为什么这些元组没有正确生成?

Why are these tuples not generated correctly?

提问人:Theodore Reimer 提问时间:11/8/2023 更新时间:11/8/2023 访问量:45

问:

我一直在做一个项目,这个项目要求我从一组(复数)数字中生成特定长度的所有可能的元组。为此,我尝试实现 Mathematica Tuples[] 命令的一个版本,但发现它没有正确生成所有元组。

经过一番挫折,我发现当我的程序生成长度为 4 的元组时,它会添加重复项而不是新元素,从而导致任何更长长度的元组出现问题。我在网上查看是否有人有任何其他类似的问题,然后找到一些其他代码来完成相同的任务,并注意到我的解决方案与他们的解决方案相似。我说不出我哪里做错了。

在更加沮丧之后,我发现如果我将元素添加到列表中,一切都可以正常工作,只有附加才是问题所在。我试图弄清楚我的原始代码出了什么问题,但一无所获。

下面是我为演示该问题而编写的代码。我绝不是一个专业的程序员,所以如果这不是完成这项任务的最惯用方式,你必须原谅我。目前,我在实际代码中使用元组ByPrepend函数,它工作正常,我真的只是希望了解元组ByAppend函数出了什么问题。同样,在第三、第五、第八和我测试过的任何其他级别,它似乎都做得很好。如果需要,我可以提供有关我的操作系统和构建的更多信息。

package main

import "fmt"

func tuplesByAppend[T any](list []T, depth int) [][]T {
    var l1 [][]T
    var l2 [][]T

    for _, v := range list {
        l1 = append(l1, []T{v})
    }

    for i := 1; i < depth; i++ {
        for _, u := range l1 {
            for _, v := range list {
                // Differs here
                next := append(u, v)
                // next is calculated properly, but added to l2 incorrectly at the fourth level only
                // at the fifth level it functions properly
                // fmt.Println(next)
                l2 = append(l2, next)
                // fmt.Println(l2)

                // it appears that at the fourth level it is writing over the previous entries
                // Printing here yields
                // [[1 1 1 1]]
                // [[1 1 1 2] [1 1 1 2]]
                // [[1 1 1 3] [1 1 1 3] [1 1 1 3]]
                // [[1 1 1 3] [1 1 1 3] [1 1 1 3] [1 1 2 1]]
                // [[1 1 1 3] [1 1 1 3] [1 1 1 3] [1 1 2 2] [1 1 2 2]]
                // and so on.
            }
        }
        l1 = l2
        l2 = [][]T{}
    }

    return l1
}

func tuplesByPrepend[T any](list []T, depth int) [][]T {
    var l1 [][]T
    var l2 [][]T

    for _, v := range list {
        l1 = append(l1, []T{v})
    }

    for i := 1; i < depth; i++ {
        for _, u := range l1 {
            for _, v := range list {
                // Differs here
                next := append([]T{v}, u...)
                l2 = append(l2, next)
            }
        }
        l1 = l2
        l2 = [][]T{}
    }

    return l1
}

func main() {
    ourlist := []int{1, 2, 3}
    ourdepth := 4
    appended := tuplesByAppend(ourlist, ourdepth)
    prepended := tuplesByPrepend(ourlist, ourdepth)

    // We should expect this slice to start [1 1 1 1] [1 1 1 2] [1 1 1 3] [1 1 2 1] ...
    // In fact, it starts                   [1 1 1 3] [1 1 1 3] [1 1 1 3] [1 1 2 3]
    fmt.Println(appended)
    // This slice is as expected
    fmt.Println(prepended)
}

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答:

4赞 Zeke Lu 11/8/2023 #1

在某些情况下,以下行无法按预期工作:

next := append(u, v)

此示例演示了发生的情况:

package main

import "fmt"

func main() {
    u := append([]int{1, 2}, 3)
    // The length is 3 but the capacity is 4.
    fmt.Printf("u %v\n  len: %d\n  cap: %d\n", u, len(u), cap(u))

    // Since u has enough capacity for the new element "4",
    // v1 will share the same underlying array.
    v1 := append(u, 4)
    fmt.Println("v1:", v1)

    // As what happened to v1, v2 will share the same underlying array too.
    // But the last element "4" in the underlying array is changed to "5".
    v2 := append(u, 5)
    fmt.Println("v2:", v2)

    // Since v1 uses the same underlying array, it sees the change in the last step.
    fmt.Println("v1:", v1)
}

若要防止它共享基础数组,请替换为以下代码:next := append(u, v)

next := make([]T, len(u)+1)
copy(next, u)
next[len(u)] = v

有关更多信息,请参见 Go Slices: usage and internals