将一个方法中的两个切片合并成一个映射。

huangapple
go评论118阅读模式
英文:

Merging 2 slices from a method into a map

问题

我有一个函数，返回两个变量，形式为切片，我想将这两个切片合并成一个键值对的映射。问题是我找不到一种方法来分别迭代每个切片，并将它们作为键和值添加。

当前的实现是遍历所有的价格，并将每个价格添加到城市键中。

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"math/rand"
  6. 	"time"
  7. )
  9. func main() {
  10. 	cities, prices := citiesAndPrices()
  11. 	towns := groupSlices(cities, prices)
  12. 	for cities := range towns {
  14. 		fmt.Println(cities, prices)
  15. 	}
  17. }
  19. func citiesAndPrices() ([]string, []int) {
  20. 	rand.Seed(time.Now().UnixMilli())
  21. 	cityChoices := []string{"Berlin", "Moscow", "Chicago", "Tokyo", "London"}
  22. 	dataPointCount := 100
  23. 	cities := make([]string, dataPointCount)
  24. 	for i := range cities {
  25. 		cities[i] = cityChoices[rand.Intn(len(cityChoices))]
  26. 	}
  27. 	prices := make([]int, dataPointCount) 
  28. 	for i := range prices {
  29. 		prices[i] = rand.Intn(100)
  30. 	}
  31. 	return cities, prices
  32. }
  34. func groupSlices([]string, []int) map[string]int {
  35. 	cities, prices := citiesAndPrices()
  36. 	towns := make(map[string]int)
  37. 	for _, cities := range cities {
  38. 		for _, prices := range prices {
  39. 			towns[cities] = prices
  40. 			break
  41. 		}
  42. 	}
  43. 	return towns
  44. }
英文:

I have a function that returns 2 variables in a form of slices and I want to merge the two slices into a map as key-value pairs. The issue is that can't find a way to iterate through each slice separately and add them as a key and a value.
The current implementation is iterating through all the prices and it's adding every price to the town key.

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"math/rand"
  6. 	"time"
  7. )
  9. func main() {
  10. 	cities, prices := citiesAndPrices()
  11. 	towns := groupSlices(cities, prices)
  12. 	for cities := range towns {
  14. 		fmt.Println(cities, prices)
  15. 	}
  17. }
  19. func citiesAndPrices() ([]string, []int) {
  20. 	rand.Seed(time.Now().UnixMilli())
  21. 	cityChoices := []string{"Berlin", "Moscow", "Chicago", "Tokyo", "London"}
  22. 	dataPointCount := 100
  23. 	cities := make([]string, dataPointCount)
  24. 	for i := range cities {
  25. 		cities[i] = cityChoices[rand.Intn(len(cityChoices))]
  26. 	}
  27. 	prices := make([]int, dataPointCount) 
  28. 	for i := range prices {
  29. 		prices[i] = rand.Intn(100)
  30. 	}
  31. 	return cities, prices
  32. }
  34. func groupSlices([]string, []int) map[string]int {
  35. 	cities, prices := citiesAndPrices()
  36. 	towns := make(map[string]int)
  37. 	for _, cities := range cities {
  38. 		for _, prices := range prices {
  39. 			towns[cities] = prices
  40. 			break
  41. 		}
  42. 	}
  43. 	return towns
  44. }

