I'm trying to use pyparsing to build a parser that will match on all text within an arbitrarily nested set of brackets. If we consider a string like this:
"[A,[B,C],[D,E,F],G] Random Middle text [H,I,J]"
What I would like is for a parser to match in a way that it returns two matches:
[
"[A,[B,C],[D,E,F],G]",
"[H,I,J]"
]
I was able to accomplish a somewhat-working version of this using a barrage of originalTextFor mashed up with nestedExpr, but this breaks when your nesting is deeper than the number of OriginalTextFor expressions.
Is there a straightforward way to only match on the outermost expression grabbed by nestedExpr, or a way to modify its logic so that everything after the first paired match is treated as plaintext rather than being parsed?
update: One thing that seems to come close to what I want to accomplish is this modified version of the logic from nestedExpr:
def mynest(opener='{', closer='}'):
content = (empty.copy()+CharsNotIn(opener+closer+ParserElement.DEFAULT_WHITE_CHARS))
ret = Forward()
ret <<= ( Suppress(opener) + originalTextFor(ZeroOrMore( ret | content )) + Suppress(closer) )
return ret
This gets me most of the way there, although there's an extra level of list wrapping in there that I really don't need, and what I'd really like is for those brackets to be included in the string (without getting into an infinite recursion situation by not suppressing them).
parser = mynest("[","]")
result = parser.searchString("[A,[B,C],[D,E,F],G] Random Middle text [H,I,J]")
result.asList()
>>> [['A,[B,C],[D,E,F],G'], ['H,I,J']]
I know I could strip these out with a simple list comprehension, but it would be ideal if I could just eliminate that second, redundant level.
Not sure why this wouldn't work:
sample = "[A,[B,C],[D,E,F],G] Random Middle text [H,I,J]"
scanner = originalTextFor(nestedExpr('[',']'))
for match in scanner.searchString(sample):
print(match[0])
prints:
'[A,[B,C],[D,E,F],G]'
'[H,I,J]'
What is the situation where "this breaks when your nesting is deeper than the number of OriginalTextFor expressions"?
Related
What is an efficient way in MATLAB to replace/insert one symbol (in series of symbols) with several others that correspond to the one that is being replaced?
For example, consider having a string Eq: Eq = 'A*exp(-((x-xc)/w)^2)'. Is there a way to replace * with .*, / with ./,\ with .\, and ^ with .^ without writing four separate strrep() lines?
Regular expressions will do the job nicely. Regular expressions simply find patterns in text. You specify what kind of pattern you are looking for by a regular expression, and the output gives you the locations of where the pattern occurred.
For our particular case, not only do we want to find where patterns occur, we also want to replace those patterns with something else. Specifically, use the function regexprep from MATLAB to replace matches in a string with something else. What you want to do is replace all *, /, \ and ^ symbols by adding a . in front of each.
How regexprep works is that the first input is the string you're looking at, the second input is a pattern that you're trying to find. In our case, we want to find any of *, /, \ and ^. To specify this pattern, you put those desired symbols in [] brackets. Regular expressions reserve \ as a special symbol to delineate characters that can be parsed as a regular expression but actually aren't. As such, you need to use \\ for the \ character and \^ for the ^ character. The third input is what you want to replace each match with. In our case, we simply want to reuse each matched character, but we add a . at the beginning of the match. This is done by doing \.$0 in the regular expression syntax. $0 means to grab the first token produced by a match... which is essentially the matched symbol from the pattern. . is also a reserved keyword using regular expressions, so we must prepend this symbol with a \ character.
Without further ado:
>> Eq = 'A*exp(-((x-xc)/w)^2)';
>> out = regexprep(Eq, '[*/\\\^]', '\.$0')
out =
A.*exp(-((x-xc)./w).^2)
The pattern we are looking for is [*/\\\^], which means that we want to find any of *, /, \ - denoted as \\ in regex, and \^ - denoted as ^ in regex. We want to find any of these symbols and replace them with the same symbol by adding a . character in front - \.$0.
As a more complicated example, let's make sure that we include all of the symbols you're looking for in a sample equation:
>> A = 'A*exp(-((x-xc)/w)^2) \ b^2';
>> out = regexprep(A, '[*/\\\^]', '\.$0')
out =
A.*exp(-((x-xc)./w).^2) .\ b.^2
I'd go with regexp as in rayryeng's answer. But here's another approach, just to provide an alternative.
ops = '*/\^'; %// operators that need a dot
ii = find(ismember(Eq, ops)); %// find where dots should be inserted
[~, jj] = sort([1:numel(Eq) ii-.5]); %// will be used to properly order the result
result = [Eq repmat('.',1,numel(ii))]; %// insert dots at the end
result = result(jj); %// properly order the result
And a variant:
ops = '*/\^'; %// operators that need a dot
ii = find(ismember(Eq, ops)); %// find where dots should be inserted
jj = sort([1:numel(Eq) ii-.5]); %// dot locations are marked with fractional part
result = Eq(ceil(jj)); %// repeat characters where the dots will be placed
result(mod(jj,1)>0) = '.'; %// place dots at indices with fractional part
The vectorize function already does almost all of what you want except that it does not convert mldivide (\) to ldivide (.\).
