> timeit.Timer("""words = ['hello', 'world', 'you', 'look', 'nice'] * 100; '"{}"'.format('", "'.join(words))""").timeit(1000)
0.018904924392700195
So it seems that format is actually quite expensive
Update 2: following @JCode's comment, adding a map to ensure that join will work, Python 2.7.12
>>> timeit.Timer("""words = ['hello', 'world', 'you', 'look', 'nice'] * 100; ', '.join('"{0}"'.format(w) for w in words)""").timeit(1000)
0.08646488189697266
>>> timeit.Timer("""words = ['hello', 'world', 'you', 'look', 'nice'] * 100; '"{}"'.format('", "'.join(map(str, words)))""").timeit(1000)
0.04855608940124512
>>> timeit.Timer("""words = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] * 100; ', '.join('"{0}"'.format(w) for w in words)""").timeit(1000)
0.17348504066467285
>>> timeit.Timer("""words = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] * 100; '"{}"'.format('", "'.join(map(str, words)))""").timeit(1000)
0.06372308731079102
6 Comments
kadrian
does this have a better performance than the one proposed by jamylak?
jamylak
@sage88 No it isn't. Preoptimization is evil, there is a 0.0000001% chance that the miniscule speed difference here is the entire bottleneck of a Python script. Also this code is much less intuitive so it is not better, it is very slightly faster. My solution is more pythonic and readable
WloHu
@marchelbling Benchmark is invalid because jamylak' solution works also for iterables of non-strings. Replace
.join(words) with .join(map(str, words)) and show us how that goes.
marchelbling
@JCode updated the benchmark. The gap is smaller but there still is a 2x gain on my machine.
WloHu
@marchelbling Keep in mind that
map in Python 2 returns a list but in Python 3 map object generator is returned. I suggested map to express the idea and I didn't know that you used Python 2. |
>>> ', '.join(['"%s"' % w for w in words])
1 Comment
Khertan
Format is available since python 2.6.
An updated version of @jamylak answer with F Strings (for python 3.6+), I've used backticks for a string used for a SQL script.
keys = ['foo', 'bar' , 'omg']
', '.join(f'`{k}`' for k in keys)
# result: '`foo`, `bar`, `omg`'
2 Comments
Marcel Flygare
Works, but is it considered to be the canonical pythonic way ?
jamylak
@MarcelFlygare Yeah
f strings seem to be the best way nowfind a faster way
'"' + '","'.join(words) + '"'
test in Python 2.7:
words = ['hello', 'world', 'you', 'look', 'nice']
print '"' + '","'.join(words) + '"'
print str(words)[1:-1]
print '"{0}"'.format('", "'.join(words))
t = time() * 1000
range10000 = range(100000)
for i in range10000:
'"' + '","'.join(words) + '"'
print time() * 1000 - t
t = time() * 1000
for i in range10000:
str(words)[1:-1]
print time() * 1000 - t
for i in range10000:
'"{0}"'.format('", "'.join(words))
print time() * 1000 - t
The resulting output is:
# "hello", "world", "you", "look", "nice"
# 'hello', 'world', 'you', 'look', 'nice'
# "hello", "world", "you", "look", "nice"
# 39.6000976562
# 166.892822266
# 220.110839844
Comments
words = ['hello', 'world', 'you', 'look', 'nice']
S = ""
for _ in range(len(words)-1):
S+=f'"{words[_]}"'+', '
S +=f'"{words[len(words)-1]}"'
print("'"+S+"'")
OUTPUT:
'"hello", "world", "you", "look", "nice"'
Comments
# Python3 without for loop
conc_str = "'{}'".format("','".join(['a', 'b', 'c']))
print(conc_str)
# "'a', 'b', 'c'"
There are two possible solution and for me we should choose wisely...
items = ['A', 'B', 'C']
# Fast since we used only string concat.
print("\"" + ("\",\"".join(items)) + "\"")
# Slow since we used loop here.
print(",".join(["\"{item}\"".format(item=item) for item in items]))
