如何计算 Swift 数组中某个元素的出现次数? [英] How to count occurrences of an element in a Swift array?
问题描述
我已经看到了一些这样的例子，但所有这些似乎都依赖于知道你想要计算哪个元素的出现次数.我的数组是动态生成的，所以我无法知道我想计算哪个元素的出现次数(我想计算所有元素的出现次数).谁能给点建议?
提前致谢
也许我应该更清楚，数组将包含多个不同的字符串(例如["FOO", "FOO", "BAR", "FOOBAR"]
在事先不知道它们是什么的情况下，如何计算 foo、bar 和 foobar 的出现次数?
Swift 3 和 Swift 2:
您可以使用 [String: Int] 类型的字典来为 [String] 中的每个项目建立计数:
let arr = [FOO"、FOO"、BAR"、FOOBAR"]var 计数:[String: Int] = [:]对于 arr { 中的项目计数[项目] = (计数[项目] ?? 0) + 1}打印(计数)//[BAR:1，FOOBAR:1，FOO:2]"；for (key, value) in counts {打印(\(键)出现\(值)次")}输出:
BAR 出现 1 次FOOBAR 出现 1 次FOO 出现 2 次Swift 4:
Swift 4 介绍 (SE-0165) 在字典查找中包含默认值的能力，并且结果值可以通过诸如 += 和 -= 之类的操作来改变，所以:
counts[item] = (counts[item] ?? 0) + 1变成:
counts[item, default: 0] += 1这使得使用 forEach 在简洁的一行中完成计数操作变得容易:
let arr = [FOO"、FOO"、BAR"、FOOBAR"]var 计数:[String: Int] = [:]arr.forEach { counts[$0, default: 0] += 1 }打印(计数)//[FOOBAR":1，FOO":2，BAR":1]"；Swift 4:reduce(into:_:)
Swift 4 引入了新版本的 reduce，它使用 inout 变量来累积结果.使用它，计数的创建真正成为一行:
let arr = [FOO"、FOO"、BAR"、FOOBAR"]let counts = arr.reduce(into: [:]) { counts, word in counts[word, default: 0] += 1 }打印(计数)//[BAR":1，FOOBAR":1，FOO":2]或者使用默认参数:
let counts = arr.reduce(into: [:]) { $0[$1, default: 0] += 1 }最后，您可以将其作为 Sequence 的扩展，以便可以在任何包含 Hashable 项目(包括 Array)的 、Sequence 上调用它ArraySlice、String 和 String.SubSequence:
extension Sequence where Element: Hashable {var 直方图:[元素:Int] {return self.reduce(into: [:]) { counts, elem in counts[elem, default: 0] += 1 }}}这个想法是从这个问题借来的，尽管我将其更改为计算属性.感谢@LeoDabus 建议扩展 Sequence 而不是 Array 以获取其他类型.
示例:
print("abacab".histogram)<块引用>
[a":3，b":2，c":1]print("Hello World!".suffix(6).histogram)<块引用>
[l":1，！":1，d":1，o":1，W":1，r":1]print([1,2,3,2,1].histogram)<块引用>
[2: 2, 3: 1, 1: 2]print([1,2,3,2,1,2,1,3,4,5].prefix(8).histogram)<块引用>
[1: 3, 2: 3, 3: 2]print(stride(from: 1, through: 10, by: 2).histogram)<块引用>
[1: 1, 3: 1, 5: 1, 7: 1, 9: 1]I've seen a few examples of this but all of those seem to rely on knowing which element you want to count the occurrences of. My array is generated dynamically so I have no way of knowing which element I want to count the occurrences of (I want to count the occurrences of all of them). Can anyone advise?
Thanks in advance
EDIT:
Perhaps I should have been clearer, the array will contain multiple different strings (e.g.
["FOO", "FOO", "BAR", "FOOBAR"]
How can I count the occurrences of foo, bar and foobar without knowing what they are in advance?
Swift 3 and Swift 2:
You can use a dictionary of type [String: Int] to build up counts for each of the items in your [String]:
let arr = ["FOO", "FOO", "BAR", "FOOBAR"]
var counts: [String: Int] = [:]
for item in arr {
counts[item] = (counts[item] ?? 0) + 1
}
print(counts) // "[BAR: 1, FOOBAR: 1, FOO: 2]"
for (key, value) in counts {
print("\(key) occurs \(value) time(s)")
}
output:
BAR occurs 1 time(s)
FOOBAR occurs 1 time(s)
FOO occurs 2 time(s)
Swift 4:
Swift 4 introduces (SE-0165) the ability to include a default value with a dictionary lookup, and the resulting value can be mutated with operations such as += and -=, so:
counts[item] = (counts[item] ?? 0) + 1
becomes:
counts[item, default: 0] += 1
That makes it easy to do the counting operation in one concise line using forEach:
let arr = ["FOO", "FOO", "BAR", "FOOBAR"]
var counts: [String: Int] = [:]
arr.forEach { counts[$0, default: 0] += 1 }
print(counts) // "["FOOBAR": 1, "FOO": 2, "BAR": 1]"
Swift 4: reduce(into:_:)
Swift 4 introduces a new version of reduce that uses an inout variable to accumulate the results. Using that, the creation of the counts truly becomes a single line:
let arr = ["FOO", "FOO", "BAR", "FOOBAR"]
let counts = arr.reduce(into: [:]) { counts, word in counts[word, default: 0] += 1 }
print(counts) // ["BAR": 1, "FOOBAR": 1, "FOO": 2]
Or using the default parameters:
let counts = arr.reduce(into: [:]) { $0[$1, default: 0] += 1 }
Finally you can make this an extension of Sequence so that it can be called on any Sequence containing Hashable items including Array, ArraySlice, String, and String.SubSequence:
extension Sequence where Element: Hashable {
var histogram: [Element: Int] {
return self.reduce(into: [:]) { counts, elem in counts[elem, default: 0] += 1 }
}
}
This idea was borrowed from this question although I changed it to a computed property. Thanks to @LeoDabus for the suggestion of extending Sequence instead of Array to pick up additional types.
Examples:
print("abacab".histogram)
["a": 3, "b": 2, "c": 1]
print("Hello World!".suffix(6).histogram)
["l": 1, "!": 1, "d": 1, "o": 1, "W": 1, "r": 1]
print([1,2,3,2,1].histogram)
[2: 2, 3: 1, 1: 2]
print([1,2,3,2,1,2,1,3,4,5].prefix(8).histogram)
[1: 3, 2: 3, 3: 2]
print(stride(from: 1, through: 10, by: 2).histogram)
[1: 1, 3: 1, 5: 1, 7: 1, 9: 1]
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