How would I return the last two characters of a string?
Scala allows you to do this in a much cleaner way than the standard String API by leveraging the collections API (for which there is an implicit conversion from a java.lang.String into an IndexedSeq[Char]):
str takeRight 2
The fantastic thing about the API of course, is that it preserves the type representation of the original "collection" (i.e. String in this case)!
you can use
.takeRight(2)
var keyword="helloStackoverFlow"
println(keyword.takeRight(2)) // ow
You can take (string length -1) that reveal last index of your string, (string length -2) will be next character from end:
str(str.length-1)+str(str.length-1)
Related
I am a beginner in programming. I have a string for example "test:1" and "test:2". And I want to remove ":1" and ":2" (including :). How can I do it using regular expression?
Hi andrew it's pretty easy. Think of a string as if it is an array of chars (letters) cause it actually IS. If the part of the string you want to delete is allways at the end of the string and allways the same length it goes like this:
var exampleString = 'test:1';
exampleString.length -= 2;
Thats it you just deleted the last two values(letters) of the string(charArray)
If you cant be shure it's allways at the end or the amount of chars to delete you'd to use the version of szymon
There are at least a few ways to do it with Groovy. If you want to stick to regular expression, you can apply expression ^([^:]+) (which means all characters from the beginning of the string until reaching :) to a StringGroovyMethods.find(regexp) method, e.g.
def str = "test:1".find(/^([^:]+)/)
assert str == 'test'
Alternatively you can use good old String.split(String delimiter) method:
def str = "test:1".split(':')[0]
assert str == 'test'
Can I convert directly between a Swift Character and its Unicode numeric value? That is:
var i:Int = ... // A plain integer index.
var myCodeUnit:UInt16 = myString.utf16[i]
// Would like to say myChar = myCodeUnit as Character, or equivalent.
or...
var j:String.Index = ... // NOT an integer!
var myChar:Character = myString[j]
// Would like to say myCodeUnit = myChar as UInt16
I can say:
myCodeUnit = String(myChar).utf16[0]
but this means creating a new String for each character. And I am doing this thousands of times (parsing text) so that is a lot of new Strings that are immediately being discarded.
The type Character represents a "Unicode grapheme cluster", which can be multiple Unicode codepoints. If you want one Unicode codepoint, you should use the type UnicodeScalar instead.
As per the swift book:
String to Code Unit
To get codeunit/ordinals for each character of the String, you can do the following:
var yourSwiftString = "甲乙丙丁"
for scalar in yourSwiftString.unicodeScalars {
print("\(scalar.value) ")
}
Code Unit to String
Because swift current does not have a way to convert ordinals/code units back to UTF, the best way I found is to still NSString. i.e. if you have int ordinals (32bit but representing the 21bit codepoints) you can use the following to convert to Unicode:
var i = 22247
var unicode_str = NSString(bytes: &i, length: 4, encoding: NSUTF32LittleEndianStringEncoding)
Obviously if you want to convert a array of ints, you'll need to pack them into a array first.
I spoke to an Apple engineer who is working on Unicode and he says they have not completed the implementation of unicode characters in strings. Are you looking at getting a code unit or a full character? Because the only and proper way to get at a full unicode character is by using a for each loop on a string. ie
for c in "hello" {
// c is a unicode character of type Character
}
But, this is not implemented as of yet.
This question already has answers here:
Is there an easy way to return a string repeated X number of times?
(21 answers)
Closed 9 years ago.
Just a curiosity I was investigating.
The matter: simply repeating (multiplying, someone would say) a string/character n times.
I know there is Enumerable.Repeat for this aim, but I was trying to do this without it.
LINQ in this case seems pretty useless, because in a query like
from X in "s" select X
the string "s" is being explored and so X is a char. The same is with extension methods, because for example "s".Aggregate(blablabla) would again work on just the character 's', not the string itself. For repeating the string something "external" would be needed, so I thought lambdas and delegates, but it can't be done without declaring a variable to assign the delegate/lambda expression to.
So something like defining a function and calling it inline:
( (a)=>{return " "+a;} )("a");
or
delegate(string a){return " "+a}(" ");
would give a "without name" error (and so no recursion, AFAIK, even by passing a possible lambda/delegate as a parameter), and in the end couldn't even be created by C# because of its limitations.
It could be that I'm watching this thing from the wrong perspective. Any ideas?
This is just an experiment, I don't care about performances, about memory use... Just that it is one line and sort of autonomous. Maybe one could do something with Copy/CopyTo, or casting it to some other collection, I don't know. Reflection is accepted too.
