遍历两个自定义数组并在变量相等时设置值Swift

Question

I have an easy question that is also hard at the same time. I have two separate structs (this can also work for classes):

struct FBTweet {
    var tweetId: Int? //set
    var tweetText: String?  //set
}

and

struct Status {
    var statusId: Int? //set
    var statusText: String? //no value
    }

I have an array of both structs var fbTweetArray: [FBTweet] = [] and var statusArray: [Status] = []

I have set every variable in to a certain value in each index in fbTweetArray but I only set the .statusId variable in each index for statusArray. For every statusArray.statusId value in statusArray, there is only one fbTweetArray.tweetId that has the same exact Int value. I am trying to make is so that if these two variables are the same then I should set set statusArray.statusText to whatever fbTweetarray.tweetText is. So for example only fbTweetArray[1].tweetid = 2346 and statusArray[4].statusId = 2346 have 2346 as their value. There for if fbTweetArray[1].tweetText = "hello friend" then statusArray[4].statusText needs to be set to "hello friend".

So far I have

func testWhat () {

    var fbTweetArray: [FBTweet] = []
    var statusArray: [Status] = []

    for  fbTweet in fbTweetArray {
        for var status in statusArray {
            if (status.statusId == fbTweet.tweetId ) {
                status.statusText = fbTweet.tweetText
            }
        }
    }
}

how do I set the for var status in the for loop back into the statusArray since it is now a var and is different than one of the indexes in var statusArray: [Status] = []

Answer 1

Your question is interesting only if both the arrays are not ordered.

To find the element from fbTweet array, you can sort it and employ binary search. Then enumerate status array and find the fbTweet object with the same identifier and modify the status object. It needs to be saved again in the array as structs get copied on write.

extension Array where Element == FBTweet {
    func binarySearchFBTweetWith(_ id:Int) -> FBTweet? {
        var range = 0..<self.count
        while range.startIndex < range.endIndex {
            let midIndex = range.startIndex + (range.endIndex - range.startIndex) / 2
            guard let tweetId = self[midIndex].tweetId else {
                continue
            }
            if tweetId == id {
                return self[midIndex]
            } else if tweetId < id {
                range = midIndex+1..<range.endIndex
            } else {
                range = range.startIndex..<midIndex
            }
        }
        return nil
    }
}

fbTweetArray.sort{($0.tweetId ?? 0) < ($1.tweetId ?? 0)}
for (index, status) in statusArray.enumerated() {
    guard let statusId = status.statusId else {continue}
    guard let fbTweet = fbTweetArray.binarySearchFBTweetWith(statusId) else {continue}
    var status = status
    status.statusText = fbTweet.tweetText
    statusArray[index] = status
}

Answer 2

Basically, you need only one for / forEach loop to achieve what you want:

var fbTweetArray: [FBTweet] = [
    FBTweet(tweetId: 1, tweetText: "1"),
    FBTweet(tweetId: 2, tweetText: "2"),
    FBTweet(tweetId: 3, tweetText: "3")
]

var statusArray: [Status] = [
    Status(statusId: 2, statusText: nil),
    Status(statusId: 1, statusText: nil),
    Status(statusId: 3, statusText: nil)
]

fbTweetArray.forEach { tweet in
    if let index = statusArray.index(where: { $0.statusId == tweet.tweetId }) {
        statusArray[index].statusText = tweet.tweetText
    }
}

print(statusArray.map { $0.statusText }) // [Optional("2"), Optional("1"), Optional("3")]

Note, that your id s in both structures can be nil . To handle this situation (if both id is nil - objects are not equal) you can write custom == func:

struct Status {
    var statusId: Int? //set
    var statusText: String? //no value

    static func == (lhs: Status, rhs: FBTweet) -> Bool {
        guard let lhsId = lhs.statusId, let rhsId = rhs.tweetId else { return false }
        return lhsId == rhsId
    }
}

...

// rewrite .index(where: ) in if condition
if let index = statusArray.index(where: { $0 == tweet }) { ... }

---

Also, there is some pro-tip. If you adopt your structs to Hashable protocol, you will be able to place FBTweet s and Status es into Set structure. The benefits of that:

If you instead store those objects in a set, you can theoretically find any one of them in constant time (O(1)) — that is, a lookup on a set with 10 elements takes the same amount of time as a lookup on a set with 10,000.

You can find more in-depth info about it in a new great article by NSHipster .

Answer 3

An alternative would be use dictionaries instead of arrays, for better performance and easier implementation (in this case). You can easily get the array of keys and values later If you need.

In this case, the Id would be the Key of the dictionary, and the text the value.