执行此操作的最佳方法是用字符串中的“-”替换字符列表。？答案

【问题标题】：What is the best way to do this replace a list of characters with '-' in a string.?执行此操作的最佳方法是用字符串中的“-”替换字符列表。？
【发布时间】：2019-07-16 19:01:55
【问题描述】：

我想用“-”替换这些符号，我知道应该有比这样做更好的方法：

if '/' in var1:
    var1= var1.replace('/', '-')
if '#' in var1:
    var1= var1.replace('#', '-')
if ';' in var1:
    var1 = var1.replace(';', '-')
if ':' in var1:
    var1= var1.replace(':', '-')

这是我尝试过的，这显然是错误的，我无法正确优化它。

str = 'Testing PRI/Sec (#434242332;PP:432:133423846,335)'
a = ['#',':',';','/']
print([str.replace(i,'-') for i in str])

replaceAll 不起作用，给我一个错误，说 str 没有该属性。

str.replaceAll("[<>]", "")

【问题讨论】：

标签： python-3.x list list-comprehension

【解决方案1】：

用str.translate()怎么样？

# make a translation table that replaces any of "#:;/" with hyphens
hyphenator = str.maketrans({c: "-" for c in "#:;/"})
# use str.translate to apply it
print("Testing PRI/Sec (#434242332;PP:432:133423846,335)".translate(hyphenator))

或者，甚至更快，使用已编译的正则表达式：

compiled_re = re.compile("|".join(re.escape(i) for i in "#:;/"))
print(compiled_re.sub("-", "Testing PRI/Sec (#434242332;PP:432:133423846,335)"))

这两种方法都比提出的其他方法快得多（至少在那个输入上）：

import re
import timeit

s = "Testing PRI/Sec (#434242332;PP:432:133423846,335)"
a = ["#", ":", ";", "/"]
hyphenator = str.maketrans({c: "-" for c in "#:;/"})


def str_translate():
    s.translate(hyphenator)


def join_generator():
    "".join("-" if ch in a else ch for ch in s)


def append_in_loop():
    temp = ""
    for i in s:
        if i in a:
            temp += "-"
        else:
            temp += i


def re_sub():
    re.sub("|".join(re.escape(i) for i in a), "-", s)

def compiled_re_sub():
    compiled_re.sub("-", s)

for method in [str_translate, join_generator, re_sub, append_in_loop, compiled_re_sub]:
    # run a million iterations and report the total time
    print("{} took a total of {}s".format(method.__name__, timeit.timeit(method)))

我的机器上的结果：

str_translate took a total of 1.1160085709998384s
join_generator took a total of 4.599312704987824s
re_sub took a total of 4.101858579088002s
append_in_loop took a total of 4.257988628000021s
compiled_re_sub took a total of 1.0353244650177658s

【讨论】：

如果你re.compile在re_sub()函数外先是正则表达式，这个正则表达式是最快的。

【解决方案2】：

s = 'Testing PRI/Sec (#434242332;PP:432:133423846,335)'
a = ['#',':',';','/']

print(''.join('-' if ch in a else ch for ch in s))

打印：

Testing PRI-Sec (-434242332-PP-432-133423846,335)

或者使用re:

s = 'Testing PRI/Sec (#434242332;PP:432:133423846,335)'
a = ['#',':',';','/']

import re
print(re.sub('|'.join(re.escape(i) for i in a), '-', s))

打印：

Testing PRI-Sec (-434242332-PP-432-133423846,335)

【讨论】：

【解决方案3】：

使用重新打包

import re
string = 'Testing PRI/Sec (#434242332;PP:432:133423846,335)'
result = re.sub('[#:;/]',"-", string)
print(result)

结果：

Testing PRI-Sec (-434242332-PP-432-133423846,335)

【讨论】：

【解决方案4】：

只需循环将每个字符添加到临时变量中，除非它在列表中“a”如果它在列表中，只需通过在变量中添加“-”来替换它。

str = 'Testing PRI/Sec (#434242332;PP:432:133423846,335)'
a = ['#',':',';','/']
temp = ''
for i in str:
    if i in a:
        temp = temp + "-"
    else:
        temp = temp + i
print(temp)

【讨论】：