在python中加入字符串列表,并将每个字符串都用引号引起来

问题:在python中加入字符串列表,并将每个字符串都用引号引起来

我有:

words = ['hello', 'world', 'you', 'look', 'nice']

我希望有:

'"hello", "world", "you", "look", "nice"'

用Python做到这一点最简单的方法是什么?

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?


回答 0

>>> words = ['hello', 'world', 'you', 'look', 'nice']
>>> ', '.join('"{0}"'.format(w) for w in words)
'"hello", "world", "you", "look", "nice"'
>>> words = ['hello', 'world', 'you', 'look', 'nice']
>>> ', '.join('"{0}"'.format(w) for w in words)
'"hello", "world", "you", "look", "nice"'

回答 1

您也可以执行一次format通话

>>> words = ['hello', 'world', 'you', 'look', 'nice']
>>> '"{0}"'.format('", "'.join(words))
'"hello", "world", "you", "look", "nice"'

更新:一些基准测试(以2009 Mbps的速度执行):

>>> 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

所以看来format实际上很贵

更新2:在@JCode的注释之后,添加了一个map以确保join可以运行,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

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

回答 2

您可以尝试以下方法:

str(words)[1:-1]

You can try this :

str(words)[1:-1]

回答 3

>>> ', '.join(['"%s"' % w for w in words])
>>> ', '.join(['"%s"' % w for w in words])

回答 4

@jamylak答案的更新版本带有F字符串(适用于python 3.6+),我已经在SQL脚本使用的字符串中使用了反引号。

keys = ['foo', 'bar' , 'omg']
', '.join(f'`{k}`' for k in keys)
# result: '`foo`, `bar`, `omg`'

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`'