问题：从字符串中删除前x个字符？
如何从字符串中删除前x个字符？例如，如果一个人有一个字符串lipsum，他们将如何删除前三个字符并得到结果sum？
How might one remove the first x characters from a string? For example, if one had a string lipsum, how would they remove the first 3 characters and get a result of sum?
回答 0
>>> text = 'lipsum'
>>> text[3:]
'sum'
有关更多信息，请参见有关字符串的官方文档，有关此符号的简要概述，请参见此SO答案。
>>> text = 'lipsum'
>>> text[3:]
'sum'
See the official documentation on strings for more information and this SO answer for a concise summary of the notation.
回答 1
另一种方法（取决于您的实际需求）：如果要弹出前n个字符并同时保存弹出的字符和修改后的字符串：
s = 'lipsum'
n = 3
a, s = s[:n], s[n:]
print(a)
# lip
print(s)
# sum
Another way (depending on your actual needs):
If you want to pop the first n characters and save both the popped characters and the modified string:
s = 'lipsum'
n = 3
a, s = s[:n], s[n:]
print(a)
# lip
print(s)
# sum
回答 2
>>> x = 'lipsum'
>>> x.replace(x[:3], '')
'sum'
>>> x = 'lipsum'
>>> x.replace(x[:3], '')
'sum'
回答 3
使用del。
例：
>>> text = 'lipsum'
>>> l = list(text)
>>> del l[3:]
>>> ''.join(l)
'sum'
Use del.
Example:
>>> text = 'lipsum'
>>> l = list(text)
>>> del l[3:]
>>> ''.join(l)
'sum'
回答 4
示例显示帐号的后3位数字。
x = '1234567890'
x.replace(x[:7], '')
o/p: '890'
Example to show last 3 digits of account number.
x = '1234567890'
x.replace(x[:7], '')
o/p: '890'
