As example: I want remove the first 2 letters from the string "ПРИВЕТ" and "HELLO." one of these are containing only two-byted unicode symbols.
Trying to use string.sub("ПРИВЕТ") and string.sub("HELLO.")
Got "РИВЕТ" and "LLO.".
string.sub() removed 2 BYTES(not chars) from these strings. So i want to know how to get the removing of the chars
Something, like utf8.sub()
The key standard function for this task is utf8.offset(s,n), which gives the position in bytes of the start of the n-th character of s.
So try this:
print(string.sub(s,utf8.offset(s,3),-1))
You can define utf8.sub as follows:
function utf8.sub(s,i,j)
i=utf8.offset(s,i)
j=utf8.offset(s,j+1)-1
return string.sub(s,i,j)
end
(This code only works for positive j. See http://lua-users.org/lists/lua-l/2014-04/msg00590.html for the general case.)
There is https://github.com/Stepets/utf8.lua, a pure lua library, which expand standard function to support utf8 string.
I found a simpler solution (the the solution using offset() didnt work for me for all cases):
function utf8.sub(s, i, j)
return utf8.char(utf8.codepoint(s, i, j))
end
Related
I have a problem with splitting string into two parts on special character.
For example:
12345#data
or
1234567#data
I have 5-7 characters in first part separated with "#" from second part, where are another data (characters,numbers, doesn't matter what)
I need to store two parts on each side of # in two variables:
x = 12345
y = data
without "#" character.
I was looking for some Lua string function like splitOn("#") or substring until character, but I haven't found that.
Use string.match and captures.
Try this:
s = "12345#data"
a,b = s:match("(.+)#(.+)")
print(a,b)
See this documentation:
First of all, although Lua does not have a split function is its standard library, it does have string.gmatch, which can be used instead of a split function in many cases. Unlike a split function, string.gmatch takes a pattern to match the non-delimiter text, instead of the delimiters themselves
It is easily achievable with the help of a negated character class with string.gmatch:
local example = "12345#data"
for i in string.gmatch(example, "[^#]+") do
print(i)
end
See IDEONE demo
The [^#]+ pattern matches one or more characters other than # (so, it "splits" a string with 1 character).
I am trying to find the Nth number in a string. Should i use string.find? If so, how? I know the arguments are the string to search and the pattern to find, but the 3rd argument (where to start) seems like it might just work.
the lua string tutorial i am looking at
thanks!
You'll want to create a function that splits your string into an array. Once you've done this, you'll be able to return whatever number position you're looking for.
function findnth(str, nth)
local array = {}
for i in string.gmatch(str, "%d+") do
table.insert(array, i)
end
return array[nth]
end
The function above works like so:
print(findnth("1 3 7 2 15 2", 4))
Returns:
2
Edit: Changed function to suit OP's specific needs.
Is there a multiline string literal syntax in Matlab or is it necessary to concatenate multiple lines?
I found the verbatim package, but it only works in an m-file or function and not interactively within editor cells.
EDIT: I am particularly after readbility and ease of modifying the literal in the code (imagine it contains indented blocks of different levels) - it is easy to make multiline strings, but I am looking for the most convenient sytax for doing that.
So far I have
t = {...
'abc'...
'def'};
t = cellfun(#(x) [x sprintf('\n')],t,'Unif',false);
t = horzcat(t{:});
which gives size(t) = 1 8, but is obviously a bit of a mess.
EDIT 2: Basically verbatim does what I want except it doesn't work in Editor cells, but maybe my best bet is to update it so it does. I think it should be possible to get current open file and cursor position from the java interface to the Editor. The problem would be if there were multiple verbatim calls in the same cell how would you distinguish between them.
I'd go for:
multiline = sprintf([ ...
'Line 1\n'...
'Line 2\n'...
]);
Matlab is an oddball in that escape processing in strings is a function of the printf family of functions instead of the string literal syntax. And no multiline literals. Oh well.
