加入 python 中的字符串列表並將每個字符串用引號引起來

Question

我有：

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

我希望有：

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

使用 Python 最簡單的方法是什么？

Answer 1

2021 年更新：在 Python3 中使用 f 字符串

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

原始答案（支持 Python 2.6+）

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

Answer 2

您也可以執行單一format調用

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

更新：一些基准測試（在 2009 mbp 上執行）：

>>> 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

所以看起來這種format實際上相當昂貴

更新 2：按照@JCode 的評論，添加map以確保join正常工作，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

Answer 3

你可以試試這個：

str(words)[1:-1]

Answer 4

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

Answer 5

@jamylak 的更新版本使用 F 字符串回答（對於 python 3.6+），我對用於 SQL 腳本的字符串使用了反引號。

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

Answer 6

找到更快的方法

'"' + '","'.join(words) + '"'

在 Python 2.7 中測試：

    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

結果輸出是：

# "hello", "world", "you", "look", "nice"
# 'hello', 'world', 'you', 'look', 'nice'
# "hello", "world", "you", "look", "nice"
# 39.6000976562
# 166.892822266
# 220.110839844

Answer 7

# Python3 without for loop
conc_str = "'{}'".format("','".join(['a', 'b', 'c']))
print(conc_str) 

# "'a', 'b', 'c'"



    

Answer 8

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"'