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
Related
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
I have the following Powershell variable
$var = "AB-0045"
I would like to increase the number in the string to become "AB-0046".
I can do:
$newNumber = [int]$var.Substring($var.length -4,4) + 1
Which will give me the desired number 46, but then I have to append that 46 as a string to a new string "AB-00".
Is there a better way to do that?
Now that you have the integer, you'll have to convert back to string formatted in the way you'd like and concatenate.
I'd recommend adding to "AB-" rather than "AB-00" in case your number goes over 100.
To pad leading zeros, you can use the -f operator.
e.g. "{0:d4}" -f 45
You'll still need to get the integer first (45 in the example) from your original string.
I tested with regex class Replace() method and string class Split() method with string formatter. Split() seems faster provided your string is always in the same format. The Replace() method does not care what happens before the last 4 numbers:
# Replace Method
[regex]::Replace($var,'\d{4}$',{([int]$args[0].Value+1).ToString('0000')})
# Split method
$a,[int]$b = $var.split('-'); "{0}-{1:0000}" -f $a,++$b
I have a spark scala dataframe which has column "Name"
I have extracted the values of that column in to scala array[string]
org_name: Array[String] = Array(SARATOGA SENIOR HIGH SCHOOL)
I want to replace whitespaces with _ and encode that value in to utf-8 (any encoding is fine as long as it replaces special chars with something else)
so if there are any special chars those will be removed. later i want to use those in file path .
var org_name = orgsFlatDF.rdd.collect
.map( _.getString(2))
This is how i am extracting those vals ^^. I haven't found any method which I can use to do that. Replace or replaceall doesn't work on array
I tried this :
org_name.replace("\\s", "")
That didn't work .
Expected output : SARATOGA_SENIOR_HIGH_SCHOOL
if name is : new $ high school it should gets converted to new_$_high_school then encoded to new_%24_high_school
There are a couple of issues with what you are asking.
Java/Scala Arrays don't have a replace method. Even if they did have a replace method, would they replace the values they hold or the characters in a String they hold?
Let's assume this line org_name.replace("\\s", "") didn't compiled and org_name is indeed a an Array[String] holding one element.
scala> val org_name=Array("SARATOGA SENIOR HIGH SCHOOL")
val org_name: Array[String] = Array(SARATOGA SENIOR HIGH SCHOOL)
scala> org_name(0).replace(" ","_")
val res15: String = SARATOGA_SENIOR_HIGH_SCHOOL
replace("\\s","_") wouldn't work because it represents a \s string. "\" represents \. That's only way you'd be able to define strings containing other escape codes like \n or \t.
PS: to transform all the string in the array use org_name.map(_.replace(" ","_")), this gives you back another another array.
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 -"
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())
}
}