To repeat a character n-times you would not use Enumerable.Repeat but just this string constructor:
string str = new string('X', 10);
To repeat a string i don't know anything better than using string.Join and Enumerable.Repeat
string foo = "Foo";
string str = string.Join("", Enumerable.Repeat(foo, 10));
edit: you could use string.Concat instead if you need no separator:
string str = string.Concat( Enumerable.Repeat(foo, 10) );
If you're trying to repeat a string, rather than a character, a simple way would be to use the StringBuilder.Insert method, which takes an insertion index and a count for the number of repetitions to use:
var sb = new StringBuilder();
sb.Insert(0, "hi!", 5);
Console.WriteLine(sb.ToString());
Otherwise, to repeat a single character, use the string constructor as I've mentioned in the comments for the similar question here. For example:
string result = new String('-', 5); // -----
For the sake of completeness, it's worth noting that StringBuilder provides an overloaded Append method that can repeat a character, but has no such overload for strings (which is where the Insert method comes in). I would prefer the string constructor to the StringBuilder if that's all I was interested in doing. However, if I was already working with a StringBuilder, it might make sense to use the Append method to benefit from some chaining. Here's a contrived example to demonstrate:
var sb = new StringBuilder("This item is ");
sb.Insert(sb.Length, "very ", 2) // insert at the end to append
.Append('*', 3)
.Append("special")
.Append('*', 3);
Console.WriteLine(sb.ToString()); // This item is very very ***special***
Is there any way to replace a character at position N in a string in Lua.
This is what I've come up with so far:
function replace_char(pos, str, r)
return str:sub(pos, pos - 1) .. r .. str:sub(pos + 1, str:len())
end
str = replace_char(2, "aaaaaa", "X")
print(str)
I can't use gsub either as that would replace every capture, not just the capture at position N.
Strings in Lua are immutable. That means, that any solution that replaces text in a string must end up constructing a new string with the desired content. For the specific case of replacing a single character with some other content, you will need to split the original string into a prefix part and a postfix part, and concatenate them back together around the new content.
This variation on your code:
function replace_char(pos, str, r)
return str:sub(1, pos-1) .. r .. str:sub(pos+1)
end
is the most direct translation to straightforward Lua. It is probably fast enough for most purposes. I've fixed the bug that the prefix should be the first pos-1 chars, and taken advantage of the fact that if the last argument to string.sub is missing it is assumed to be -1 which is equivalent to the end of the string.
But do note that it creates a number of temporary strings that will hang around in the string store until garbage collection eats them. The temporaries for the prefix and postfix can't be avoided in any solution. But this also has to create a temporary for the first .. operator to be consumed by the second.
It is possible that one of two alternate approaches could be faster. The first is the solution offered by Paŭlo Ebermann, but with one small tweak:
function replace_char2(pos, str, r)
return ("%s%s%s"):format(str:sub(1,pos-1), r, str:sub(pos+1))
end
This uses string.format to do the assembly of the result in the hopes that it can guess the final buffer size without needing extra temporary objects.
But do beware that string.format is likely to have issues with any \0 characters in any string that it passes through its %s format. Specifically, since it is implemented in terms of standard C's sprintf() function, it would be reasonable to expect it to terminate the substituted string at the first occurrence of \0. (Noted by user Delusional Logic in a comment.)
A third alternative that comes to mind is this:
function replace_char3(pos, str, r)
return table.concat{str:sub(1,pos-1), r, str:sub(pos+1)}
end
table.concat efficiently concatenates a list of strings into a final result. It has an optional second argument which is text to insert between the strings, which defaults to "" which suits our purpose here.
My guess is that unless your strings are huge and you do this substitution frequently, you won't see any practical performance differences between these methods. However, I've been surprised before, so profile your application to verify there is a bottleneck, and benchmark potential solutions carefully.
You should use pos inside your function instead of literal 1 and 3, but apart from this it looks good. Since Lua strings are immutable you can't really do much better than this.
Maybe
"%s%s%s":format(str:sub(1,pos-1), r, str:sub(pos+1, str:len())
is more efficient than the .. operator, but I doubt it - if it turns out to be a bottleneck, measure it (and then decide to implement this replacement function in C).
With luajit, you can use the FFI library to cast the string to a list of unsigned charts:
local ffi = require 'ffi'
txt = 'test'
ptr = ffi.cast('uint8_t*', txt)
ptr[1] = string.byte('o')
What is the simplest method to remove the last character from the end of a String in Scala?
I find Rubys String class has some very useful methods like chop. I would have used "oddoneoutz".headOption in Scala, but it is depreciated. I don't want to get into the overly complex:
string.slice(0, string.length - 1)
Please someone tell me there is a nice simple method like chop for something this common.
How about using dropRight, which works in 2.8:-
"abc!".dropRight(1)
Which produces "abc"
string.init // padding for the minimum 15 characters
val str = "Hello world!"
str take (str.length - 1) mkString
If you want the most efficient solution than just use:
str.substring(0, str.length - 1)
string.reverse.substring(1).reverse
That's basically chop, right? If you're longing for a chop method, why not write your own StringUtils library and include it in your projects until you find a suitable, more generic replacement?
Hey, look, it's in commons.
Apache Commons StringUtils.
If you want just to remove the last character use .dropRight(1). Alternatively, if you want to remove a specific ending character you may want to use a match pattern as
val s: String = "hello!"
val sClean: String = s.takeRight(1) match {
case "!" => s.dropRight(1)
case _ => s
}