Python：字符串替换索引
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Python: String replace index
提问by David Davó
I mean, i want to replace str[9:11]for another string.
If I do str.replace(str[9:11], "###")It doesn't work, because the sequence [9:11] can be more than one time.
If str is "cdabcjkewabcef"i would get "cd###jkew###ef"but I only want to replace the second.
我的意思是，我想替换str[9:11]另一个字符串。如果我这样做str.replace(str[9:11], "###")它不起作用，因为序列 [9:11] 可以不止一次。如果 str 是"cdabcjkewabcef"我会得到"cd###jkew###ef"但我只想替换第二个。
回答by janbrohl
you can do
你可以做
s="cdabcjkewabcef"
snew="".join((s[:9],"###",s[12:]))
which should be faster than joining like snew=s[:9]+"###"+s[12:]on large strings
这应该比snew=s[:9]+"###"+s[12:]在大字符串上加入要快
回答by Damián Rafael Lattenero
You can achieve this by doing:
您可以通过执行以下操作来实现此目的：
yourString = "Hello"
yourIndexToReplace = 1 #e letter
newLetter = 'x'
yourStringNew="".join((yourString[:yourIndexToReplace],newLetter,yourString[yourIndexToReplace+1:]))
回答by veerabhadra butte
str = "cdabcjkewabcef"
print((str[::-1].replace('cba','###',1))[::-1])
回答by Israel Unterman
Given txt and s - the string you want to replace:
给定 txt 和 s - 要替换的字符串：
txt.replace(s, "***", 1).replace(s, "###").replace("***", s)
Another way:
其它的办法：
txt[::-1].replace(s[::-1], "###", 1)[::-1]
回答by ospahiu
You can use join()with sub-strings.
您可以使用join()子字符串。
s = 'cdabcjkewabcef'
sequence = '###'
indicies = (9,11)
print sequence.join([s[:indicies[0]-1], s[indicies[1]:]])
>>> 'cdabcjke###cef'
回答by User_Targaryen
Here is a sample code:
这是一个示例代码：
word = "astalavista"
index = 0
newword = ""
addon = "xyz"
while index < 8:
newword = newword + word[index]
index += 1
ind = index
i = 0
while i < len(addon):
newword = newword + addon[i]
i += 1
while ind < len(word):
newword = newword + word[ind]
ind += 1
print newword
