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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***
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'
I have an abstract class Writer which allows clients to write into something. Could be the screen, could be a file. Now, I try to create a derived class to write into a string.
I have two problems with the denoted line in method write(...):
It's probably very inefficient. Is there something like a string buffer in Matlab?
It writes escape sequences like \n plain into the string, instead of taking their actual meaning.
How can I get the denoted line properly?
Code:
classdef StringTextWriter < Writer
properties
str;
end
methods
function this = StringTextWriter()
% Init the write-target which is a string in our case.
% (Other Writer classes would maybe open a file.)
this.str = '';
end
function write(this, val)
% Write to writer target.
% (Other Writer classes would maybe use fprinf here for file write.)
% ?????????????????????????????
this.str = [this.str val]; % How to do this properly?
% ?????????????????????????????
end
end
end
To answer your questions point by point:
The closest notion to a string buffer would be a string cell. Instead of:
str = '';
str = [strbuf, 'abc\n'];
str = [strbuf, 'def\n'];
str = [strbuf, 'ghi\n'];
%// and so on...
one may use
strbuf = {};
strbuf{end+1} = 'abc\n';
strbuf{end+1} = 'def\n';
strbuf{end+1} = 'ghi\n';
%// and so on...
str = sprintf([strbuf{:}]); %// careful about percent signs and single quotes in string
the drawback being that you have to reconstruct the string every time you ask for it. This can be alleviated by setting a modified flag every time you add strings to the end of strbuf, resetting it every time you concatenate the strings, and memoizing the result of concatenation in the last line (rebuild if modified, or last result if not).
Further improvement can be achieved by choosing a better strategy for growing the strbuf cell array; probably this would be effective if you have a lot of write method calls.
The escape sequences are really linked to the <?>printf family and not to the string literals, so MATLAB in general doesn't care about them, but sprintf in particular might.
I'm starting to like the Swift string formatting since it uses variable names in the string rather than ambiguous formatting tags like "%#"
I want to load a large string from a file that has Swift-style formatting in it (like this)
Now is the time for all good \(who) to come to babble incoherently.
Then I want to feed the contents of that String variable into a statement that lest me replace
\(who)
with the contents of the constant/variable who at runtime.
The code below works with a string constant as the formatting string.
let who = "programmers"
let aString = "Now is the time for all good \(who) to come to babble incoherently."
That code does formatting of a quoted string that appears in-line in my code.
Instead I want something like the code
let formatString = "Now is the time for all good %# to come to babble incoherently."
aString = String(format: formatString, who)
But where I can pass in a Swift-style format string in a constant/variable I read from a file.
Is that possible? I didn't have any luck searching for it since I wasn't exactly sure what search terms to use.
I can always use C-style string formatting and the String class' initWithFormat method if I have to...
I don't think there's a way to do this. String interpolation is implemented via conforming to the StringInterpolationConvertible protocol, and presumably you're hoping to tap into that in the same way you can tap into the methods required by StringLiteralConvertible, a la:
let someString = toString(42)
// this is the method String implements to conform to StringLiteralConvertible
let anotherString = String(stringLiteral: someString)
// anotherString will be "42"
print(anotherString)
Unfortunately, you can't do quite the same trick with StringInterpolationConvertible. Seeing how the protocol works may help:
struct MyString: Printable {
let actualString: String
var description: String { return actualString }
}
extension MyString: StringInterpolationConvertible {
// first, this will get called for each "segment"
init<T>(stringInterpolationSegment expr: T) {
println("Processing segment: " + toString(expr))
actualString = toString(expr)
}
// here is a type-specific override for Int, that coverts
// small numbers into words:
init(stringInterpolationSegment expr: Int) {
if (0..<4).contains(expr) {
println("Embigening \(expr)")
let numbers = ["zeo","one","two","three"]
actualString = numbers[expr]
}
else {
println("Processing segment: " + toString(expr))
actualString = toString(expr)
}
}
// finally, this gets called with an array of all of the
// converted segments
init(stringInterpolation strings: MyString...) {
// strings will be a bunch of MyString objects
actualString = "".join(strings.map { $0.actualString })
}
}
let number = 3
let aString: MyString = "Then shalt thou count to \(number), no more, no less."
println(aString)
// prints "Then shalt thou count to three, no more, no less."
So, while you can call String.init(stringInterpolation:) and String.init(stringInterpolationSegment:) directly yourself if you want (just try String(stringInterpolationSegment: 3.141) and String(stringInterpolation: "blah", "blah")), this doesn't really help you much. What you really need is a facade function that coordinates the calls to them. And unless there's a handy pre-existing function in the standard library that does exactly that which I've missed, I think you're out of luck. I suspect it's built into the compiler.
You could maybe write your own to achieve your goal, but a lot of effort since you'd have to break up the string you want to interpolate manually into bits and handle it yourself, calling the segment init in a loop. Also you'll hit problems with calling the combining function, since you can't splat an array into a variadic function call.
I don't think so. The compiler needs to be able to resolve the interpolated variable at compile time.
I'm not a Swift programmer, specifically, but I think you can workaround it to something pretty close to what you want using a Dictionary and standard string-replacing and splitting methods:
var replacement = [String: String]()
replacement["who"] = "programmers"
Having that, you can try to find the occurrences of "\(", reading what is next and prior to a ")", (this post can help with the split part, this one, with the replacing part), finding it in the dictionary, and reconstructing your string from the pieces you get.
this one works like a charm:
let who = "programmers"
let formatString = "Now is the time for all good %# to come to babble incoherently."
let aString = String(format: formatString, who)
I'm trying to do a function in Lua that swaps the characters in a string.
Can somebody help me ?
Here is an example:
Input = "This LIBRARY should work with any string!"
Result = "htsil biaryrs ohlu dowkrw ti hna ytsirgn!"
Note: The space is also swapped
Thank You Very Much :)
The simplest and clearest solution is this:
Result = Input:gsub("(.)(.)","%2%1")
This should do it:
input = "This LIBRARY should work with any string!"
function swapAlternateChars(str)
local t={}
-- Iterate through the string two at a time
for i=1,#str,2 do
first = str:sub(i,i)
second = str:sub(i+1,i+1)
t[i] = second
t[i+1] = first
end
return table.concat(t)
end
print(input)
print(swapAlternateChars(input))
Prints:
This LIBRARY should work with any string!
hTsiL BIARYRs ohlu dowkrw ti hna ytsirgn!
If you need the output as lower case you could always end it with:
output = swapAlternateChars(input)
print(string.lower(output))
Note, in this example, I'm not actually editing the string itself, since strings in Lua are immutable. Here's a read: Modifying a character in a string in Lua
I've used a table to avoid overhead from concatenating to a string because each concatenation may allocate a new string in memory.
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')