如何找到数组中每个项目的字符数

How do I find the character count of each item in an array

我正在编写一个函数,用于打印字典中字符数超过 8 的字符串值。以下是我目前所拥有的,但我不确定如何制定我的 where 条件以使其看起来数组中每个字符串值的字符数。

var stateCodes = ["NJ": "New Jersey", "CO": "Colorado", "WI": "Wisconsin", "OH": "Ohio"]

func printLongState (_ dictionary: [String: String]) -> (Array<Any>) {
let fullStateNames = Array(stateCodes.values)

for _ in fullStateNames where fullStateNames.count > 8 {
    print(fullStateNames)
    return fullStateNames
}

return fullStateNames
}

printLongState(stateCodes)

如果你想用for循环,那么你可以这样做。

func printLongState (_ dictionary: [String: String]) -> (Array<Any>) {
    var fullStateNames = [String]()
    for (_, value) in dictionary where value.characters.count > 8 {
        fullStateNames.append(value)
    }
    return fullStateNames
}

但这不是 SwiftSwift 中的方法你可以做的是你可以使用 flatMap 和你的 Dictionary 来制作 [=15= 的数组] 或使用 dictionary.values.filter

将 flatMap 与字典一起使用

func printLongState (_ dictionary: [String: String]) -> (Array<Any>) {  
    return dictionary.flatMap { .characters.count > 8 ?  : nil }
}
// Call it like this way.
var stateCodes = ["NJ": "New Jersey", "CO": "Colorado", "WI": "Wisconsin", "OH": "Ohio"] 
print(printLongState(stateCodes)) //["Wisconsin", "New Jersey"]

dictionary.values

上使用过滤器
func printLongState (_ dictionary: [String: String]) -> (Array<Any>) {  
    return dictionary.values.filter { [=12=].characters.count > 8 }
}

只是 filter 你的结果而不是使用 for-loop:

如果你想 return 字典使用以下:

func printLongState (_ dictionary: [String: String]) -> (Array<Any>) {
    let overEightChars = stateCodes.filter({ [=10=].value.characters.count > 8 })
    return overEightChars
}

如果您想 return 一个字符串数组,请使用以下内容:

func printLongState (_ dictionary: [String: String]) -> (Array<Any>) {
    return dictionary.values.filter { [=11=].characters.count > 8 }
}

尝试像这样将 filtercharacters.count 一起使用:

var states = ["NJ": "New Jersey", "CO": "Colorado", "WI": "Wisconsin", "OH": "Ohio"]

states.filter({ (_, value) -> Bool in
    return value.characters.count > 8
}).map({ (_, value) in
    print(value)
})