答案1

得分: 0

你必须将groupSlice函数返回为map[string][]int类型。

这是一个提示：

Seed方法与Rand.Seed方法不同，它在并发使用时是安全的。来源

因此，你可以只调用一次Seed方法，而不是在每次调用函数时都调用它。这将加快你的应用程序的速度。

我认为这是你想要的代码。

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"math/rand"
  6. 	"time"
  7. )
  9. func main() {
  10. 	rand.Seed(time.Now().UnixNano())
  11. 	cities, prices := citiesAndPrices()
  12. 	towns := groupSlices(cities, prices)
  13. 	fmt.Println()
  14. 	total := 0
  15. 	for c, p := range towns {
  16. 		fmt.Println(c, p)
  17. 		total += len(p)
  18. 	}
  19. 	fmt.Println()
  20. 	fmt.Println("total must 200, found :", total) // total must be 200
  21. }
  23. func citiesAndPrices() ([]string, []int) {
  24. 	cityChoices := []string{"Berlin", "Moscow", "Chicago", "Tokyo", "London"}
  25. 	dataPointCount := 100
  26. 	cities := make([]string, dataPointCount)
  27. 	prices := make([]int, dataPointCount)
  28. 	for i := range cities {
  29. 		cities[i] = cityChoices[rand.Intn(len(cityChoices))]
  30. 		prices[i] = rand.Intn(100)
  31. 	}
  32. 	fmt.Println("inside citiesAndPrices func")
  33. 	fmt.Println(cities)
  34. 	fmt.Println(prices)
  35. 	fmt.Println("done run citiesAndPrices func\n")
  36. 	return cities, prices
  37. }
  39. func groupSlices(c1 []string, p1 []int) map[string][]int {
  40. 	c2, p2 := citiesAndPrices()
  41. 	towns := make(map[string][]int)
  42. 	for i, t := range c1 {
  43. 		// this is for cities1 and price1
  44. 		if towns[t] == nil {
  45. 			towns[t] = make([]int, 0)
  46. 		}
  47. 		towns[t] = append(towns[t], p1[i])
  49. 		// this is for cities2 and price2
  50. 		if towns[c2[i]] == nil {
  51. 			towns[c2[i]] = make([]int, 0)
  52. 		}
  53. 		towns[c2[i]] = append(towns[c2[i]], p2[i])
  54. 	}
  55. 	return towns
  56. }
英文:

You must return groupSlice function as map[string][]int

This is a tips for you:

> Seed, unlike the Rand.Seed method, is safe for concurrent use. source.

So You can call it once, not on every call function. This will speed up your app.

I think this is the code that what you want.

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"math/rand"
  6. 	"time"
  7. )
  9. func main() {
  10. 	rand.Seed(time.Now().UnixNano())
  11. 	cities, prices := citiesAndPrices()
  12. 	towns := groupSlices(cities, prices)
  13. 	fmt.Println()
  14. 	total := 0
  15. 	for c, p := range towns {
  16. 		fmt.Println(c, p)
  17. 		total += len(p)
  18. 	}
  19. 	fmt.Println()
  20. 	fmt.Println("total must 200, found :", total) // total must be 200
  21. }
  23. func citiesAndPrices() ([]string, []int) {
  24. 	cityChoices := []string{"Berlin", "Moscow", "Chicago", "Tokyo", "London"}
  25. 	dataPointCount := 100
  26. 	cities := make([]string, dataPointCount)
  27. 	prices := make([]int, dataPointCount)
  28. 	for i := range cities {
  29. 		cities[i] = cityChoices[rand.Intn(len(cityChoices))]
  30. 		prices[i] = rand.Intn(100)
  31. 	}
  32. 	fmt.Println("inside citiesAndPrices func")
  33. 	fmt.Println(cities)
  34. 	fmt.Println(prices)
  35. 	fmt.Println("done run citiesAndPrices func\n")
  36. 	return cities, prices
  37. }
  39. func groupSlices(c1 []string, p1 []int) map[string][]int {
  40. 	c2, p2 := citiesAndPrices()
  41. 	towns := make(map[string][]int)
  42. 	for i, t := range c1 {
  43. 		// this is for cities1 and price1
  44. 		if towns[t] == nil {
  45. 			towns[t] = make([]int, 0)
  46. 		}
  47. 		towns[t] = append(towns[t], p1[i])
  49. 		// this is for cities2 and price2
  50. 		if towns[c2[i]] == nil {
  51. 			towns[c2[i]] = make([]int, 0)
  52. 		}
  53. 		towns[c2[i]] = append(towns[c2[i]], p2[i])
  54. 	}
  55. 	return towns
  56. }

huangapple
  • 本文由 发表于 2022年2月25日 05:05:22
  • 转载请务必保留本文链接：https://go.coder-hub.com/71258244.html
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