Aam*_*nan 36 python performance
我生成所有可能的三个字母关键字e.g. aaa, aab, aac.... zzy, zzz
下面是我的代码:
alphabets = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
keywords = []
for alpha1 in alphabets:
for alpha2 in alphabets:
for alpha3 in alphabets:
keywords.append(alpha1+alpha2+alpha3)
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能否以更加流畅有效的方式实现此功能?
agf*_*agf 85
keywords = itertools.product(alphabets, repeat = 3)
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请参阅文档itertools.product
.如果您需要字符串列表,请使用
keywords = [''.join(i) for i in itertools.product(alphabets, repeat = 3)]
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alphabets
也不需要是一个列表,它可以只是一个字符串,例如:
from itertools import product
from string import ascii_lowercase
keywords = [''.join(i) for i in product(ascii_lowercase, repeat = 3)]
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如果您只想要小写的ascii字母,它将起作用.
Joh*_*ooy 15
您也可以使用map而不是list comprehension(这是map仍然比LC更快的情况之一)
>>> from itertools import product
>>> from string import ascii_lowercase
>>> keywords = map(''.join, product(ascii_lowercase, repeat=3))
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列表理解的这种变化也比使用更快 ''.join
>>> keywords = [a+b+c for a,b,c in product(ascii_lowercase, repeat=3)]
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from itertools import combinations_with_replacement
alphabets = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
for (a,b,c) in combinations_with_replacement(alphabets, 3):
print a+b+c
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