I've ended up doing two things. First, make CR() and LF() functions that just return processed \r and \n respectively, so you can use them as pseudo-literals in your code. I prefer doing this way rather than sending entire strings through sprintf(), because there might be other backslashes in there you didn't want processed as escape sequences (e.g. if some of your strings came from function arguments or input read from elsewhere).
function out = CR()
out = char(13); % # sprintf('\r')
function out = LF()
out = char(10); % # sprintf('\n');
Second, make a join(glue, strs) function that works like Perl's join or the cellfun/horzcat code in your example, but without the final trailing separator.
function out = join(glue, strs)
strs = strs(:)';
strs(2,:) = {glue};
strs = strs(:)';
strs(end) = [];
out = cat(2, strs{:});
And then use it with cell literals like you do.
str = join(LF, {
'abc'
'defghi'
'jklm'
});
You don't need the "..." ellipses in cell literals like this; omitting them does a vertical vector construction, and it's fine if the rows have different lengths of char strings because they're each getting stuck inside a cell. That alone should save you some typing.
Bit of an old thread but I got this
multiline = join([
"Line 1"
"Line 2"
], newline)
I think if makes things pretty easy but obviously it depends on what one is looking for :)
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')
I've a question about Fortran 77 and I've not been able to find a solution.
I'm trying to store an array of strings defined as the following:
character matname(255)*255
Which is an array of 255 strings of length 255.
Later I read the list of names from a file and I set the content of the array like this:
matname(matcount) = mname
EDIT: Actually mname value is hardcoded as mname = 'AIR' of type character*255, it is a parameter of a function matadd() which executes the previous line. But this is only for testing, in the future it will be read from a file.
Later on I want to print it with:
write(*,*) matname(matidx)
But it seems to print all the 255 characters, it prints the string I assigned and a lot of garbage.
So that is my question, how can I know the length of the string stored?
Should I have another array with all the lengths?
And how can I know the length of the string read?
Thanks.
You can use this function to get the length (without blank tail)
integer function strlen(st)
integer i
character st*(*)
i = len(st)
do while (st(i:i) .eq. ' ')
i = i - 1
enddo
strlen = i
return
end
Got from here: http://www.ibiblio.org/pub/languages/fortran/ch2-13.html
PS: When you say: matname(matidx) it gets the whole string(256) chars... so that is your string plus blanks or garbage
The function Timotei posted will give you the length of the string as long as the part of the string you are interested in only contains spaces, which, if you are assigning the values in the program should be true as FORTRAN is supposed to initialize the variables to be empty and for characters that means a space.
However, if you are reading in from a file you might pick up other control characters at the end of the lines (particularly carriage return and/or line feed characters, \r and/or \n depending on your OS). You should also toss those out in the function to get the correct string length. Otherwise you could get some funny print statements as those characters are printed as well.
Here is my version of the function that checks for alternate white space characters at the end besides spaces.
function strlen(st)
integer i,strlen
character st*(*)
i = len(st)
do while ((st(i:i).eq.' ').or.(st(i:i).eq.'\r').or.
+ (st(i:i).eq.'\n').or.(st(i:i).eq.'\t'))
i = i - 1
enddo
strlen = i
return
end
If there are other characters in the "garbage" section this still won't work completely.
Assuming that it does work for your data, however, you can then change your write statement to look like this:
write(*,*) matname(matidx)(1:strlen(matname(matidx)))
and it will print out just the actual string.
As to whether or not you should use another array to hold the lengths of the string, that is up to you. the strlen() function is O(n) whereas looking up the length in a table is O(1). If you find yourself computing the lengths of these static strings often, it may improve performance to compute the length once when they are read in, store them in an array and look them up if you need them. However, if you don't notice the slowdown, I wouldn't worry about it.
Depending on the compiler that you are using, you may be able to use the trim() intrinsic function to remove any leading/trailing spaces from a string, then process it as you normally would, i.e.
character(len=25) :: my_string
my_string = 'AIR'
write (*,*) ':', trim(my_string), ':'
should print :AIR:.
Edit:
Better yet, it looks like there is a len_trim() function that returns the length of a string after it has been trimmed.
intel and Compaq Visual Fortran have the intrinsic function LEN_TRIM(STRING) which returns the length without trailing blanks or spaces.
If you want to suppress leading blanks or spaces, use "Adjust Left" i.e. ADJUSTF(STRING)
In these FORTRANs I also note a useful feature: If you pass a string in to a function or subroutine as an argument, and inside the subroutine it is declared as CHARACTER*(*), then
using the LEN(STRING) function in the subroutine retruns the actual string length passed in, and not the length of the string as declared in the calling program.
Example:
CHARACTER*1000 STRING
.
.
CALL SUBNAM(STRING(1:72)
SUBROUTINE SYBNAM(STRING)
CHARACTER*(*) STRING
LEN(STRING) will be 72, not 1000