英文:
Go lang slice columns from 2d array?
问题
我很好奇，你如何从一个二维数组中创建一个列切片？
我有一个用于井字游戏棋盘的数组，我试图创建一个列切片，但我的切片都是相同的。
/* 先尝试让行和列正常工作 */
func() isWin() bool {
win := make([]char, SIZE*2)
for i:= range BOARD {
fmt.Println("行")
win[i] = check(BOARD[i][0:SIZE])
fmt.Println("列")
win[i+SIZE] = check(BOARD[0:SIZE][i])
}
return false
}
func() check(slice []char) (char) {
fmt.Println(slice)
return "-"
}
我给它以下输入：
[E E E E]
[E E E E]
[X O E E]
[X O E E]
但我得到的返回值是：
行
[X O E E]
列
[X O E E]
但我想要的返回值是：
行
[X O E E]
列
[E E X X]
我该如何创建这个切片？
英文:
I'm curious, how do you create a column slice from a 2d Array?
I have an array for a game board for Tic-Tac-Toe, and I'm trying to create a column slice, but my slices are coming out identical.
/* Just trying to get rows and columns working first */
func() isWin() bool {
win := make([]char, SIZE*2)
for i:= range BOARD {
fmt.Println("Row")
win[i] = check(BOARD[i][0:SIZE])
fmt.Println("Column")
win[i+SIZE] = check(BOARD[0:SIZE][i])
}
return false
}
func() check(slice []char) (char) {
fmt.Println(slice)
return "-"
}
I give it the following input:
[E E E E]
[E E E E]
[X O E E]
[X O E E]
And I get a return of
Row
[X O E E]
Column
[X O E E]
But I want a return of
Row
[X O E E]
Column
[E E X X]
How do I make this slice?
答案1
得分: 8
你想要用切片语法实现的功能是不可能的。你认为x[i][0:n]会获取到第i行的所有列。实际上，它返回的是第i行的第0到n列。
你需要使用循环来获取一列：
func boardColumn(board [][]char, columnIndex int) (column []char) {
column = make([]char, 0)
for _, row := range board {
column = append(column, row[columnIndex])
}
return
}
你标记为“Row”的代码实际上是我期望的获取列的代码：
win[i] = check(BOARD[0:SIZE][i])
从语义上讲，切片的索引表示：从零到SIZE，取所有元素的第i个元素。
但是，正如前面提到的，Go的切片工作方式不同。自己试试：
x := [][]int{{1,2,3},{4,5,6}}
fmt.Println(x[0:2]) // [[1,2,3],[4,5,6]]
fmt.Println(x[0:2][0]) // [1,2,3]
如你在上面的示例的第二行中所看到的，x[0:2]返回整个二维切片。因此，取第一个元素返回的是该二维切片的第一行。
英文:
What you want to do is not possible with the slice syntax. You think that x[i][0:n] will get you all the columns of row i. What it actually does is returning the columns 0 to n from row i.
You have to use a loop to get a column:
func boardColumn(board [][]char, columnIndex int) (column []char) {
column = make([]char, 0)
for _, row := range board {
column = append(column, row[columnIndex])
}
return
}
The code you label with 'Row' is actually the code I would have expected to deliver the columns:
win[i] = check(BOARD[0:SIZE][i])
Semantically the slice indices say: take the ith element of all elements from zero to SIZE.
But, as already mentioned, Go slices work differently. Try for yourself:
x := [][]int{{1,2,3},{4,5,6}}
fmt.Println(x[0:2]) // [[1,2,3],[4,5,6]]
fmt.Println(x[0:2][0]) // [1,2,3]
As you can see in line two of the example above, x[0:2] returns the full 2d slice. Consequently
taking the first element returns the first row of that 2d slice.
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