I'm trying to figure out why the following works on the first string cluster (character) but not on a second one. Perhaps the endIndex cannot be applied on another String?
let part = "A"
let full = "ABC"
print(full[part.startIndex ... part.startIndex]) // "A"
print(full[part.endIndex ... part.endIndex]) // "" <- ???
print(full[part.endIndex ... full.index(after: part.endIndex)]) // "B"
bSecond should hold "B", but instead is empty. But the proof that one string index works on another is that the last statement works.
EDIT:
Assuming full.hasPrefix(part) is true.
Swift puzzles.
You cannot use the indices of one string to subscript a different
string. That may work by chance (in your first example) or not
(in your second example), or crash at runtime.
In this particular case, part.endIndex (which is the "one past the end position" for the part string) returns
String.UnicodeScalarView.Index(_position: 1), _countUTF16: 0)
with _countUTF16: (which is the "count of this extended grapheme cluster in UTF-16 code units") being zero, i.e. it describes
a position (in the unicode scalar view) with no extent. Then
full[part.endIndex ... part.endIndex]
returns an empty string. But that is an implementation detail
(compare StringCharacterView.swift). The real answer is just "you can't do that".
A safe way to obtain the intended (?) result is
let part = "A"
let full = "ABC"
if let range = full.range(of: part) {
print(full[range]) // A
if range.upperBound != full.endIndex {
print(full[range.upperBound...range.upperBound]) // B
}
}
Related
Situation: I was solving LeetCode 3. Longest Substring Without Repeating Characters, when I use the Dictionary using Swift the result is Time Limit Exceeded that failed to last test case, but using the same notion of code with C++ it acctually passed with runtime just fine. I thought in swift Dictionary is same thing as UnorderdMap.
Some research: I found some resources said use NSDictionary over regular one but it requires reference type instead of Int or Character etc.
Expected result: fast performance in accessing Dictionary in Swift
Question: I know there are better answer for the question, but the main goal here is Is there a effiencient to access and write to Dictionary or someting we can use to substitude.
func lengthOfLongestSubstring(_ s: String) -> Int {
var window:[Character:Int] = [:] //swift dictionary is kind of slow?
let array = Array(s)
var res = 0
var left = 0, right = 0
while right < s.count {
let rightChar = array[right]
right += 1
window[rightChar, default: 0] += 1
while window[rightChar]! > 1 {
let leftChar = array[left]
window[leftChar, default: 0] -= 1
left += 1
}
res = max(res, right - left)
}
return res
}
Because complexity of count in String is O(n), so that you should save count in a variable. You can read at chapter
Strings and Characters in Swift Book
Extended grapheme clusters can be composed of multiple Unicode scalars. This means that different characters—and different representations of the same character—can require different amounts of memory to store. Because of this, characters in Swift don’t each take up the same amount of memory within a string’s representation. As a result, the number of characters in a string can’t be calculated without iterating through the string to determine its extended grapheme cluster boundaries. If you are working with particularly long string values, be aware that the count property must iterate over the Unicode scalars in the entire string in order to determine the characters for that string.
The count of the characters returned by the count property isn’t always the same as the length property of an NSString that contains the same characters. The length of an NSString is based on the number of 16-bit code units within the string’s UTF-16 representation and not the number of Unicode extended grapheme clusters within the string.
I'm trying to work through a problem at the moment which is currently doing the rounds on the internet. The problem is: Given an array of characters, find the first non repeating character. I had a go at it and solved it but I was curious about how other people solved it so I did some looking around and found this answer:
let characters = ["P","Q","R","S","T","P","R","A","T","B","C","P","P","P","P","P","C","P","P","J"]
var counts: [String: Int] = [:]
for character in characters {
counts[character] = (counts[character] ?? 0) + 1
}
let nonRepeatingCharacters = characters.filter({counts[$0] == 1})
let firstNonRepeatingCharacter = nonRepeatingCharacters.first!
print(firstNonRepeatingCharacter) //"Q"
Source: Finding the first non-repeating character in a String using Swift
What I don't understand about this solution, is why it always returns Q, when there are other elements "S" "A" "B" and "J" that could be put first when the filter is applied to the dictionary. My understanding of dictionaries is that they are unordered, and when you make one they change from run to run. So if I make one:
let dictionary:[String:Int] = ["P": 9, "C": 8, "E": 1]
And then print 'dictionary', the ordering will be different. Given this, can anyone explain why the solution above works and maintains the order in which the dictionary elements were added?
You are not looking correctly at the code. The filter is not applied to a dictionary. It is applied to the array (characters), which has a defined order. The dictionary is used only to store counts.
I was playing with swift's Data in the following a small code:
var d = Data(count: 10)
d[5] = 3
let d2 = d[5..<8]
print("\(d2[0])")
To my surprise, this code throws exception on print() while the following code does not:
var d = Data(count: 10)
d[5] = 3
let d2 = d.subdata(in: 5..<8)
print("\(d2[0])")
I somehow understand why this happens, but I don't get why this is designed like this. When I use subdata() I get a whole copy of range, so indexing is valid from 0. But when I use range subscribe [], I get access to the requested range while indexing is the same as before. So in my first example d2[5] is 3.
