I tried to use regexprep to solve a problem - I'm given a string, that represents a function; it contains a patterns like these: 'sin(arcsin(f))' where f - any substring; and I need to replace it with simple 'f_2'. I successfully used regexprep unless I face with such string:
str = 'sin(arcsin(sin(arcsin(f_2))))*x^2';
str = regexprep(str, 'sin\(arcsin\((\w*)\)\)','$1');
it returns
str =
sin(arcsin(f_2))*x^2
But I want it to be
str =
f_2*x^2
Is there any way to solve it (except obvious solution with for-loops).
I was not able to test this, but I thinkg I found an expression that you can call multiple times to do what you asked for; each time it will "strip" one sin(arcsin()) pair out of your equation. Once it stops changing, you're done.
(.*)sin\(arcsin\((.*(\(.*?\))*)(\)\).*$)
Here is some Matlab code that shows how this might work:
str = 'sin(arcsin(sin(arcsin(f_2))))*x^2';
regex = (.*)sin\(arcsin\((.*(\(.*?\))*)(\)\).*$);
oldlength = 0
newlength = length(str)
while (newlength != oldlength)
oldlength = newlength;
str = regexprep(str, regex,'$1$2');
newlength = length(str);
end
As I said - I could not test this. Let me know if you have any problems with this.
Demo of the regular expression:
http://regex101.com/r/bR9gC7
Change your pattern to search for 1 or more (+) nested sin(arcsin( occurrences:
str = 'sin(arcsin(sin(arcsin(f_2))))*x^2';
str2 = regexprep(str, '(sin\(arcsin\()+(\w*)(\)\))+','$2')
str2 =
f_2*x^2
Related
I am using regexpi to find a string in a phrase. But I also encountered with something different which I never intended.
Let's say the words I need to find are anandalak and nandaki.
str1 = {'anandalak'};
str2 = {'nanda'};
button = {'nanda'};
Both of the following return me logical 1:
~cellfun('isempty',regexpi(str1,button))
~cellfun('isempty',regexpi(str2,button))
How can I avoid this? I need logical 0 in first case and logical 1 in the second.
You probably need to use the word-boundaries(\<\>) in order to get the match which you require.
You may try:
str1 = {'anandalak'}
str2 = {'nanda'}
button = {'\<nanda\>'} % Notice this
~cellfun(#isempty,regexpi(str1,button)) % Returns ans = 0 No match
~cellfun(#isempty,regexpi(str2,button)) % Return ans = 1 Exact match
You can find the sample run result of the above implementation in here.
I have to replace strings in a hash. I have:
hash = {"{STAY_ID}"=>"30030303", "{USER_NAME}"=>"test"}
And I have to replace it here:
str = "www.domain.com?person={STAY_ID}&user={USER_NAME}"
#=> www.domain.com?person=30030303&user=test
Also, it should work when there is a string with at least one match:
str = "www.domain.com?person={STAY_ID}"
#=> www.domain.com?person=30030303
I need some method/solution that can handle any situation like above.
Something great about the gsub method is that it can actually take a hash of mappings as the second argument, which it then uses to replace matched values. Therefore, if you regex any text between curly braces, you can do something like this.
str = "www.domain.com?person={STAY_ID}&user={USER_NAME}"
hash = {
"{STAY_ID}"=>"30030303",
"{USER_NAME}"=>"test"
}
str.gsub(/{(.*?)}/, hash) #www.domain.com?person=30030303&user=test
And then ya done!
I think, regex is not readable solution. You can use simple gsub method:
str = "www.domain.com?person={STAY_ID}&user={USER_NAME}"
hash = {"{STAY_ID}"=>"30030303", "{USER_NAME}"=>"test"}
result_str = hash.inject(str.dup) do |acc, (key, value)|
acc = acc.gsub(key, value)
end
result_str # www.domain.com?person=30030303&user=test
I have strings like:
str1 = eval(sum(feat(57),feat(57),feat(66))/feat(57));
str2 = eval(sum(feat(47),feat(55),feat(86)));
str3 = eval(feat(47)/sum(feat(51),feat(52),feat(53)));
str4 = eval(feat(63)/sum(feat(57):feat(66)));
I want to write a regex to get out as:
str1_output = (feat(57),feat(57),feat(66))
str2_output = (feat(47),feat(55),feat(86))
str3_output = (feat(51),feat(52),feat(53))
str4_output = (feat(57):feat(66))
I tried in the following way:
output = re.findall(re.compile(r"sum.*"),str_name)
This is giving correct output except str1.
