I am trying use String(format:, ) for reading some characters from left or right, do we have something for this job?
for example reading 2 characters from left would be: "AB" like this: "%2L#"
my code:
let stringOfText = String(format: "%#", "ABCDEF")
String(format:) is usually to transform a different value type into a string.
Since you already have a string, you don't really need this method.
try:
https://developer.apple.com/documentation/swift/string/2894830-prefix
Related
I have the following string:
{\"Id\":\"135\",\"Type\":0}
The number in the Id field will vary, but will always be an integer with no comma separator. I'm not sure how to get just that value from that string given that it's string data type and not real "XML". I was toying with the replace() function, but the special characters are making it more complex than it seems it needs to be.
is there a way to convert that to XML or something that I can reference the Id value directly?
Maybe use a regular expression, e.g.
import re
txt = "{\"Id\":\"135\",\"Type\":0}"
x = re.search('"Id":"([0-9]+)"', txt)
if x:
print(x.group(1))
gives
135
It is assumed here that the ids are numeric and consist of at least one digit.
Non-regex answer as you asked
\" is an escape sequence in python.
So if {\"Id\":\"135\",\"Type\":0} is a raw string and if you put it into a python variable like
a = '{\"Id\":\"135\",\"Type\":0}'
gives
>>> a
'{"Id":"135","Type":0}'
OR
If the above string is python string which has \" which is already escaped, then do a.replace("\\","") which will give you the string without \.
Now just load this string into a dict and access element Id like below.
import json
d = json.loads(a)
d['Id']
Output :
135
I am using Swift 4.2. I am getting extraneous characters when formatting one string (s1) from another string(s0) using the %# format code.
I have searched extensively for details of string formatting but have come up with only partial answers including the code in the second line below. I need to be able to format s1 so that I can customize output from a Swift process. I ask this because I have not found an answer while searching for ways to format a string from a string.
I tried the following three statements:
let s0:[String] = ["abcdef"]
let s1:[String] = [String(format:"%#",s0)]
print(s1)
...
The output is shown below. It may not be clear, here, but there are four leading spaces to the left of the abcdef string.
["(\n abcdef\n)"]
How can I format s1 so it does not include the brackets, the \n escape characters, and the leading spaces?
The issue here is you are using an array but a string in s0.
so the following index will help you.
let s0:[String] = ["abcdef"]
let s1:[String] = [String(format:" %#",s0[0])]
I am getting extraneous characters when formatting one string (s1) from another string (s0) ...
The s0 is not a string. It is an array of strings (i.e. the square brackets of [String] indicate an array and is the same as saying Array<String>). And your s1 is also array, but one that that has one element, whose value is the string representation of the entire s0 array of strings. That’s obviously not what you intended.
How can I format s1 so it does not include the brackets, the \n escape characters, and the leading spaces?
You’re getting those brackets because s1 is an array. You’re getting the string with the \n and spaces because its first value is the string representation of yet another array, s0.
So, if you’re just trying to format a string, s0, you can do:
let s0: String = "abcdef"
let s1: String = String(format: "It is ‘%#’", s0)
Or, if you really want an array of strings, you can call String(format:) for each using the map function:
let s0: [String] = ["abcdef", "ghijkl"]
let s1: [String] = s0.map { String(format: "It is ‘%#’", $0) }
By the way, in the examples above, I didn’t use a string format of just %#, because that doesn’t accomplish anything at all, so I assumed you were formatting the string for a reason.
FWIW, we generally don’t use String(format:) very often. Usually we do “string interpolation”, with \( and ):
let s0: String = "abcdef"
let s1: String = "It is ‘\(s0)’"
Get rid of all the unneccessary arrays and let the compiler figure out the types:
let s0 = "abcdef" // a string
let s1 = String(format:"- %# -",s0) // another string
print(s1) // prints "- abcdef -"
In sales column i have values with pound sign £1200. It is not readable by Data frame in scala, please help me for the same. i want column value in double, 1200. I am using below method but its not working.
def getRemovedDollarValue = udf(
(actualSales: String) => {
val actualSalesDouble = actualSales
.replace(",", "")
.replace("$", "")
.replace("\\u00A3","")
.replace("\\U00A3","")
.replaceAll("\\s", "_").trim().toDouble
java.lang.Double.parseDouble(actualSalesDouble.toString)
}
)
You need write: .replace("\u00A3","") instead of escaping .replace("\\u00A3","").
But I prefer just: .replace("£", "") - it is more readable.
I think the proposed solutions and comments all work but don't address the confusion behind why your code isn't working.
From the Pattern docs:
Thus the strings "\u2014" and "\\u2014", while not equal, compile into the same pattern, which matches the character with hexadecimal value 0x2014.
replace and replaceAll are both replacing all occurrences in a String, but only replaceAll is taking in a regular expression. You're passing in "\\u00A3" which will work as a pattern, but not a unicode literal due to the added backslash. As already suggested, either use replace with a unicode literal or the actual symbol, or change to replaceAll.
All I want to do is convert a single Character to uppercase without the overhead of converting to a String and then calling .uppercased(). Is there any built-in way to do this, or a way for me to call the toupper() function from C without any bridging? I really don't think I should have to go out of my way for something so simple.
To call the C toupper() you need to get the Unicode code point of the Character. But Character has no method for getting its code point (a Character may consist of multiple code points), so you have to convert the Character into a String to obtain any of its code points.
So you really have to convert to String to get anywhere. Unless you store the character as a UnicodeScalar instead of a Character. In this case you can do this:
assert(unicodeScalar.isASCII) // toupper argument must be "representable as an unsigned char"
let uppercase = UnicodeScalar(toupper(CInt(unicodeScalar.value)))
But this isn't really more readable than simply using String:
let uppercase = Character(String(character).uppercased())
just add this to your program
extension Character {
//converts a character to uppercase
func convertToUpperCase() -> Character {
if(self.isUppercase){
return self
}
return Character(self.uppercased())
}
}
I am creating an iPhone app and I need to convert a single digit number into an integer.
My code has a variable called char that has a type Character, but I need to be able to do math with it, therefore I think I need to convert it to a string, however I cannot find a way to do that.
In the latest Swift versions (at least in Swift 5) there is a more straighforward way of converting Character instances. Character has property wholeNumberValue which tries to convert a character to Int and returns nil if the character does not represent and integer.
let char: Character = "5"
if let intValue = char.wholeNumberValue {
print("Value is \(intValue)")
} else {
print("Not an integer")
}
With a Character you can create a String. And with a String you can create an Int.
let char: Character = "1"
if let number = Int(String(char)) {
// use number
}
The String middleman type conversion isn’t necessary if you use the unicodeScalars property of Swift 4.0’s Character type.
let myChar: Character = "3"
myChar.unicodeScalars.first!.value - Unicode.Scalar("0")!.value // 3: UInt32
This uses a trick commonly seen in C code of subtracting the value of the char ’0’ literal to convert from ascii values to decimal values. See this site for the conversions: https://www.asciitable.com
Also there are some implicit unwraps in my answer. To avoid those, you can validate that you have a decimal digit with CharacterSet.decimalDigits, and/or use guard lets around the first property. You can also subtract 48 directly rather than converting ”0” through Unicode.Scalar.