Question

使用系统库，效率非常低：

foo = "foo bar car haz can bar foo"
repl = {('foo', 'bar'), ('car', 'bar'), ('foo', 'haz')}

for rep in repl:
    foo = foo.replace(rep)

替换批次replace应该在O（| foo | + | repl |）而不是O（| foo |×| repl |）中解决此问题，如上述方法。

您能想到实现这种更有效解决方案的简洁方法吗？

Answer 1

使用正则表达式

import re

foo = "foo bar car haz can bar foo"

re.sub(r"(foo|car)", "bar", foo)

> 'bar bar bar haz can bar bar'

实际上在这种情况下速度较慢：

import timeit

timeit.timeit('re.sub(r"(foo|car)", "bar", "foo bar car haz can bar foo")', 'import re', number=10000)

> 0.051492929458618164

def test():
    foo = "foo bar car haz can bar foo"
    repl = [('foo', 'bar'), ('car', 'bar'), ('foo', 'haz')]
    for rep in repl:
        foo = foo.replace(*rep)

timeit.timeit("test()", setup="from __main__ import test", number=10000)

> 0.026629924774169922

批量字符串`replace`？

1 个答案: