Question

您将获得存储在变量字符串中的字符串数组。创建一个名为countsStrings的新数组，其中包含type（String，Int）的值。每个元组都包含一个来自strings数组的字符串，后跟一个整数，表示它在strings数组中出现的次数。每个字符串只应在countsStrings数组中出现一次。

我还没有解决问题，因为它已经解决了。我只是想了解它。我可以理解的部分内容和我无法理解的部分内容。我无法理解for循环中的部分。

 var a =  ["tuples", "are", "awesome", "tuples", "are", "cool","tuples",  "tuples", "tuples", "shades"]

 var y: [(String,Int)] = []

  for z in a{ 
  var x = false

  for i in 0..<y.count {
  if (y[i].0 == z) {
  y[i].1 += 1
  x = true
   }
   }
   if x == false {
   y.append((z,1))
    }
    }
    print(y)

打印

[（“tuples”，5），（“are”，2），（“awesome”，1），（“cool”，1），（“shades”，1）] cool“，1），（“阴影”，1）]

Answer 1

如果遇到这样的问题，请先尝试重命名变量。它有助于理解问题。

例如，我拿了你的代码，只是为了清晰起见改变了变量名。现在告诉我是否还有一些你不能得到的东西。

let words = ["tuples", "are", "awesome", "tuples", "are", "cool","tuples",  "tuples", "tuples", "shades"]

var tuples: [(String, Int)] = []

for word in words {
    var isAlreadyInTupleArray = false

    // Loop trough the existing tuples and updates the number of apparition if the word is found
    for (index, tuple) in tuples.enumerated() {
        let tupleWord:String = tuple.0
        let numberOfAppearances:Int = tuple.1

        if tupleWord == word {
            tuples[index].1 += 1

            isAlreadyInTupleArray = true
        }
    }

    // In the case the word was not in the existing tuples, we append a new tuple
    if isAlreadyInTupleArray == false {
        tuples.append((word, 1))
    }
}
print(tuples)

Answer 2

let arr = ["tuples", "are", "awesome", "tuples", "are", "cool","tuples",  "tuples", "tuples", "shades"]
 var counts:[String:Int] = [:]

 for item in arr {
    counts[item] = (counts[item] ?? 0) + 1
 }
 for (key, value) in counts {
    print("\(key) occurs \(value) time(s)")
 }

如何使用元组计算子字符串（单词）出现在字符串中的次数

2 个答案: