This question already has answers here:
Any way to replace characters on Swift String?
(23 answers)
Closed 7 years ago.
I'm trying to create a simple iOS app that takes user input ( a city ) and searches a website for that city, and then will display the forecasts for that city.
What I'm currently stuck on and unable to find much documentation that isn't overwhelming is how I can be sure that the user input will translate well to a URL if there are more then one words in the name of the city.
aka if a user inputs Salt Lake City into my text field, how can I write an if else statement that determines the amount of spaces, and if the amount of spaces is greater than 0 will convert those spaces to "-".
So far I've tried creating an array out of the string, but still can't figure out how I can append a - to each element in the array. I don't think it's possible.
Does anyone know how I can do what I'm trying to do? Or am I approaching it the incorrect way?
Here's a poor first attempt. I know this doesn't work, but hopefully it explains it a bit more of what I'm trying to accomplish than my text above.
var cityText = "Salt Lake City"
let cityArray = cityText.componentsSeparatedByString(" ")
let combineDashUrl = cityArray[0] + "-" + cityArray[1] + "-" + cityArray[2]
print(combineDashUrl)
Assuming there are never multiple spaces in a row you should be able to use stringByReplacingOccurrencesOfString.
let cityText = "Salt Lake City"
let newCityText = cityText.stringByReplacingOccurrencesOfString(
" ",
withString: "-")
Replacing variable numbers of spaces with a dash would be more complicated. I'd probably use regular expressions for that.
You can use map over the array of characters to transform spaces into hyphens.
let city = "Salt Lake City"
let hyphenatedCity = String(city.characters.map{$0 == " " ? "-" : $0})
Related
I'm trying to identify the longest three substrings from a string using SPARQL and the Wikidata Query Service and then rank
the substrings within a string by length
the strings by the lengths of any of those longest substrings .
I managed to identify the first and second substring from a string and could of course just create similar additional lines to tackle the problem, but this seems ugly and inefficient, so I am wondering if anyone here knows of a better way to get there.
This is a simplified version of the code, though I have left some auxiliary variables in that I am using for tracking progress on the way. You can try it here.
Clarification in response to this comment: if it is necessary to treat this query as a subquery and to feed it with results from another subquery, that's fine with me. To get an idea of the kinds of use I have in mind, see this demo.
SELECT * WHERE {
{
VALUES (?title) {
("What are the longest three words in this string?")
("A really complicated title")
("OneWordTitleInCamelCase")
("Thanks for your help!")
}
}
BIND(STRLEN(REPLACE(?title, " ", "")) AS ?titlelength)
BIND(STRBEFORE(?title, " ") AS ?substring1)
BIND(STRLEN(REPLACE(?substring1, " ", "")) AS ?substring1length)
BIND(STRAFTER(?title, " ") AS ?postfix)
BIND(STRLEN(REPLACE(?postfix, " ", "")) AS ?postfixlength)
BIND(STRBEFORE(?postfix, " ") AS ?substring2)
BIND(STRLEN(REPLACE(?substring2, " ", "")) AS ?substring2length)
}
ORDER BY DESC(?substring1length)
Expected results:
longsubstring substringlength
OneWordTitleInCamelCase 23
complicated 11
longest 7
really 6
string 6
Thanks 6
title 5
three 5
your 4
help 4
Actual results:
title titlelength substring1 substring1length postfix postfixlength substring2 substring2length
Thanks for your help! 18 Thanks 6 for your help! 12 for 3
What are the longest three words in this string? 40 What 4 are the longest three words in this string? 36 are 3
A really complicated title 23 A 1 really complicated title 22 really 6
OneWordTitleInCamelCase 23 0 0 0
looking for a push in the right direction on a minor non problem but more curiosity driven search.
I'm trying to take a ton of text which has been "encrypted" with a plain as day key using uppercase, lowercase & numbers.
ie.
Array('1'=>'h', '0'=>'L', '3'=>'H',....
