mirror of
https://github.com/unknwon/the-way-to-go_ZH_CN.git
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Merge pull request #33 from chidouhu/master
add 08.2.md 08.3.md 08.4.md 08.5.md 08.6.md
This commit is contained in:
72
eBook/08.2.md
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72
eBook/08.2.md
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#8.2 删除元素
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测试map1中是否存在key1:
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在例子8.1中,我们已经见过可以使用val1 = map1[key1]的方法获取key1对应的值val1。如果map中不存在key1,val1就是一个值类型的空值。
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这就会给我们带来困惑了:现在我们没法区分到底是key1不存在还是它对应的value就是空值。
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为了解决这个问题,我们可以这么用: val1, isPresent = map1[key1]
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isPresent返回一个bool值:如果key1存在于map1,val1就是key1对应的value值,并且isPresent为true;如果key1不存在,val1就是一个空值,并且isPresent会返回false。
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如果你只是想判断某个key是否存在而不关心它对应的值到底是多少,你可以这么做:
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_, ok := map1[key1] // 如果key1存在则ok == true,否在ok为false
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或者和if混合使用:
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if _, ok := map1[key1]; ok {
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// ...
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}
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从map1中删除key1:
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直接: delete(map1, key1)
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如果key1不存在,该操作不会产生错误。
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示例 8.4 [map_testelement.go](exmaples/chapter_8/map_testelement.go)
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package main
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import "fmt"
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func main() {
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var value int
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var isPresent bool
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map1 := make(map[string]int)
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map1["New Delhi"] = 55
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map1["Beijing"] = 20
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map1["Washington"] = 25
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value, isPresent = map1["Beijing"]
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if isPresent {
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fmt.Printf("The value of \"Beijin\" in map1 is: %d\n", value)
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} else {
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fmt.Printf("map1 does not contain Beijing")
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}
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value, isPresent = map1["Paris"]
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fmt.Printf("Is \"Paris\" in map1 ?: %t\n", isPresent)
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fmt.Printf("Value is: %d\n", value)
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// delete an item:
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delete(map1, "Washington")
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value, isPresent = map1["Washington"]
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if isPresent {
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fmt.Printf("The value of \"Washington\" in map1 is: %d\n", value)
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} else {
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fmt.Println("map1 does not contain Washington")
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}
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}
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输出结果:
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The value of "Beijing" in map1 is: 20
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Is "Paris" in map1 ?: false
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Value is: 0
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map1 does not contain Washington
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##链接
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- [目录](directory.md)
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- 上一节:[声明,初始化和make](08.1.md)
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- 下一节:[for循环构造方法](08.3.md)
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59
eBook/08.3.md
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59
eBook/08.3.md
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#8.3 for循环构造方法
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可以使用for循环构造map:
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for key, value := range map1 {
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...
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}
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第一个返回值key是map中的key值,第二个返回值则是该key对应的value值;这两个都是仅for循环内部可见的局部变量。其中第一个返回值key值是一个可选元素。如果你只关心值,可以这么使用
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for _, value := range map1 {
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...
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}
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如果只想获取key,你可以这么使用:
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for key := range map1 {
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fmt.Printf("key is: %d\n", key)
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}
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示例 8.5 [maps_forrange.go](exmaples/chapter_8/maps_forrange.go)
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package main
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import "fmt"
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func main() {
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map1 := make(map[int]float32)
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map1[1] = 1.0
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map1[2] = 2.0
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map1[3] = 3.0
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map1[4] = 4.0
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for key, value := range map1 {
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fmt.Printf("key is: %d - value is: %f\n", key, value)
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}
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}
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输出结果:
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key is: 3 - value is: 3.000000
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key is: 1 - value is: 1.000000
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key is: 4 - value is: 4.000000
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key is: 2 - value is: 2.000000
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注意到map不是按照key的顺序排列的,也不是按照value的序排列的。
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问题 8.1: 下面这段代码的输出是什么?
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capitals := map[string] string {"France":"Paris", "Italy":"Rome", "Japan":"Tokyo" }
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for key := range capitals {
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fmt.Println("Map item: Capital of", key, "is", capitals[key])
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}
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练习 8.1: map_days.go
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创建一个map来保存每周7天的名字,将它们打印出来并且测试是否存在tuesday和hollyday。
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##链接
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- [目录](directory.md)
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- 上一节:[删除元素](08.2.md)
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- 下一节:[maps分片](08.4.md)
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37
eBook/08.4.md
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37
eBook/08.4.md
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#8.3 map分片
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假设我们想获取一个map的分片,我们必须使用两次make()方法,第一次分配slice,第二次分配slice的每个map元素(参见下面的例子8.3)。
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示例 8.3 [maps_forrange.go](exmaples/chapter_8/maps_forrange.go)
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package main
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import "fmt"
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func main() {
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// Version A:
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items := make([]map[int]int, 5)
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for i:= range items {
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items[i] = make(map[int]int, 1)
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items[i][1] = 2
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}
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fmt.Printf("Version A: Value of items: %v\n", items)
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// Version B: NOT GOOD!
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items2 := make([]map[int]int, 5)
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for _, item := range items2 {
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item = make(map[int]int, 1) // item is only a copy of the slice element.
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item[1] = 2 // This 'item' will be lost on the next iteration.
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}
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fmt.Printf("Version B: Value of items: %v\n", items2)
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}
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输出结果:
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Version A: Value of items: [map[1:2] map[1:2] map[1:2] map[1:2] map[1:2]]
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Version B: Value of items: [map[] map[] map[] map[] map[]]
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需要注意的是,应当像A版本那样通过索引使用slice的map项。在B版本中获得的项只是map值的一个拷贝而已,所以真正的map元素没有得到初始化。
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##链接
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- [目录](directory.md)
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- 上一节:[for循环构造方法](08.3.md)
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- 下一节:[map排序](08.5.md)
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59
eBook/08.5.md
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59
eBook/08.5.md
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#8.5 map排序
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map默认是无序的,不管是按照key还是按照value默认都不排序(参见8.3节)
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如果你想为map排序,需要将key(或者value)拷贝到一个slice,再对slice排序(使用sort包,参见7.6.6),然后可以使用slice的for-range方法打印出所有的key和value。
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下面有一个示例:
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示例 8.6 [sort_map.go](exmaples/chapter_8/sort_map.go)
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// the telephone alphabet:
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package main
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import (
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"fmt"
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"sort"
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)
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var (
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barVal = map[string]int{"alpha": 34, "bravo": 56, "charlie": 23,
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"delta": 87, "echo": 56, "foxtrot": 12,
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"golf": 34, "hotel": 16, "indio": 87,
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"juliet": 65, "kili": 43, "lima": 98}
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)
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func main() {
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fmt.Println("unsorted:")
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for k, v := range barVal {
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fmt.Printf("Key: %v, Value: %v / ", k, v)
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}
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keys := make([]string, len(barVal))
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i := 0
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for k, _ := range barVal {
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keys[i] = k
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i++
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}
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sort.Strings(keys)
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fmt.Println()
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fmt.Println("sorted:")
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for _, k := range keys {
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fmt.Printf("Key: %v, Value: %v / ", k, barVal[k])
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}
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}
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输出结果:
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unsorted:
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Key: bravo, Value: 56 / Key: echo, Value: 56 / Key: indio, Value: 87 / Key: juliet, Value: 65 / Key: alpha, Value: 34 / Key: charlie, Value: 23 / Key: delta, Value: 87 / Key: foxtrot, Value: 12 / Key: golf, Value: 34 / Key: hotel, Value: 16 / Key: kili, Value: 43 / Key: lima, Value: 98 /
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sorted:
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Key: alpha, Value: 34 / Key: bravo, Value: 56 / Key: charlie, Value: 23 / Key: delta, Value: 87 / Key: echo, Value: 56 / Key: foxtrot, Value: 12 / Key: golf, Value: 34 / Key: hotel, Value: 16 / Key: indio, Value: 87 / Key: juliet, Value: 65 / Key: kili, Value: 43 / Key: lima, Value: 98 / [fangjun@st01-dstream-0001.st01.baidu.com go]$ sz -be sort_map.go
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但是如果你想要一个排序的列表你最好使用结构体slice,这样会更有效:
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type struct {
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key string
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value int
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}
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##链接
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- [目录](directory.md)
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- 上一节:[maps分片](08.4.md)
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- 下一节:[倒置map](08.6.md)
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44
eBook/08.6.md
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44
eBook/08.6.md
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@@ -0,0 +1,44 @@
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#8.6 倒置map
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这里倒置是指调换key和value。如果map的值类型可以作为key且所有的value是唯一的,那么通过下面的方法可以简单的做到倒置:
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示例 8.7 [invert_map.go](exmaples/chapter_8/invert_map.go)
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package main
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import (
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"fmt"
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)
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var (
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barVal = map[string]int{"alpha": 34, "bravo": 56, "charlie": 23,
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"delta": 87, "echo": 56, "foxtrot": 12,
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"golf": 34, "hotel": 16, "indio": 87,
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"juliet": 65, "kili": 43, "lima": 98}
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)
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func main() {
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invMap := make(map[int]string, len(barVal))
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for k, v := range barVal {
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invMap[v] = k
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}
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fmt.Println("inverted:")
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for k, v := range invMap {
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fmt.Printf("Key: %v, Value: %v / ", k, v)
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}
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fmt.Println()
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}
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输出结果:
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inverted:
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Key: 34, Value: golf / Key: 23, Value: charlie / Key: 16, Value: hotel / Key: 87, Value: delta / Key: 98, Value: lima / Key: 12, Value: foxtrot / Key: 43, Value: kili / Key: 56, Value: bravo / Key: 65, Value: juliet /
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如果原始value值不唯一那么这么做肯定会出错;为了保证不出错,当遇到不唯一的key时应当立刻停止,这样可能会导致没有包含原map的所有键值对!一种解决方法就是仔细检查唯一性并且使用多值map,比如使用`map[int][]string`类型。
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练习 8.2: map_drinks.go
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构造一个将英文饮料名映射为法语(或者任意你的母语)的集合;先打印所有的饮料,然后打印原名和翻译后的名字。接下来按照英文名排序后再打印出来。
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##链接
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- [目录](directory.md)
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- 上一节:[map排序](08.5.md)
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- 下一节:[包](09.0.md)
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23
eBook/examples/chapter_8/invert_map.go
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23
eBook/examples/chapter_8/invert_map.go
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@@ -0,0 +1,23 @@
