Python：是否有 BASIC 中的 mid、right 和 left 等价物？
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Python: Is there an equivalent of mid, right, and left from BASIC?
提问by pythonprogrammer
I want to do something like this:
我想做这样的事情：
>>> mystring = "foo"
>>> print(mid(mystring))
Help!
帮助！
采纳答案by Andy W
slices to the rescue :)
切片救援:)
def left(s, amount):
return s[:amount]
def right(s, amount):
return s[-amount:]
def mid(s, offset, amount):
return s[offset:offset+amount]
回答by frnhr
If I remember my QBasic, right, left and mid do something like this:
如果我记得我的 QBasic，右、左和中，做这样的事情：
>>> s = '123456789'
>>> s[-2:]
'89'
>>> s[:2]
'12'
>>> s[4:6]
'56'
http://www.angelfire.com/scifi/nightcode/prglang/qbasic/function/strings/left_right.html
http://www.angelfire.com/scifi/nightcode/prglang/qbasic/function/strings/left_right.html
回答by Gareth Lock
These work great for reading left / right "n" characters from a string, but, at least with BBC BASIC, the LEFT$()and RIGHT$()functions allowed you to change the left / right "n" characters too...
这些非常适合从字符串中读取左/右“n”个字符，但是，至少在 BBC BASIC 中，LEFT$()和RIGHT$()函数也允许您更改左/右“n”个字符......
E.g.:
例如：
10 a$="00000"
20 RIGHT$(a$,3)="ABC"
30 PRINT a$
Would produce:
会产生：
00ABC
Edit : Since writing this, I've come up with my own solution...
编辑：自从写这篇文章以来，我想出了我自己的解决方案......
def left(s, amount = 1, substring = ""):
if (substring == ""):
return s[:amount]
else:
if (len(substring) > amount):
substring = substring[:amount]
return substring + s[:-amount]
def right(s, amount = 1, substring = ""):
if (substring == ""):
return s[-amount:]
else:
if (len(substring) > amount):
substring = substring[:amount]
return s[:-amount] + substring
To return n characters you'd call
要返回您要调用的 n 个字符
substring = left(string,<n>)
Where defaults to 1 if not supplied. The same is true for right()
如果未提供，则默认为 1。right() 也是如此
To change the left or right n characters of a string you'd call
要更改您要调用的字符串的左侧或右侧 n 个字符
newstring = left(string,<n>,substring)
This would take the first n characters of substring and overwrite the first n characters of string, returning the result in newstring. The same works for right().
这将取 substring 的前 n 个字符并覆盖 string 的前 n 个字符，以 newstring 形式返回结果。对 right() 也一样。
回答by Stanton Attree
Thanks Andy W
谢谢安迪 W
I found that the mid() did not quite work as I expected and I modified as follows:
我发现 mid() 并没有像我预期的那样工作，我修改如下：
def mid(s, offset, amount):
return s[offset-1:offset+amount-1]
I performed the following test:
我进行了以下测试：
print('[1]23', mid('123', 1, 1))
print('1[2]3', mid('123', 2, 1))
print('12[3]', mid('123', 3, 1))
print('[12]3', mid('123', 1, 2))
print('1[23]', mid('123', 2, 2))
Which resulted in:
这导致：
[1]23 1
1[2]3 2
12[3] 3
[12]3 12
1[23] 23
Which was what I was expecting. The original mid() code produces this:
这正是我所期待的。原来的 mid() 代码产生这个：
[1]23 2
1[2]3 3
12[3]
[12]3 23
1[23] 3
But the left() and right() functions work fine. Thank you.
但是 left() 和 right() 函数工作正常。谢谢你。
回答by Stanton Attree
You can use this method also it will act like that
你可以使用这个方法，它也会像那样
thadari=[1,2,3,4,5,6]
#Front Two(Left)
print(thadari[:2])
[1,2]
#Last Two(Right)# edited
print(thadari[-2:])
[5,6]
#mid
mid = len(thadari) //2
lefthalf = thadari[:mid]
[1,2,3]
righthalf = thadari[mid:]
[4,5,6]
Hope it will help
希望它会有所帮助
回答by Jim Sullivan
This is Andy's solution. I just addressed User2357112's concern and gave it meaningful variable names. I'm a Python rookie and preferred these functions.
这是安迪的解决方案。我刚刚解决了 User2357112 的问题，并给了它有意义的变量名。我是 Python 菜鸟，更喜欢这些功能。
def left(aString, howMany):
if howMany <1:
return ''
else:
return aString[:howMany]
def right(aString, howMany):
if howMany <1:
return ''
else:
return aString[-howMany:]
def mid(aString, startChar, howMany):
if howMany < 1:
return ''
else:
return aString[startChar:startChar+howMany]
回答by Scott
There are built-in functions in Python for "right" and "left", if you are looking for a boolean result.
如果您正在寻找布尔结果，Python 中有用于“右”和“左”的内置函数。
str = "this_is_a_test"
left = str.startswith("this")
print(left)
> True
right = str.endswith("test")
print(right)
> True
