使用go中的范围选择2D切片的2D子切片 [英] selecting 2D sub-slice of a 2D-slice using ranges in go

问题描述

选择切片的2D子切片时，我得到了令人惊讶的结果.考虑下面的二维int数组

  a:= [] [] int {{0，1，2，3}，{1,2,3,4}，{2，3，4，5}，{3，4，5，6}，} 

要使用我要使用的范围来选择左上3x3 2D切片

  b:= a [0:2] [0:2] 

我希望结果是

  [[0 1 2] [1 2 3] [2 3 4]] 

但是第二个索引范围似乎没有任何作用，而是返回以下内容:

  [[0 1 2 3] [1 2 3 4] [2 3 4 5]] 

我想念什么?您是否可以简单地不选择尺寸> 1的子切片?

解决方案

您无法一步一步完成自己想做的事情.切片和数组不是二维的，它们只是组成一个多维对象而构成.参见二维数组的内存表示如何

因此，使用slice表达式，您仅得到一个将容纳完整"行的子集的slice，其类型将相同: [] [] int .如果再次对其进行切片，则只需再次对行的切片进行切片.

还请注意，切片表达式中的较高索引是互斥的，因此 a [0:2] 仅具有2行，因此应使用 a [0:3] 或简单地 a [:3] .

要获得所需的内容，您必须像这样分别对行进行切片:

  b:= a [0:3]对于i:=范围b {b [i] = b [i] [0:3]}fmt.Println(b) 

这将输出(在游乐场上尝试):

  [[0 1 2] [1 2 3] [2 3 4]] 

或更短:

  b:= a [:3]对于i，bi:=范围b {b [i] = bi [:3]} 

I'm getting a surprising result when selecting a 2D sub-slice of a slice. Consider the following 2D int array

a := [][]int{
    {0, 1, 2, 3},
    {1, 2, 3, 4},
    {2, 3, 4, 5},
    {3, 4, 5, 6},
}

To select the top left 3x3 2D slice using ranges I would use

b := a[0:2][0:2]

I would expect the result to be

[[0 1 2] [1 2 3] [2 3 4]]

however the second index range doesn't seem to have any effect, and returns the following instead:

[[0 1 2 3] [1 2 3 4] [2 3 4 5]]

What am I missing? Can you simply not select a sub-slice like this where the dimension > 1 ?

解决方案

You can't do what you want in a single step. Slices and arrays are not 2-dimensional, they are just composed to form a multi-dimensional object. See How is two dimensional array's memory representation

So with a slice expression, you just get a slice that will hold a subset of the "full" rows, and its type will be the same: [][]int. If you slice it again, you just slicing the slice of rows again.

Also note that the higher index in a slice expression is exclusive, so a[0:2] will only have 2 rows, so you should use a[0:3] or simply a[:3] instead.

To get what you want, you have to slice the rows individually like this:

b := a[0:3]
for i := range b {
    b[i] = b[i][0:3]
}
fmt.Println(b)

This will output (try it on the Go Playground):

[[0 1 2] [1 2 3] [2 3 4]]

Or shorter:

b := a[:3]
for i, bi := range b {
    b[i] = bi[:3]
}

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