How to count a character occurrence in string in Nim, mainly using its native statements prior go to module ? eg.
var
str = "Hello World"
c : int
c = numChar( "o", str ) # <- illustration only ?
The earlier answer is correct but if you do not want to import any modules you can write your own procedure:
proc count_char(value: string = "Hello World", ch: char = 'o'): int =
var cnt_c: int = 0
for c in value:
if c == ch:
cnt_c += 1
result = cnt_c
var
val: string = "Mother Goose"
ch: char = 'o'
echo $count_char(val, ch)
PS: Unrelated - Need syntax highlight for nim-lang on SO.
Use the count function from strutils:
import std/strutils
let str = "Hello World"
let count = count(str, 'o')
assert count = 1
There’s also a string overload for counting sub strings as well.
So I got stuck on a coding challenge that I almost knew the answer too. And I think I have to use the subString call in Swift 4 to get it 100%. I want to reverse every OTHER word in a string, but ignore or keep the punctuation in its original place( index ).
var sample = "lets start. And not worry about proper sentences."
func reverseString(inputString: String) -> String {
let oldSentence = sample.components(separatedBy: " ")
var newSentence = ""
for index in 0...oldSentence.count - 1 {
let word = oldSentence[index]
if newSentence != "" {
newSentence += " "
}
if index % 2 == 1 {
let reverseWord = String(word.reversed())
newSentence += reverseWord
} else {
newSentence += word
}
}
return newSentence
}
reverseString(inputString: sample)
And this would be the expected output.
"lets trats. And ton worry tuoba proper secnetnes."
Notice the punctuation is not reversed.
You shouldn't use components(separatedBy: ) to split a string in words. See this article for the reason. Use enumerateSubstrings and pass in the appropriate option:
func reverseString(inputString: String) -> String {
var index = 1
var newSentence = inputString
inputString.enumerateSubstrings(in: inputString.startIndex..., options: .byWords) { substr, range, _, stop in
guard let substr = substr else { return }
if index % 2 == 0 {
newSentence = newSentence.replacingCharacters(in: range, with: String(substr.reversed()))
}
index += 1
}
return newSentence
}
print(reverseString(inputString: "lets start. And not worry about proper sentences."))
// lets trats. And ton worry tuoba proper secnetnes.
print(reverseString(inputString: "I think, therefore I'm"))
// I kniht, therefore m'I
let amount = "73.45"
I want four different constants (strings, not characters) with each number in this string. Ideally:
let amount1 = amount[0] // 7
let amount2 = amount[1] // 3
let amount3 = amount[3] // 4
let amount4 = amount[4] // 5
I have searched and can't find anything that works, I either get the whole string or a character of the string. Any advice would be helpful--new to xcode and swift
You can always get the characters using
let characters = amount.characters
to get strings instead of characters, you can:
let amount1 = String(characters[0])
to do it for all digits
let amounts = amount.characters.map {
return String($0)
}
To filter out the separator, you can
let amounts = amount.characters.map {
return String($0)
}.filter {
$0 != "."
}
Note that if you have the input number localized, you should check NSLocale for the correct decimal separator or just remove all non-numeric characters. One way to do that is using:
let amounts = amount.characters.filter {
$0 >= "0" && $0 <= "9"
}.map {
String($0)
}
You can put your digits into separate variables then but I would advise against it:
let amount1 = amounts[0]
let amount2 = amounts[1]
let amount3 = amounts[2]
let amount4 = amounts[3]
I need to extract numbers from string and put them into a new array in Swift.
var str = "I have to buy 3 apples, 7 bananas, 10eggs"
I tried to loop each characters and I have no idea to compare between Characters and Int.
Swift 3/4
let string = "0kaksd020dk2kfj2123"
if let number = Int(string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()) {
// Do something with this number
}
You can also make an extension like:
extension Int {
static func parse(from string: String) -> Int? {
return Int(string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined())
}
}
And then later use it like:
if let number = Int.parse(from: "0kaksd020dk2kfj2123") {
// Do something with this number
}
First, we split the string so we can process the single items. Then we use NSCharacterSet to select the numbers only.
import Foundation
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let strArr = str.split(separator: " ")
for item in strArr {
let part = item.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
if let intVal = Int(part) {
print("this is a number -> \(intVal)")
}
}
Swift 4:
let string = "I have to buy 3 apples, 7 bananas, 10eggs"
let stringArray = string.components(separatedBy: CharacterSet.decimalDigits.inverted)
for item in stringArray {
if let number = Int(item) {
print("number: \(number)")
}
}
Using the "regex helper function" from Swift extract regex matches:
func matchesForRegexInText(regex: String!, text: String!) -> [String] {
let regex = NSRegularExpression(pattern: regex,
options: nil, error: nil)!
let nsString = text as NSString
let results = regex.matchesInString(text,
options: nil, range: NSMakeRange(0, nsString.length))
as! [NSTextCheckingResult]
return map(results) { nsString.substringWithRange($0.range)}
}
you can achieve that easily with
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let numbersAsStrings = matchesForRegexInText("\\d+", str) // [String]
let numbersAsInts = numbersAsStrings.map { $0.toInt()! } // [Int]
println(numbersAsInts) // [3, 7, 10]
The pattern "\d+" matches one or more decimal digit.
Of course the same can be done without the use of a helper function
if you prefer that for whatever reason:
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let regex = NSRegularExpression(pattern: "\\d+", options: nil, error: nil)!
let nsString = str as NSString
let results = regex.matchesInString(str, options: nil, range: NSMakeRange(0, nsString.length))
as! [NSTextCheckingResult]
let numbers = map(results) { nsString.substringWithRange($0.range).toInt()! }
println(numbers) // [3, 7, 10]
Alternative solution without regular expressions:
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let digits = "0123456789"
let numbers = split(str, allowEmptySlices: false) { !contains(digits, $0) }
.map { $0.toInt()! }
println(numbers) // [3, 7, 10]
let str = "Hello 1, World 62"
let intString = str.componentsSeparatedByCharactersInSet(
NSCharacterSet
.decimalDigitCharacterSet()
.invertedSet)
.joinWithSeparator("")
That will get you a string with all the number then you can just do this:
let int = Int(intString)
Just make sure you unwrap it since let int = Int(intString) is an optional.
For me makes more sense to have it as a String extension, probably it's a matter of tastes:
extension String {
func parseToInt() -> Int? {
return Int(self.components(separatedBy: CharacterSet.decimalDigits.inverted).joined())
}
}
So can be used like this:
if let number = "0kaksd020dk2kfj2123".parseToInt() {
// Do something with this number
}
Adapting from #flashadvanced's answer,
I found that the following is shorter and simpler for me.
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let component = str.componentsSeparatedByCharactersInSet(NSCharacterSet.decimalDigitCharacterSet().invertedSet)
let list = component.filter({ $0 != "" }) // filter out all the empty strings in the component
print(list)
Tried in in the play ground and it works
Hope it helps :)
Swift 2.2
let strArr = str.characters.split{$0 == " "}.map(String.init)
for item in strArr {
let components = item.componentsSeparatedByCharactersInSet(NSCharacterSet.decimalDigitCharacterSet().invertedSet)
let part = components.joinWithSeparator("")
if let intVal = Int(part) {
print("this is a number -> \(intVal)")
}
}
// This will only work with single digit numbers. Works with “10eggs” (no space between number and word
var str = "I have to buy 3 apples, 7 bananas, 10eggs"
var ints: [Int] = []
for char:Character in str {
if let int = "\(char)".toInt(){
ints.append(int)
}
}
The trick here is that you can check if a string is an integer (but you can’t check if a character is).
By looping though every character of the string, use string interpolation to create a string from the character and check if that string cas be casted as a integer.
If it can be, add it to the array.
// This will work with multi digit numbers. Does NOT work with “10 eggs” (has to have a space between number and word)
var str = "I have to buy 3 apples, 7 bananas, 10 eggs"
var ints: [Int] = []
var strArray = split(str) {$0 == " "}
for subString in strArray{
if let int = subString.toInt(){
ints.append(int)
}
}
Here we split the string at any space and create an array of every substring that is in the long string.
We again check every string to see if it is (or can be casted as) an integer.
Thanks for everyone who answered to my question.
I was looking for a block of code which uses only swift grammar, because I'm learning grammar only now..
I got an answer for my question.Maybe it is not an easier way to solve, but it uses only swift language.
var article = "I have to buy 3 apples, 7 bananas, 10 eggs"
var charArray = Array(article)
var unitValue = 0
var total = 0
for char in charArray.reverse() {
if let number = "\(char)".toInt() {
if unitValue==0 {
unitValue = 1
}
else {
unitValue *= 10
}
total += number*unitValue
}
else {
unitValue = 0
}
}
println("I bought \(total) apples.")
Swift 5:
extension String {
var allNumbers: [Int] {
let numbersInString = self.components(separatedBy: .decimalDigits.inverted).filter { !$0.isEmpty }
return numbersInString.compactMap { Int($0) }
}
}
You can get all numbers like
var str = "I have to buy 3 apples, 7 bananas, 10eggs"
// numbers = [3, 7, 10]
numbers = str.allNumbers
That is what i am trying to do:
var i = 0
var string = "abcdef"
for value in string
{
value.[Put value of variable i here] = "a"
i++
}
How can i insert the value of i in the code?
Easiest is probably just convert it to an NSMutableString:
let string = "abcdef".mutableCopy() as NSMutableString
println( "\(string)")
for var i = 0; i < string.length; ++i {
string.replaceCharactersInRange(NSMakeRange(i, 1), withString: "a")
}
println( "\(string)")
Yes, it's a bit ugly but it works.
A much cleaner way is to use Swifts map function:
var string = "abcdef"
let result = map(string) { (c) -> Character in
"a"
}
println("\(result)") // aaaaaa
You should just be able to use the following but this doesn't compile:
map(string) { "a" }
In you comments you mention you want to split up the string on a space, you can just use this for that:
let stringWithSpace = "abcdef 012345"
let splitString = stringWithSpace.componentsSeparatedByString(" ")
println("\(splitString[0])") // abcdef
println("\(splitString[1])") // 012345