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package main
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import (
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"fmt"
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)
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var (
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barVal = map[string]int{"alpha": 34, "bravo": 56, "charlie": 23,
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"delta": 87, "echo": 56, "foxtrot": 12,
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"golf": 34, "hotel": 16, "indio": 87,
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"juliet": 65, "kili": 43, "lima": 98}
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)
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func main() {
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invMap := make(map[int]string, len(barVal))
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for k, v := range barVal {
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invMap[v] = k
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}
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fmt.Println("inverted:")
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for k, v := range invMap {
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fmt.Printf("Key: %v, Value: %v / ", k, v)
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}
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fmt.Println()
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}
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31
eBook/examples/chapter_8/map_testelement.go
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31
eBook/examples/chapter_8/map_testelement.go
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@@ -0,0 +1,31 @@
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package main
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import "fmt"
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func main() {
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var value int
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var isPresent bool
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map1 := make(map[string]int)
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map1["New Delhi"] = 55
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map1["Beijing"] = 20
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map1["Washington"] = 25
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value, isPresent = map1["Beijing"]
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if isPresent {
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fmt.Printf("The value of \"Beijin\" in map1 is: %d\n", value)
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} else {
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fmt.Printf("map1 does not contain Beijing")
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}
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value, isPresent = map1["Paris"]
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fmt.Printf("Is \"Paris\" in map1 ?: %t\n", isPresent)
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fmt.Printf("Value is: %d\n", value)
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// delete an item:
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delete(map1, "Washington")
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value, isPresent = map1["Washington"]
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if isPresent {
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fmt.Printf("The value of \"Washington\" in map1 is: %d\n", value)
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} else {
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fmt.Println("map1 does not contain Washington")
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}
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}
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13
eBook/examples/chapter_8/maps_forrange.go
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13
eBook/examples/chapter_8/maps_forrange.go
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@@ -0,0 +1,13 @@
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package main
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import "fmt"
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func main() {
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map1 := make(map[int]float32)
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map1[1] = 1.0
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map1[2] = 2.0
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map1[3] = 3.0
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map1[4] = 4.0
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for key, value := range map1 {
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fmt.Printf("key is: %d - value is: %f\n", key, value)
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}
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}
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20
eBook/examples/chapter_8/slice_maps.go
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20
eBook/examples/chapter_8/slice_maps.go
Normal file
@@ -0,0 +1,20 @@
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package main
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import "fmt"
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func main() {
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// Version A:
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items := make([]map[int]int, 5)
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for i:= range items {
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items[i] = make(map[int]int, 1)
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items[i][1] = 2
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}
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fmt.Printf("Version A: Value of items: %v\n", items)
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// Version B: NOT GOOD!
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items2 := make([]map[int]int, 5)
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for _, item := range items2 {
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item = make(map[int]int, 1) // item is only a copy of the slice element.
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item[1] = 2 // This 'item' will be lost on the next iteration.
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}
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fmt.Printf("Version B: Value of items: %v\n", items2)
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}
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32
eBook/examples/chapter_8/sort_map.go
Normal file
32
eBook/examples/chapter_8/sort_map.go
Normal file
@@ -0,0 +1,32 @@
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// the telephone alphabet:
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package main
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import (
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"fmt"
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"sort"
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)
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var (
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barVal = map[string]int{"alpha": 34, "bravo": 56, "charlie": 23,
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"delta": 87, "echo": 56, "foxtrot": 12,
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"golf": 34, "hotel": 16, "indio": 87,
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"juliet": 65, "kili": 43, "lima": 98}
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)
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func main() {
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fmt.Println("unsorted:")
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for k, v := range barVal {
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fmt.Printf("Key: %v, Value: %v / ", k, v)
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}
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keys := make([]string, len(barVal))
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i := 0
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for k, _ := range barVal {
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keys[i] = k
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i++
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}
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sort.Strings(keys)
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fmt.Println()
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fmt.Println("sorted:")
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for _, k := range keys {
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fmt.Printf("Key: %v, Value: %v / ", k, barVal[k])
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}
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}
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