用于计算字数的 Swift 字符串中的字数 [英] Number of words in a Swift String for word count calculation
问题描述
我想制作一个程序来找出字符串中有多少个单词,用空格、逗号或其他字符分隔.然后把总数加起来.
I want to make a procedure to find out how many words are there in a string, separated by space, or comma, or some other character. And then add up the total later.
我正在制作一个平均计算器,所以我想要数据的总数,然后将所有单词相加.
I'm making an average calculator, so I want the total count of data and then add up all the words.
推荐答案
更新:Xcode 10.2.x • Swift 5 或更高版本
使用 Foundation 方法 enumerateSubstrings(in: Range)
并将 .byWords
设置为选项:
Using Foundation method enumerateSubstrings(in: Range)
and setting .byWords
as options:
let sentence = "I want to an algorithm that could help find out how many words are there in a string separated by space or comma or some character. And then append each word separated by a character to an array which could be added up later I'm making an average calculator so I want the total count of data and then add up all the words. By words I mean the numbers separated by a character, preferably space Thanks in advance"
var words: [Substring] = []
sentence.enumerateSubstrings(in: sentence.startIndex..., options: .byWords) { _, range, _, _ in
words.append(sentence[range])
}
print(words) // "["I", "want", "to", "an", "algorithm", "that", "could", "help", "find", "out", "how", "many", "words", "are", "there", "in", "a", "string", "separated", "by", "space", "or", "comma", "or", "some", "character", "And", "then", "append", "each", "word", "separated", "by", "a", "character", "to", "an", "array", "which", "could", "be", "added", "up", "later", "I\\'m", "making", "an", "average", "calculator", "so", "I", "want", "the", "total", "count", "of", "data", "and", "then", "add", "up", "all", "the", "words", "By", "words", "I", "mean", "the", "numbers", "separated", "by", "a", "character", "preferably", "space", "Thanks", "in", "advance"]\n"
print(words.count) // 79
或者使用原生 Swift 5 新的 Character
属性 isLetter
和 split 方法:
Or using native Swift 5 new Character
property isLetter
and the split method:
let words = sentence.split { !$0.isLetter }
print(words) // "["I", "want", "to", "an", "algorithm", "that", "could", "help", "find", "out", "how", "many", "words", "are", "there", "in", "a", "string", "separated", "by", "space", "or", "comma", "or", "some", "character", "And", "then", "append", "each", "word", "separated", "by", "a", "character", "to", "an", "array", "which", "could", "be", "added", "up", "later", "I", "m", "making", "an", "average", "calculator", "so", "I", "want", "the", "total", "count", "of", "data", "and", "then", "add", "up", "all", "the", "words", "By", "words", "I", "mean", "the", "numbers", "separated", "by", "a", "character", "preferably", "space", "Thanks", "in", "advance"]\n"
print(words.count) // 80
<小时>
扩展 StringProtocol
以支持子字符串:
extension StringProtocol {
var words: [SubSequence] {
return split { !$0.isLetter }
}
var byWords: [SubSequence] {
var byWords: [SubSequence] = []
enumerateSubstrings(in: startIndex..., options: .byWords) { _, range, _, _ in
byWords.append(self[range])
}
return byWords
}
}
<小时>
sentence.words // ["I", "want", "to", "an", "algorithm", "that", "could", "help", "find", "out", "how", "many", "words", "are", "there", "in", "a", "string", "separated", "by", "space", "or", "comma", "or", "some", "character", "And", "then", "append", "each", "word", "separated", "by", "a", "character", "to", "an", "array", "which", "could", "be", "added", "up", "later", "I", "m", "making", "an", "average", "calculator", "so", "I", "want", "the", "total", "count", "of", "data", "and", "then", "add", "up", "all", "the", "words", "By", "words", "I", "mean", "the", "numbers", "separated", "by", "a", "character", "preferably", "space", "Thanks", "in", "advance"]
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