But I wonder why it is designed like this? I don't want to make a copy of my data by using subdata() method. I just wanted to access a portion of my data with better indexing.
This is especially creates unexpected behaviors if you pass it to a function. For example, following code creates unexpected results and exceptions and you may not find out easily why:
func testit(idata: Data) {
if idata.count > 0 {
print("\(idata.count)")
print("\(idata[0])")
}
}
//...
var d = Data(count: 10)
d[5] = 3
let d2 = d[5..<8]
testit(idata: d2)
This code is really strange. Because if you debug your code, you see that print("\(idata.count)") prints 3 as size of idata which is correct, but accessing it with idata[0] creates exception.
Is there any reason for this design? I was expecting that I could access resulting Data from subscribe starting index 0 while it is not true. Can I do this without using subdata() which creates copy of data or using additional arguments to pass base of data slice?
d[5..<8] returns Data.Slice – which happens to be Data. Generally, slices share the indices with their base collection, as documented in Slice.
One possible reason for this design decision is that it guarantees that subscripting a slice is a O(1) operation (adding an offset for accessing the base collection is not necessarily O(1), e.g. not for strings.)
It is also convenient, as in this example to locate the text after the second occurrence of a character in a string:
let string = "abcdefgabcdefg"
// Find first occurrence of "d":
if let r1 = string.range(of: "d") {
// Find second occurrence of "d":
if let r2 = string[r1.upperBound...].range(of: "d") {
print(string[r2.upperBound...]) // efg
}
}
As a consequence, you must never assume that the indices of a collection are zero-based (unless documented, as for Array.startIndex). Use startIndex to get the first index, or first to get the first element.
Say I have the following code...
let x = "ABCDE"
// 'x' is a String
var y = x[1...3]
// 'y' is a Substring that equals "BCD"
If you only have access to y, is it possible to access x, or specifically parts of x which are outside the range of y? (i.e. can you access 'A' or 'E', or grow the range of y?)
So here's what Apple says:
Important
Don’t store substrings longer than you need them to perform a specific
operation. A substring holds a reference to the entire storage of the
string it comes from, not just to the portion it presents, even when
there is no other reference to the original string. Storing substrings
may, therefore, prolong the lifetime of string data that is no longer
otherwise accessible, which can appear to be memory leakage.
Now I find their use of the word "otherwise" in the last sentence rather interesting. It seems to me to keep the door open on this question - could a substring be manipulated to be expanded to include memory on either side that we know still exists as part of the original string?
So here's what I'd think is a fair test:
let x = "ABCDEFGH"
let substr = x.prefix(3)
var substrIndex = substr.startIndex
substr.formIndex(&substrIndex, offsetBy: 4) // offset beyond the substring
let prefix = substr.prefix(through:substrIndex)
print(prefix)
So what'cha think that would print?
Actually we never get to the print. We get a runtime fatal error instead.
Thread 1: Fatal error: Operation results in an invalid index
BTW, even trying the following results in an EXC_BAD_ACCESS crash:
let x = "ABCDEFGH"
var substr = x.prefix(3)
withUnsafePointer(to: &substr)
{ substrPointer in
let z = substrPointer.advanced(by: 3)
print(z.pointee)
}
So I don't think there's a way to get to the rest of the string if you just have a substring... from within Substring or String classes anyhow, or even dealing with unsafe pointers. I'm sure there's a way using direct memory access, for Apple claims the rest of the String's memory is there... but you'd probably have to fall back to C or C++.
From having following string: 12#17#15 I want to get an array with the numbers: [12, 17, 15].
I've tried following approach, but firstly I still get an error (Cannot convert value of type '[Double?]' to expected argument type 'Double?'), and obviously, I prefer to do it all on one map instead of such chain. Why do these types differ..? I'd say they should be matching...
let substrings = records?.split(separator: "#", maxSplits: Int.max, omittingEmptySubsequences: true).map(String.init).map(Double.init)
let objects = substrings.map {value in Model(value: value ?? 0)}
Unless there's some technique I've never heard of, you're not using map correctly.
Here's an example of the code you want:
let string = "12#17#15"
let objects = string.split(separator: "#").map {Double($0) ?? 0}
in Swift, map does something to every entry of an array, and then results in some sort of output. What's going on here is that first just doing a simple split (I'm going to assume you don't actually need the upper limit of an Int for the max results, but you can re-add that if you wish), and then initing a Double with each substring (which you call with $0). If trying to create that Double fails, then I'm coalescing it to a 0 instead of a nil.
If you don't want the Doubles that fail and return nil to be zero, then use flatmap {$0} instead
I would use flatMap instead of map, as Double init with String can return optional.
let records = "12#17#15"
let substrings = records.split(separator: "#").flatMap {Double($0)}
print(substrings) // [12.0, 17.0, 15.0]