英文:
How to efficiently parallelize array list and control the parallelism?
问题
我有一个resourceId
数组,我需要并行循环遍历它。然后为每个资源生成一个URL,并将其放入一个映射中,其中键是resourceId
,值是URL。
我找到了下面的代码来完成这个任务,但我不确定这是否是正确的方法。我在这里使用sizedwaitgroup来并行化resourceId
列表。在向映射写入数据时,我还使用了锁。我确定这不是高效的代码,因为使用锁,然后使用sizedwaitgroup会有一些性能问题。
有什么更好和更高效的方法来做到这一点吗?我应该在这里使用通道吗?我想控制并行度,而不是运行与resourceId
列表长度相同的goroutine。如果任何resourceId
的URL生成失败,我希望将其记录为该resourceId
的错误,但不会中断其他并行运行以获取其他resourceId
的URL。
例如:如果有10个资源,其中2个失败,则记录这2个的错误,并且映射应该包含剩余8个的条目。
// 并行运行20个线程
swg := sizedwaitgroup.New(20)
var mutex = &sync.Mutex{}
start := time.Now()
m := make(map[string]*customerPbV1.CustomerResponse)
for _, resources := range resourcesList {
swg.Add()
go func(resources string) {
defer swg.Done()
customerUrl, err := us.GenerateUrl(clientId, resources, appConfig)
if err != nil {
errs.NewWithCausef(err, "Could not generate the url for %s", resources)
}
mutex.Lock()
m[resources] = customerUrl
mutex.Unlock()
}(resources)
}
swg.Wait()
elapsed := time.Since(start)
fmt.Println(elapsed)
**注意:**上面的代码将从多个读取线程以高吞吐量调用,因此需要具有良好的性能。
英文:
I have a resourceId
array which I need loop in parallel. And generate URL
for each resource and then put inside a map which is key (resourcId) and value is url.
I got below code which does the job but I am not sure if this is the right way to do it. I am using sizedwaitgroup here to parallelize the resourceId
list. And also using lock on map while writing the data to it. I am sure this isn't efficient code as using lock and then using sizedwaitgroup will have some performance problem.
What is the best and efficient way to do this? Should I use channels here? I want to control the parallelism on how much I should have instead of running length of resourceId
list. If any resourceId
url generation fails, I want to log that as an error for that resourceId
but do not disrupt other go routine running in parallel to get the url generated for other resourceId
.
For example: If there are 10 resources, and 2 fails then log error for those 2 and map should have entry for remaining 8.
// running 20 threads in parallel
swg := sizedwaitgroup.New(20)
var mutex = &sync.Mutex{}
start := time.Now()
m := make(map[string]*customerPbV1.CustomerResponse)
for _, resources := range resourcesList {
swg.Add()
go func(resources string) {
defer swg.Done()
customerUrl, err := us.GenerateUrl(clientId, resources, appConfig)
if err != nil {
errs.NewWithCausef(err, "Could not generate the url for %s", resources)
}
mutex.Lock()
m[resources] = customerUrl
mutex.Unlock()
}(resources)
}
swg.Wait()
elapsed := time.Since(start)
fmt.Println(elapsed)
Note: Above code will be called at high throughput from multiple reader threads so it needs to perform well.