Stumbling around my brain trying to think if there was a way to build a dictionary with the value / key as has ben provided can I enter the encrypted text and reference the dictionary for the answer to output decrypted text?
Assuming the mapping is 1:1, ie, 1 character maps to 1 character, ie no numbers are greater than 9. This should work:
let cypher = ["1": "h",
"0": "L"] as [Character: Character]
//Add more here as needed.
let yourText = "014"
let decypheredText = yourText.map { char in
return cypher[char] ?? "?" //Untranslatable things mpa to ?
}.joined()
This question already has answers here:
Number of words in a Swift String for word count calculation
(7 answers)
Closed 5 years ago.
Edit: there is already a question similar to this one but it's for numbers separated by a specific character (Get no. Of words in swift for average calculator). Instead this question is about to get the number of real words in a text, separated in various ways: a line break, some line breaks, a space, more than a space etc.
I would like to get the number of words in a string with Swift 3.
I'm using this code but I get imprecise result because the number is get counting the spaces and new lines instead of the effective number of words.
let str = "Architects and city planners,are \ndesigning buildings to create a better quality of life in our urban areas."
// 18 words, 21 spaces, 2 lines
let components = str.components(separatedBy: .whitespacesAndNewlines)
let a = components.count
print(a)
// 23 instead of 18
Consecutive spaces and newlines aren't coalesced into one generic whitespace region, so you're simply getting a bunch of empty "words" between successive whitespace characters. Get rid of this by filtering out empty strings:
let components = str.components(separatedBy: .whitespacesAndNewlines)
let words = components.filter { !$0.isEmpty }
print(words.count) // 17
The above will print 17 because you haven't included , as a separation character, so the string "planners,are" is treated as one word.
You can break that string up as well by adding punctuation characters to the set of separators like so:
let chararacterSet = CharacterSet.whitespacesAndNewlines.union(.punctuationCharacters)
let components = str.components(separatedBy: chararacterSet)
let words = components.filter { !$0.isEmpty }
print(words.count) // 18
Now you'll see a count of 18 like you expect.
This question already has answers here:
Any way to replace characters on Swift String?
(23 answers)
Closed 5 years ago.
I have a string var m = "I random don't like confusing random code." I want to delete all instances of the substring random within string m, returning string parsed with the deletions completed.
The end result would be: parsed = "I don't like confusing code."
How would I go about doing this in Swift 3.0+?
It is quite simple enough, there is one of many ways where you can replace the string "random" with empty string
let parsed = m.replacingOccurrences(of: "random", with: "")
Depend on how complex you want the replacement to be (remove/keep punctuation marks after random). If you want to remove random and optionally the space behind it:
var m = "I random don't like confusing random code."
m = m.replacingOccurrences(of: "random ?", with: "", options: [.caseInsensitive, .regularExpression])
I have a word that is being displayed into a label. Could I program it, where it will only show the last 2 characters of the word, or the the first 3 only? How can I do this?
Swift's string APIs can be a little confusing. You get access to the characters of a string via its characters property, on which you can then use prefix() or suffix() to get the substring you want. That subset of characters needs to be converted back to a String:
let str = "Hello, world!"
// first three characters:
let prefixSubstring = String(str.characters.prefix(3))
// last two characters:
let suffixSubstring = String(str.characters.suffix(2))
I agree it is definitely confusing working with String indexing in Swift and they have changed a little bit from Swift 1 to 2 making googling a bit of a challenge but it can actually be quite simple once you get a hang of the methods. You basically need to make it into a two-step process:
1) Find the index you need
2) Advance from there
For example:
let sampleString = "HelloWorld"
let lastThreeindex = sampleString.endIndex.advancedBy(-3)
sampleString.substringFromIndex(lastThreeindex) //prints rld
let secondIndex = sampleString.startIndex.advancedBy(2)
sampleString.substringToIndex(secondIndex) //prints He