英文:
Randomize order of a MongoDB query in Go
问题
这是我的查询代码:
c := session.DB("searchV").C("video")
var results []BadVideo
err5 := c.Find(nil).All(&results)
fmt.Println("请求完成")
if err5 != nil {
panic(err5)
}
var i = 0
for _, badvideo := range results {
}
我想要随机化浏览查询结果的顺序,以便对每个请求的项进行操作...
所以每次运行时,我以不同的顺序浏览它。
英文:
Here is my query :
c := session.DB("searchV").C("video")
var results []BadVideo
err5 := c.Find(nil).All(&results)
fmt.Println("request done")
if err5 != nil {
panic(err5)
}
var i = 0
for _,badvideo := range results {
}
I would like to randomize the order of browsing the items of the query for making operation on each item of the request ...
So each time I run it, I browse it in a different order.
答案1
得分: 1
手动洗牌
这是一个简单的洗牌算法,用于对 []BadVideo
切片进行洗牌(随机化):
func shuffle(r []BadVideo) {
for i := len(r) - 1; i > 0; i-- {
j := rand.Intn(i + 1)
r[i], r[j] = r[j], r[i]
}
}
所以在加载结果后,只需对其调用 shuffle(results)
。
为了测试,我将使用一个简单的 BadVideo
类型:
type BadVideo struct {
Name string
}
func main() {
rand.Seed(time.Now().UnixNano())
results := []BadVideo{{"a"}, {"b"}, {"c"}, {"d"}, {"e"}}
shuffle(results)
fmt.Println(results)
}
输出结果(在 Go Playground 上尝试):
[{c} {d} {b} {e} {a}]
工作原理:
为了对切片进行洗牌,shuffle()
函数会随机选择切片中的一个元素,并将其放置在每个索引位置上。它通过向下迭代所有元素,并从剩余的切片中随机选择一个元素(包括当前选择的元素的索引,因为随机顺序也包括元素“保持原位”的情况),然后使用随机索引将该元素与所选的随机元素交换位置。循环继续直到 i > 0
(而不是 i >= 0
),因为如果只剩下一个元素,就不需要将其与自身交换位置。
使用 rand.Perm()
shuffle()
的另一种变体可以利用 rand.Perm()
,它返回一个包含随机排序数字的切片。我们可以使用这些随机数来指示如何重新排序结果:
func shuffle(r []BadVideo) {
r2 := append([]BadVideo(nil), r...)
for i, j := range rand.Perm(len(r)) {
r[i] = r2[j]
}
}
在 Go Playground 上尝试这个变体。
这里需要注意的一点是:在重新排序之前,我们必须保存原始切片(创建一个副本),以便在将结果写入切片时,可以根据随机索引选择原始元素。我通过将完整切片附加到 nil
切片来创建了一个副本。
英文:
Manual shuffling
Here's a simple shuffle algorithm, which shuffles (randomizes) a []BadVido
slice:
func shuffle(r []BadVideo) {
for i := len(r) - 1; i > 0; i-- {
j := rand.Intn(i + 1)
r[i], r[j] = r[j], r[i]
}
}
So after you loaded your results, simply call shuffle(results)
on it.
For testing, I will use a simple BadVideo
type:
type BadVideo struct {
Name string
}
func main() {
rand.Seed(time.Now().UnixNano())
results := []BadVideo{{"a"}, {"b"}, {"c"}, {"d"}, {"e"}}
shuffle(results)
fmt.Println(results)
}
Output (try it on the Go Playground):
[{c} {d} {b} {e} {a}]
How it works:
To shuffle a slice, the shuffle()
function randomly selects one element from the slice for each index. It does it like iterating over all elements downward, and selects a random element from the remaining slice (including index of the element we're currently selecting, because random orders also include ones where an element "stays in place"), and using a random index to swaps the element with the chosen random one. The loop goes until i > 0
(and not until i >=0
), because if only 1 element left, no need to swap it with itself.
Using rand.Perm()
Another variant of shuffle()
could take advantage of rand.Perm()
which returns a slice containing shuffled numbers. We can use these random numbers to tell how to reorder the results:
func shuffle(r []BadVideo) {
r2 := append([]BadVideo(nil), r...)
for i, j := range rand.Perm(len(r)) {
r[i] = r2[j]
}
}
Try this variant on the Go Playground.
One thing to note here: before we do the reordering, we have to save the original slice (make a copy of it), so we can select the original elements specified by random indices when writing the results in the slice. I created a copy by appending the complete slice to a nil
slice.
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