chars_to_be_replaced = "ihgr"
new_char = "b"
my_dict = {k: new_char for k in chars_to_be_replaced}
s = "im hungry"
new_s = ''.join(my_dict.get(x, x) for x in s)
print(new_s) # bm bunbby
''.join(my_dict.get(x, x) for x in s):对于原始字符串中的每个字母,它将尝试获取其字典值,除非它不存在,否则将返回原始字符串。
NOTE: You can speed it up (a bit) by passing a list to join instead of a generator:
这个怎么样:
''.join(my_dict.get(x, x) for x in s)
:对于原始字符串中的每个字母,它将尝试获取其字典值,除非它不存在,否则将返回原始字符串。^{pr2}$
您可以使用^{}
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