答案1
得分: 4
我不确定sizedwaitgroup
是什么,也没有解释,但总体而言,这种方法看起来并不像Go语言的典型方式。至于“最佳”方法,这是一个主观问题,但在Go语言中最典型的方法可能是以下这样的:
func main() {
wg := new(sync.WaitGroup)
start := time.Now()
numWorkers := 20
m := make(map[string]*customerPbV1.CustomerResponse)
work := make(chan string)
results := make(chan result)
for i := 0; i < numWorkers; i++ {
wg.Add(1)
go worker(work, results)
}
go func() {
for _, resources := range resourcesList {
work <- resources
}
close(work)
}()
go func() {
wg.Wait()
close(results)
}()
for result := range results {
m[result.resources] = result.response
}
elapsed := time.Since(start)
fmt.Println(elapsed)
}
type result struct {
resources string
response *customerPbV1.CustomerResponse
}
func worker(ch chan string, r chan result) {
for w := range ch {
customerUrl, err := us.GenerateUrl(clientId, w, appConfig)
if err != nil {
errs.NewWithCausef(err, "Could not generate the url for %s", resources)
continue
}
r <- result{w, customerUrl}
}
}
(不过,根据名称,我会假设errs.NewWithCause
实际上并不处理错误,而是返回一个错误。在这种情况下,当前的代码会将错误丢弃,一个正确的解决方案是为处理错误添加一个额外的chan error
:
func main() {
wg := new(sync.WaitGroup)
start := time.Now()
numWorkers := 20
m := make(map[string]*customerPbV1.CustomerResponse)
work := make(chan string)
results := make(chan result)
errors := make(chan error)
for i := 0; i < numWorkers; i++ {
wg.Add(1)
go worker(work, results, errors)
}
go func() {
for _, resources := range resourcesList {
work <- resources
}
close(work)
}()
go func() {
wg.Wait()
close(results)
close(errors)
}()
go func() {
for err := range errors {
// 处理错误
}
}()
for result := range results {
m[result.resources] = result.response
}
elapsed := time.Since(start)
fmt.Println(elapsed)
}
type result struct {
resources string
response *customerPbV1.CustomerResponse
}
func worker(ch chan string, r chan result, errs chan error) {
for w := range ch {
customerUrl, err := us.GenerateUrl(clientId, w, appConfig)
if err != nil {
errs <- errs.NewWithCausef(err, "Could not generate the url for %s", resources)
continue
}
r <- result{w, customerUrl}
}
}
英文:
I'm not sure what sizedwaitgroup
is and it's not explained, but overall this approach doesn't look very typical of Go. For that matter, "best" is a matter of opinion, but the most typical approach in Go would be something along these lines:
func main() {
wg := new(sync.WaitGroup)
start := time.Now()
numWorkers := 20
m := make(map[string]*customerPbV1.CustomerResponse)
work := make(chan string)
results := make(chan result)
for i := 0; i < numWorkers; i++ {
wg.Add(1)
go worker(work, results)
}
go func() {
for _, resources := range resourcesList {
work <- resources
}
close(work)
}()
go func() {
wg.Wait()
close(results)
}()
for result := range results {
m[result.resources] = result.response
}
elapsed := time.Since(start)
fmt.Println(elapsed)
}
type result struct {
resources string
response *customerPbV1.CustomerResponse
}
func worker(ch chan string, r chan result) {
for w := range ch {
customerUrl, err := us.GenerateUrl(clientId, w, appConfig)
if err != nil {
errs.NewWithCausef(err, "Could not generate the url for %s", resources)
continue
}
r <- result{w, customerUrl}
}
}
(Though, based on the name, I would assume errs.NewWithCause
doesn't actually handle errors, but returns one, in which case the current code is dropping them on the floor, and a proper solution would have an additional chan error
for handling errors:
func main() {
wg := new(sync.WaitGroup)
start := time.Now()
numWorkers := 20
m := make(map[string]*customerPbV1.CustomerResponse)
work := make(chan string)
results := make(chan result)
errors := make(chan error)
for i := 0; i < numWorkers; i++ {
wg.Add(1)
go worker(work, results, errors)
}
go func() {
for _, resources := range resourcesList {
work <- resources
}
close(work)
}()
go func() {
wg.Wait()
close(results)
close(errors)
}()
go func() {
for err := range errors {
// Do something with err
}
}()
for result := range results {
