问题

我想从1到6中随机选择一个数字，但是如果程序选择了1，我希望确保在选择2-6之前不再使用1。我想这样做是为了有机会遍历所有选项，而不是连续几次选择相同的选项，因为只有6个选项。有什么建议吗？

// 为食谱选择随机数
rand.Seed(time.Now().UTC().UnixNano())
myrand := random(1, 6)
fmt.Println(myrand)

...

// 处理它的函数
// 随机数函数
func random(min, max int) int {
    return rand.Intn(max-min) + min
}

每次运行时，我的程序使用myrand的部分与if语句相关联，并与当天将被选择的食谱相关联。

// 为食谱选择随机数
rand.Seed(time.Now().UTC().UnixNano())
myrand := random(1, 6)
fmt.Println(myrand)

// 以下是随机替换的测试
list := rand.Perm(6)
for i, _ := range list {
    fmt.Printf("%d\n", list[i]+1)
}

// 食谱选择逻辑
if myrand == 1 {
    fmt.Println(1)
    printRecipeOfTheDay(recipe1)
} else if myrand == 2 {
    fmt.Println(2)
    printRecipeOfTheDay(recipe2)
} else if myrand == 3 {
    fmt.Println(3)
    printRecipeOfTheDay(recipe3)
} else if myrand == 4 {
    fmt.Println(4)
}
}

我想知道我是否做错了什么？我不确定如何在每次运行时使用列表来生成一个新的数字（myrand），并且程序将能够记住上一次运行时的上一个数字（myrand）是什么？

<details>
<summary>英文:</summary>

I&#39;m looking to pick a random number from 1-6 but if for example 1 is picked by the program I want to make sure it is not again used until 2-6 has been picked. I want to do this in order to have a chance to go through all options instead of having the same option 2 or maybe 3 times in a row since there are only 6 options. Any ideas please?

    //choose random number for recipe
	rand.Seed(time.Now().UTC().UnixNano())
	myrand := random(1, 6)
	fmt.Println(myrand)

...

function that processes it

    //random number function
    func random(min, max int) int {
	    return rand.Intn(max-min) + min
   }


The part of my program that uses myrand from each run is used in a if statement and is tied to the recipe that will be picked for the day

	//choose random number for recipe
	rand.Seed(time.Now().UTC().UnixNano())
	myrand := random(1, 6)
	fmt.Println(myrand)

	//test below of random replacement
	list := rand.Perm(6)
	for i, _ := range list {
		fmt.Printf(&quot;%d\n&quot;, list[i]+1)
	}

	//logic for recipe to choose
	if myrand == 1 {
		fmt.Println(1)
		printRecipeOfTheDay(recipe1)
	} else if myrand == 2 {
		fmt.Println(2)
		printRecipeOfTheDay(recipe2)
	} else if myrand == 3 {
		fmt.Println(3)
		printRecipeOfTheDay(recipe3)
	} else if myrand == 4 {
		fmt.Println(4)
	}
}

I&#39;m wondering if I&#39;m doing something incorrectly ? I&#39;m not sure how I can use the list on every run to generate a new number (myrand) and the program will be able to remember what the last number (myrand) was when it ran the previous time?

</details>


# 答案1
**得分**: 3

你可以使用`rand.Perm(6)`，然后循环遍历结果。

例如，以下是一个打印1-6随机顺序且不重复的代码示例：

```go
package main
import "fmt"
import "math/rand"
 
func main(){
    list := rand.Perm(6)
    for i, _ := range list {
        fmt.Printf("%d\n", list[i]+1)
    }
}

请注意，我们在每个数字上都加了1，因为Perm(6)返回的是0-5，而你需要的是1-6。在使用完所有六个数字之后，再次调用rand.Perm(6)以获取六个新的数字。

package main
import "fmt"
import "math/rand"
func main(){
list := rand.Perm(6);
for i, _ := range list {
fmt.Printf("%d\n", list[i]+1);
}
}