删除特定索引处的字符 - python
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Remove char at specific index - python
提问by tgunn
I have a string that has two "0" (str) in it and I want to remove only the "0" (str) at index 4
我有一个字符串，其中有两个“0”（str），我只想删除索引 4 处的“0”（str）
I have tried calling .replace but obviously that removes all "0", and I cannot find a function that will remove the char at position 4 for me.
我试过调用 .replace 但显然删除了所有“0”，我找不到一个函数可以为我删除位置 4 的字符。
Anyone have a hint for me?
有人对我有提示吗？
采纳答案by Martijn Pieters
Use slicing, rebuilding the string minus the index you want to remove:
使用切片，重建字符串减去要删除的索引：
newstr = oldstr[:4] + oldstr[5:]
回答by root
as a sidenote, replacedoesn't have to move all zeros. If you just want to remove the first specify countto 1:
作为旁注，replace不必移动所有零。如果您只想删除第一个指定count为 1：
'asd0asd0'.replace('0','',1)
Out:
出去：
'asdasd0'
'asdasd0'
回答by Jon Clements
Slicing works (and is the preferred approach), but just an alternative if more operations are needed (but then converting to a list wouldn't hurt anyway):
切片有效（并且是首选方法），但如果需要更多操作，这只是一种替代方法（但无论如何转换为列表都不会受到伤害）：
>>> a = '123456789'
>>> b = bytearray(a)
>>> del b[3]
>>> b
bytearray(b'12356789')
>>> str(b)
'12356789'
回答by mclafee
Another option, using list comprehension and join:
另一种选择，使用列表理解和连接：
''.join([_str[i] for i in xrange(len(_str)) if i != 4])
回答by HARRY47
Try this code:
试试这个代码：
s = input()
a = int(input())
b = s.replace(s[a],'')
print(b)
回答by ?ngelo Polotto
This is my generic solution for any string sand any index i:
这是我对任何字符串s和任何索引的通用解决方案 i：
def remove_at(i, s):
return s[:i] + s[i+1:]
回答by Shishir
def remove_char(input_string, index):
first_part = input_string[:index]
second_part - input_string[index+1:]
return first_part + second_part
s = 'aababc'
index = 1
remove_char(s,index)
ababc
zero-based indexing
基于零的索引
回答by bhupen.chn
rem = lambda x, unwanted : ''.join([ c for i, c in enumerate(x) if i != unwanted])
rem('1230004', 4)
'123004'