m[result.resources] = result.response
}
elapsed := time.Since(start)
fmt.Println(elapsed)
}
type result struct {
resources string
response *customerPbV1.CustomerResponse
}
func worker(ch chan string, r chan result, errs chan error) {
for w := range ch {
customerUrl, err := us.GenerateUrl(clientId, w, appConfig)
if err != nil {
errs <- errs.NewWithCausef(err, "Could not generate the url for %s", resources)
continue
}
r <- result{w, customerUrl}
}
}
答案2
得分: 1
我已经为你翻译了代码和注释,请阅读以下内容:
package main
import (
"errors"
"fmt"
"log"
"math/rand"
"runtime"
"strconv"
"sync"
"time"
)
type Result struct {
resource string
val int
err error
}
/*
将 Result 结构体更改为以下内容
result 结构体将收集创建映射所需的所有信息
type Result struct {
resources string
customerUrl *customerPbV1.CustomerResponse
err error
}
*/
// const numWorker = 8
func main() {
now := time.Now()
rand.Seed(time.Now().UnixNano())
m := make(map[string]int)
// m := make(map[string]*customerPbV1.CustomerResponse) // 更改为此行
numWorker := runtime.NumCPU()
fmt.Println(numWorker)
chanResult := make(chan Result)
go func() {
for i := 0; i < 20; i++ {
/*
customerUrl, err := us.GenerateUrl(clientId, resources, appConfig)
假设 i 是资源
chanResult <- Result {resource: strconv.Itoa(i)}
*/
chanResult <- Result{ // 这将阻塞,直到 chanResult 在第 68 行被消耗
resource: strconv.Itoa(i),
}
}
close(chanResult)
}()
var wg sync.WaitGroup
cr := make(chan Result)
wg.Add(numWorker)
go func() {
wg.Wait()
close(cr) // 注意:不要忘记关闭 cr
}()
go func() {
for i := 0; i < numWorker; i++ { // 此循环将运行 goroutine
go func(x int) {
for job := range chanResult { // 在第 49 行解除 chan 阻塞
log.Println("worker", x, "working on", job.resource)
x, err := query(job.resource) // TODO: customerUrl, err := us.GenerateUrl(clientId, resources, appConfig)
cr <- Result{ // 发送到通道,将阻塞,直到它被消耗。消耗在主 goroutine 中的 "line 84"
resource: job.resource,
val: x,
err: err,
}
}
wg.Done()
}(i)
}
}()
counterTotal := 0
counterSuccess := 0
for res := range cr { // 将在第 71 行解除通道阻塞
if res.err != nil {
log.Printf("发现错误 %s。堆栈跟踪:%s", res.resource, res.err)
} else {
m[res.resource] = res.val // 注意:保存到映射
counterSuccess++
}
counterTotal++
}
log.Printf("%d/%d 个作业已运行", counterSuccess, counterTotal)
fmt.Println("最终结果:", m)
fmt.Println("m 的长度:", len(m))
fmt.Println(runtime.NumGoroutine())
fmt.Println(time.Since(now))
}
func query(s string) (int, error) {
time.Sleep(time.Second)
i, err := strconv.Atoi(s)
if err != nil {
return 0, err
}
if i%3 == 0 {
return 0, errors.New("i 被 3 整除")
}
ms := i + 500 + rand.Intn(500)
return ms, nil
}
playground: https://go.dev/play/p/LeyE9n1hh81
英文:
I have create example code with comment.
please read the comment.
> note: query function will sleep in 1 second.
package main
import (
"errors"
"fmt"
"log"
"math/rand"
"runtime"
"strconv"
"sync"
"time"
)
type Result struct {
resource string
val int
err error
}
/*
CHANGE Result struct to this
result struct will collect all you need to create map
type Result struct {
resources string
customerUrl *customerPbV1.CustomerResponse
err error
}
*/
// const numWorker = 8
func main() {
now := time.Now()
rand.Seed(time.Now().UnixNano())
m := make(map[string]int)
// m := make(map[string]*customerPbV1.CustomerResponse) // CHANGE TO THIS
numWorker := runtime.NumCPU()
fmt.Println(numWorker)
chanResult := make(chan Result)
go func() {
for i := 0; i < 20; i++ {
/*
customerUrl, err := us.GenerateUrl(clientId, resources, appConfig)
we asume i is resources
chanResult <- Result {resource: strconv.Itoa(i)}
*/
chanResult <- Result{ // this will block until chanResult is consume in line 68
resource: strconv.Itoa(i),
}
}
close(chanResult)
}()
var wg sync.WaitGroup
cr := make(chan Result)
wg.Add(numWorker)
go func() {
wg.Wait()
close(cr) // NOTE: don't forget to close cr
}()
go func() {
for i := 0; i < numWorker; i++ { // this for loop will run goroutine
go func(x int) {
for job := range chanResult { // unblock chan on line 49
log.Println("worker", x, "working on", job.resource)
