I want to get upper"T"..
how to get upper string!
str = "Test Version"
print(str.upper())
print(str[3])
It's not clear what you are asking.
But from context I am guessing you would like to make the second non-capitalised "t" in the string uppercase. I'm also going to assume you are using python 3 given your use of upper().
If you just want to get the "t" (and not change the string itself):
upper_T = str[3].upper()
If you want to create a string from the original you may be running into the fact that strings in python are immutable. You therefore must create a new string.
One way do this:
str2 = list(str)
str2[3] = str[3].upper()
str2 = ''.join(str2)
Related
Background
I have the following string:
var MyString = 'Test1⏎Test2⏎Test3⏎Test4'
⏎ = line feed = \n
What I'm trying to do
I want to create a List which is a list of lines. Basically every item that is followed by a \n would become an entry in the list.
I want the base string MyString to become shortened to reflect what pieces of the string have been moved to the List
The reason I want to leave a residual MyString is that new data might come in later that might be considered part of the same line, so I do not want to commit the data to the List until there is a carriage return seen
What the result of all this would be
So in my above example, only Test1 Test2 Test3 are followed by \n but not Test4
Output List would be: [Test1, Test2, Test3]
MyString would become: Test4
What I've tried and failed with
I tried using LineSplitter but it seems to want to take Test4 as a separate entry as well
final lines = const LineSplitter().convert(MyString);
for (final daLine in lines) {
MyList.add(daLine);
}
And it creates [Test1, Test2, Test3, Test4]
A solution would be to just .removeLast() on the list that you split.
String text = 'Test1\nTest2\nTest3\nTest4';
List<String> list = text.split('\n');
text = list.removeLast();
print(list); // [Test1, Test2, Test3]
print(text); // Test4
To me you are combining two questions. Every language I know has built-in ways to split a string on a char, including newline chars. The distinct thing you want is a split function that doesn't include the last entry.
You may be combining your answers as well :) Is there some resource constraint or streamed input that prevents you from just building the list, then popping off the final entry?
If yes:
I think you have to build your own split. Look at the implementation code for LineSplitter(), and make something similar except which leaves the final entry.
If no:
simply call
MyString = MyList.removeLast();
after your for-loop.
I'm using AutoHotkey for this as the code is the most understandable to me. So I have a document with numbers and text, for example like this
120344 text text text
234000 text text
and the desired output is
12:03:44 text text text
23:40:00 text text
I'm sure StrReplace can be used to insert the colons in, but I'm not sure how to specify the position of the colons or ask AHK to 'find' specific strings of 6 digit numbers. Before, I would have highlighted the text I want to apply StrReplace to and then press a hotkey, but I was wondering if there is a more efficient way to do this that doesn't need my interaction. Even just pointing to the relevant functions I would need to look into to do this would be helpful! Thanks so much, I'm still very new to programming.
hfontanez's answer was very helpful in figuring out that for this problem, I had to use a loop and substring function. I'm sure there are much less messy ways to write this code, but this is the final version of what worked for my purposes:
Loop, read, C:\[location of input file]
{
{ If A_LoopReadLine = ;
Continue ; this part is to ignore the blank lines in the file
}
{
one := A_LoopReadLine
x := SubStr(one, 1, 2)
y := SubStr(one, 3, 2)
z := SubStr(one, 5)
two := x . ":" . y . ":" . z
FileAppend, %two%`r`n, C:\[location of output file]
}
}
return
Assuming that the "timestamp" component is always 6 characters long and always at the beginning of the string, this solution should work just fine.
String test = "012345 test test test";
test = test.substring(0, 2) + ":" + test.substring(2, 4) + ":" + test.substring(4, test.length());
This outputs 01:23:45 test test test
Why? Because you are temporarily creating a String object that it's two characters long and then you insert the colon before taking the next pair. Lastly, you append the rest of the String and assign it to whichever String variable you want. Remember, the substring method doesn't modify the String object you are calling the method on. This method returns a "new" String object. Therefore, the variable test is unmodified until the assignment operation kicks in at the end.
Alternatively, you can use a StringBuilder and append each component like this:
StringBuilder sbuff = new StringBuilder();
sbuff.append(test.substring(0,2));
sbuff.append(":");
sbuff.append(test.substring(2,4));
sbuff.append(":");
sbuff.append(test.substring(4,test.length()));
test = sbuff.toString();
You could also use a "fancy" loop to do this, but I think for something this simple, looping is just overkill. Oh, I almost forgot, this should work with both of your test strings because after the last colon insert, the code takes the substring from index position 4 all the way to the end of the string indiscriminately.
It used to be you could use substring to get a portion of a string. That has been deprecated in favor on string index. But I can't seem to make a string index out of integers.
var str = "hellooo"
let newindex = str.index(after: 3)
str = str[newindex...str.endIndex]
No matter what the string is, I want the second 3 characters. So and str would contain "loo". How can I do this?
Drop the first three characters and the get the remaining first three characters
let str = "helloo"
let secondThreeCharacters = String(str.dropFirst(3).prefix(3))
You might add some code to handle the case if there are less than 6 characters in the string
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 -"
I'm trying to remove the last numbers of an IP address string in Swift so I can loop through IP addresses. For instance if my variable = 192.168.1.123, I would like to trim the string to equal 192.169.1.
I'm not sure how to do this since some IP addresses will end in 1, 2 or 3 digits. I couldn't figure out how to trim back to a certain character.
I have a solution (In your case only). You can try it
let str = "192.168.1.123"
var arr = str.components(separatedBy: ".")
arr.removeLast()
let newstr = arr.joined(separator: ".") + "."
You can find the range of the last .:
let ip = "192.168.1.123"
let lastdot = ip.range(of: ".", options: .backwards)!
let base = ip[...lastdot.lowerBound]
This code assumes there is at least one . in the string. If not it will crash. That is easily fixed with proper use of if let.
base will be a Substring so depending on what you do next, you may need to wrap that as:
let base = String(ip[...lastdot.lowerBound])
Whether explicitly converting to String depends on whether subsequent methods require String or StringProtocol. Converting to String copies over the storage again, which is costly and unnecessary for many operations, but may be required in some cases.