我想要统计以下字符串中字母、数字和特殊字符的数量:
let phrase = "The final score was 32-31!"
我尝试了:
for tempChar in phrase {
if (tempChar >= "a" && tempChar <= "z") {
letterCounter++
}
// etc.
但我一直出错。我尝试了各种各样的变化-仍然出错-例如:
找不到接受所提供参数的'<='的重载
我想要统计以下字符串中字母、数字和特殊字符的数量:
let phrase = "The final score was 32-31!"
我尝试了:
for tempChar in phrase {
if (tempChar >= "a" && tempChar <= "z") {
letterCounter++
}
// etc.
但我一直出错。我尝试了各种各样的变化-仍然出错-例如:
找不到接受所提供参数的'<='的重载
对于Swift 5,请参见rustylepord的回答.
更新Swift 3:
let letters = CharacterSet.letters
let digits = CharacterSet.decimalDigits
var letterCount = 0
var digitCount = 0
for uni in phrase.unicodeScalars {
if letters.contains(uni) {
letterCount += 1
} else if digits.contains(uni) {
digitCount += 1
}
}
(旧版本的 Swift 的先前回答)
一种可能的 Swift 解决方案:
var letterCounter = 0
var digitCount = 0
let phrase = "The final score was 32-31!"
for tempChar in phrase.unicodeScalars {
if tempChar.isAlpha() {
letterCounter++
} else if tempChar.isDigit() {
digitCount++
}
}
更新:上述解决方案只适用于ASCII字符集中的字符,即不识别Ä、é或ø作为字母。下面的替代解决方案使用了Foundation
框架中的NSCharacterSet
,可以根据Unicode字符类测试字符:
let letters = NSCharacterSet.letterCharacterSet()
let digits = NSCharacterSet.decimalDigitCharacterSet()
var letterCount = 0
var digitCount = 0
for uni in phrase.unicodeScalars {
if letters.longCharacterIsMember(uni.value) {
letterCount++
} else if digits.longCharacterIsMember(uni.value) {
digitCount++
}
}
更新2:自从Xcode 6 beta 4发布以来,第一种解决方案已经不再适用,因为Swift中的isAlpha()
和相关的(仅ASCII)方法已被移除。第二个解决方案仍然适用。
使用unicodeScalars的值
let phrase = "The final score was 32-31!"
var letterCounter = 0, digitCounter = 0
for scalar in phrase.unicodeScalars {
let value = scalar.value
if (value >= 65 && value <= 90) || (value >= 97 && value <= 122) {++letterCounter}
if (value >= 48 && value <= 57) {++digitCounter}
}
println(letterCounter)
println(digitCounter)
对于Swift 5,您可以针对简单字符串执行以下操作,但要注意处理像"1️⃣" ,"④"这样的字符,因为这些字符也会被视为数字。
let phrase = "The final score was 32-31!"
var numberOfDigits = 0;
var numberOfLetters = 0;
var numberOfSymbols = 0;
phrase.forEach {
if ($0.isNumber) {
numberOfDigits += 1;
}
else if ($0.isLetter) {
numberOfLetters += 1
}
else if ($0.isSymbol || $0.isPunctuation || $0.isCurrencySymbol || $0.isMathSymbol) {
numberOfSymbols += 1;
}
}
print(#"\#(numberOfDigits) || \#(numberOfLetters) || \#(numberOfSymbols)"#);
如果您只需要一个信息(字母、数字或符号),您可以在一行中完成:
let phrase = "The final score was 32-31!"
let count = phrase.filter{ $0.isLetter }.count
print(count) // "16\n"
但是多次执行 phrase.filter
效率较低,因为它需要遍历整个字符串。
extension String {
var letterCount : Int {
return self.unicodeScalars.filter({ CharacterSet.letters.contains($0) }).count
}
var digitCount : Int {
return self.unicodeScalars.filter({ CharacterSet.decimalDigits.contains($0) }).count
}
}
或者编写一个函数来获取输入的任何CharacterSet
的计数
extension String {
func characterCount(for set: CharacterSet) -> Int {
return self.unicodeScalars.filter({ set.contains($0) }).count
}
}
用法:
let phrase = "the final score is 23-13!"
let letterCount = phrase.characterCount(for: .letters)
isalpha()
等函数在C中非常著名,详见https://developer.apple.com/library/ios/documentation/System/Conceptual/ManPages_iPhoneOS/man3/ctype.3.html#//apple_ref/doc/man/3/ctype。因此,我只需要想办法从Swift中使用它们就可以了。但请注意,这仅适用于ASCII字符。它不能识别来自“外语”(如“ä”,“è”或“ø”)的字母。 - Martin R+
和-
视为数字。 - Christian Dietrich