x, err := query(job.resource) // TODO: customerUrl, err := us.GenerateUrl(clientId, resources, appConfig)
cr <- Result{ // send to channel, will block until it consume. Consume is in MAIN goroutine "line 84"
resource: job.resource,
val: x,
err: err,
}
}
wg.Done()
}(i)
}
}()
counterTotal := 0
counterSuccess := 0
for res := range cr { // will unblock channel in line 71
if res.err != nil {
log.Printf("error found %s. stack trace: %s", res.resource, res.err)
} else {
m[res.resource] = res.val // NOTE: save to map
counterSuccess++
}
counterTotal++
}
log.Printf("%d/%d of total job run", counterSuccess, counterTotal)
fmt.Println("final :", m)
fmt.Println("len m", len(m))
fmt.Println(runtime.NumGoroutine())
fmt.Println(time.Since(now))
}
func query(s string) (int, error) {
time.Sleep(time.Second)
i, err := strconv.Atoi(s)
if err != nil {
return 0, err
}
if i%3 == 0 {
return 0, errors.New("i divided by 3")
}
ms := i + 500 + rand.Intn(500)
return ms, nil
}
playground : https://go.dev/play/p/LeyE9n1hh81
答案3
得分: 0
这是一个纯通道解决方案(playground)。
我认为性能主要取决于GenerateUrl
或者在我的代码中是generateURL
。
另外,我想指出的一件事是,这个正确的术语应该是并发而不是并行。
package main
import (
"errors"
"log"
"strconv"
"strings"
)
type result struct {
resourceID, url string
err error
}
func generateURL(resourceID string) (string, error) {
if strings.HasPrefix(resourceID, "error-") {
return "", errors.New(resourceID)
}
return resourceID, nil
}
func main() {
// 这是资源的ID
resources := make([]string, 10000)
for i := 0; i < 10000; i++ {
s := strconv.Itoa(i)
if i%10 == 0 {
resources[i] = "error-" + s
} else {
resources[i] = "id-" + s
}
}
numOfChannel := 20
// 我们通过这个通道将结果发送到resourceMap
ch := make(chan result, 10)
// 这些通道是用来接收资源ID的goroutine的
channels := make([]chan string, numOfChannel)
// 在处理完所有资源后,这个通道用于通知goroutine退出
done := make(chan struct{})
for i := range channels {
c := make(chan string)
channels[i] = c
go func() {
for {
select {
case rid := <-c:
u, err := generateURL(rid)
ch <- result{rid, u, err}
case _, ok := <-done:
if !ok {
break
}
}
}
}()
}
go func() {
for i, r := range resources {
channels[i%numOfChannel] <- r
}
}()
resourceMap := make(map[string]string)
i := 0
for p := range ch {
if p.err != nil {
log.Println(p.resourceID, p.err)
} else {
resourceMap[p.resourceID] = p.url
}
i++
if i == len(resources)-1 {
break
}
}
close(done)
}
英文:
Here is a pure channel solution (playground).
I think the performance really depends on the GenerateUrl
or in my code generateURL
.
Also one more thing I would like to point out is that correct term for this is concurrency not parallelism.
package main
import (
"errors"
"log"
"strconv"
"strings"
)
type result struct {
resourceID, url string
err error
}
func generateURL(resourceID string) (string, error) {
if strings.HasPrefix(resourceID, "error-") {
return "", errors.New(resourceID)
}
return resourceID, nil
}
func main() {
// This is the resource IDs
resources := make([]string, 10000)
for i := 0; i < 10000; i++ {
s := strconv.Itoa(i)
if i%10 == 0 {
resources[i] = "error-" + s
} else {
resources[i] = "id-" + s
}
}
numOfChannel := 20
// We send result through this channel to the resourceMap
ch := make(chan result, 10)
// These are the channels that go routine receives resource ID from
channels := make([]chan string, numOfChannel)
// After processing all resources, this channel is used to signal the go routines to exit
done := make(chan struct{})
for i := range channels {
c := make(chan string)
channels[i] = c
go func() {
for {
select {
case rid := <-c:
u, err := generateURL(rid)
ch <- result{rid, u, err}
case _, ok := <-done:
if !ok {
break
}
}
}
}()
}
go func() {
for i, r := range resources {
channels[i%numOfChannel] <- r
}
}()
resourceMap := make(map[string]string)
i := 0
for p := range ch {
if p.err != nil {
log.Println(p.resourceID, p.err)
} else {
resourceMap[p.resourceID] = p.url
}
i++
if i == len(resources)-1 {
break
}
}
close(done)
}
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