By "efficient," do you mean fewer lines of code or faster? Regular expressions are almost always slower than other approaches and less readable. I don't think they're necessary or a good choice in this case. If you only need to convert your string once, then speed is less of a concern than readability (strrep will still be faster). If you need to do it many times, this simple code that you alluded to is 4–5 times faster than regexrep for short strings like your example (and much faster for longer strings):
out = strrep(Eq,'*','.*');
out = strrep(out,'/','./');
out = strrep(out,'\','.\');
out = strrep(out,'^','.^');
If you want one line, use:
out = strrep(strrep(strrep(strrep(Eq,'*','.*'),'/','./'),'\','.\'),'^','.^');
which will also be slightly faster still. Or create your own version of vectorize and call that.
Where regular expressions shine is in more complex cases, e.g., if your string is already partially vectorized: Eq = 'A.*exp(-((x-xc)/w)^2)'. Even still, the vectorize function just uses strrep and then calls strfind to "remove any possible '..*', '../', etc." and replace them with the proper element-wise operators because it's faster (symbolic math strings can get very large, for example).
I'm stuck with something that usually is pretty easily in other programming languages.
I want to test whether a string is inside another one in R. For example I tried:
match("Diagnosi Prenatale,Esercizio Fisico", "Diagnosi Prenatale")
pmatch("Diagnosi Prenatale,Esercizio Fisico", "Diagnosi Prenatale")
grep("Diagnosi Prenatale,Esercizio Fisico", "Diagnosi Prenatale")
And none worked. To make it work I should fist split the first string with strsplit and extract the first element.
NOTE: I'd like to do this on a vector of strings to receive a yes/no vector, so in the function I wrote should go a vector not a single string. But of course if the single string doesn't work, image a full vector of them...
Any ideas?
Try grepl
grepl("Diagnosi Prenatale","Diagnosi Prenatale,Esercizio Fisico" )
[1] TRUE
You can also do this with character vectors, for example:
x <- c("Diagnosi Prenatale,Esercizio Fisico", "Diagnosi Prenatale")
grepl("Diagnosi Prenatale",x)
#[1] TRUE TRUE
I want to search for a query (a string) in a subject (another string).
The query may appear in whole or in parts, but will not be rearranged. For instance, if the query is 'da', and the subject is 'dura', it is still a match.
I am not allowed to use string functions like strfind or find.
The constraints make this actually quite straightforward with a single loop. Imagine you have two indices initially pointing at the first character of both strings, now compare them - if they don't match, increment the subject index and try again. If they do, increment both. If you've reached the end of the query at that point, you've found it. The actual implementation should be simple enough, and I don't want to do all the work for you ;)
If this is homework, I suggest you look at the explanation which precedes the code and then try for yourself, before looking at the actual code.
The code below looks for all occurrences of chars of the query string within the subject string (variables m; and related ii, jj). It then tests all possible orders of those occurrences (variable test). An order is "acceptable" if it contains all desired chars (cond1) in increasing positions (cond2). The result (variable result) is affirmative if there is at least one acceptable order.
subject = 'this is a test string';
query = 'ten';
m = bsxfun(#eq, subject.', query);
%'// m: test if each char of query equals each char of subject
[ii jj] = find(m);
jj = jj.'; %'// ii: which char of query is found within subject...
ii = ii.'; %'// jj: ... and at which position
test = nchoosek(1:numel(jj),numel(query)).'; %'// test all possible orders
cond1 = all(jj(test) == repmat((1:numel(query)).',1,size(test,2)));
%'// cond1: for each order, are all chars of query found in subject?
cond2 = all(diff(ii(test))>0);
%// cond2: for each order, are the found chars in increasing positions?
result = any(cond1 & cond2); %// final result: 1 or 0
The code could be improved by using a better approach as regards to test, i.e. not testing all possible orders given by nchoosek.
Matlab allows you to view the source of built-in functions, so you could always try reading the code to see how the Matlab developers did it (although it will probably be very complex). (thanks Luis for the correction)
Finding a string in another string is a basic computer science problem. You can read up on it in any number of resources, such as Wikipedia.
Your requirement of non-rearranging partial matches recalls the bioinformatics problem of mapping splice variants to a genomic sequence.
You may solve your problem by using a sequence alignment algorithm such as Smith-Waterman, modified to work with all English characters and not just DNA bases.
Is this question actually from bioinformatics? If so, you should tag it as such.
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')