在Python快速矩阵转置 [英] Fast matrix transposition in Python
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问题描述
有没有快速的方法来制作一个矩形的二维矩阵Python中的换位(非涉及任何库导入)。?
我说,如果我有一个数组
X = [1,2,3]
[4,5,6]
我需要一个数组Y,它应该是X的转版,所以
Y = [1,4],
〔2.5〕,
[3,6]
解决方案
简单: Y = ZIP(* X)
>>> X = [[1,2,3],[4,5,6]]
>>> Y = ZIP(* X)
>>> ÿ
[(1,4),(2,5),(3,6)]
编辑:来回答什么呢ZIP(* X)的意思,这里是蟒蛇人工为例意见的问题:
>>>范围(3,6)#独立参数正常呼叫
[3,4,5]
>>>的args = [3,6]
>>>范围(*参数)与参数#打来的电话清单解压
[3,4,5]
所以,当 X
是 [1,2,3],[4,5,6]]
,压缩(* X)
是压缩([1,2,3],[4,5,6]),
Is there any fast method to make a transposition of a rectangular 2D matrix in Python (non-involving any library import).?
Say, if I have an array
X=[ [1,2,3],
[4,5,6] ]
I need an array Y which should be a transposed version of X, so
Y=[ [1,4],
[2,5],
[3,6] ]
解决方案
Simple: Y=zip(*X)
>>> X=[[1,2,3], [4,5,6]]
>>> Y=zip(*X)
>>> Y
[(1, 4), (2, 5), (3, 6)]
EDIT: to answer questions in the comments about what does zip(*X) mean, here is an example from python manual:
>>> range(3, 6) # normal call with separate arguments
[3, 4, 5]
>>> args = [3, 6]
>>> range(*args) # call with arguments unpacked from a list
[3, 4, 5]
So, when X
is [[1,2,3], [4,5,6]]
, zip(*X)
is zip([1,2,3], [4,5,6])
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