通过创建一个元组切片 [英] Create a slice using a tuple
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问题描述
有没有在Python任何方式使用元组作为指数分得一杯羹?
以下是无效的:
>>>一个=范围(20)
>>> B =(5,12)#我的切片索引
>>>一个[B]#无效
>>> A [切片(B)]#无效
>>> A [B [0]:B [1]]#是一个尴尬的语法
[5,6,7,8,9,10,11]
>>> B1,B2 = b的
>>>一个[B1:B2]#看起来有点清洁
[5,6,7,8,9,10,11]
这似乎是一个合理的Python的语法,让我感到惊讶,我不能这样做。
(更新)
并将该溶液证明是:
>>>一个[切片(* B)]
[5,6,7,8,9,10,11]
解决方案
您可以使用Python * ARGS
语法如下:
>>>一个=范围(20)
>>> B =(5,12)
>>>一个[切片(* B)]
[5,6,7,8,9,10,11]
基本上,你告诉Python来把它解析 B
成单独的元素和各个元素传递给片()
函数作为单独的参数。
Is there any way in python to use a tuple as the indices for a slice? The following is not valid:
>>> a = range(20)
>>> b = (5, 12) # my slice indices
>>> a[b] # not valid
>>> a[slice(b)] # not valid
>>> a[b[0]:b[1]] # is an awkward syntax
[5, 6, 7, 8, 9, 10, 11]
>>> b1, b2 = b
>>> a[b1:b2] # looks a bit cleaner
[5, 6, 7, 8, 9, 10, 11]
It seems like a reasonably pythonic syntax so I am surprised that I can't do it.
(update) And the solution turns out to be:
>>> a[slice(*b)]
[5, 6, 7, 8, 9, 10, 11]
解决方案
You can use Python's *args
syntax for this:
>>> a = range(20)
>>> b = (5, 12)
>>> a[slice(*b)]
[5, 6, 7, 8, 9, 10, 11]
Basically, you're telling Python to unpack the tuple b
into individual elements and pass each of those elements to the slice()
function as individual arguments.
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