Please suggest me a way to find out the desired output.
I guess you could try
sum\((?:\([^()]*\)|.)*?\)
It matches sum and the following matching pair of parentheses, and whatever are between them.
Example at regex101.
Regards.
I was wondering if there was a way to do pattern matching in Octave / matlab? I know Maple 10 has commands to do this but not sure what I need to do in Octave / Matlab. So if a number was 12341234123412341234 the pattern match would be 1234. I'm trying to find the shortest pattern that upon repetiton generates the whole string.
Please note: the numbers (only numbers will be used) won't be this simple. Also, I won't know the pattern ahead of time (that's what I'm trying to find). Please see the Maple 10 example below which shows that the pattern isn't known ahead of time but the command finds the pattern.
Example of Maple 10 pattern matching:
ns:=convert(12341234123412341234,string);
ns := "12341234123412341234"
StringTools:-PrimitiveRoot(ns);
"1234"
How can I do this in Octave / Matlab?
Ps: I'm using Octave 3.8.1
To find the shortest pattern that upon repetition generates the whole string, you can use regular expressions as follows:
result = regexp(str, '^(.+?)(?=\1*$)', 'match');
Some examples:
>> str = '12341234123412341234';
>> result = regexp(str, '^(.+?)(?=\1*$)', 'match')
result =
'1234'
>> str = '1234123412341234123';
>> result = regexp(str, '^(.+?)(?=\1*$)', 'match')
result =
'1234123412341234123'
>> str = 'lullabylullaby';
>> result = regexp(str, '^(.+?)(?=\1*$)', 'match')
result =
'lullaby'
>> str = 'lullaby1lullaby2lullaby1lullaby2';
>> result = regexp(str, '^(.+?)(?=\1*$)', 'match')
result =
'lullaby1lullaby2'
I'm not sure if this can be accomplished with regular expressions. Here is a script that will do what you need in the case of a repeated word called pattern.
It loops through the characters of a string called str, trying to match against another string called pattern. If matching fails, the pattern string is extended as needed.
EDIT: I made the code more compact.
str = 'lullabylullabylullaby';
pattern = str(1);
matchingState = false;
sPtr = 1;
pPtr = 1;
while sPtr <= length(str)
if str(sPtr) == pattern(pPtr) %// if match succeeds, keep looping through pattern string
matchingState = true;
pPtr = pPtr + 1;
pPtr = mod(pPtr-1,length(pattern)) + 1;
else %// if match fails, extend pattern string and start again
if matchingState
sPtr = sPtr - 1; %// don't change str index when transitioning out of matching state
end
matchingState = false;
pattern = str(1:sPtr);
pPtr = 1;
end
sPtr = sPtr + 1;
end
display(pattern);
The output is:
pattern =
lullaby
Note:
This doesn't allow arbitrary delimiters between occurrences of the pattern string. For example, if str = 'lullaby1lullaby2lullaby1lullaby2';, then
pattern =
lullaby1lullaby2
This also allows the pattern to end mid-way through a cycle without changing the result. For example, str = 'lullaby1lullaby2lullaby1'; would still result in
pattern =
lullaby1lullaby2
To fix this you could add the lines
if pPtr ~= length(pattern)
pattern = str;
end
Another approach is as follows:
determine length of string, and find all possible factors of the string length value
for each possible factor length, reshape the string and check
for a repeated substring
To find all possible factors, see this solution on SO. The next step can be performed in many ways, but I implement it in a simple loop, starting with the smallest factor length.
function repeat = repeats_in_string(str);
ns = numel(str);
nf = find(rem(ns, 1:ns) == 0);
for ii=1:numel(nf)
repeat = str(1:nf(ii));
if all(ismember(reshape(str,nf(ii),[])',repeat));
break;
end
end
This problem is a great Rorschach test for your approach to problem solving. I'll add a signal engineering solution, which should be simple since the signal is expected to be perfectly repetitive, assuming this holds: find the shortest pattern that upon repetition generates the whole string.
In the following str fed to the function is actually a column vector of floats, not a string, the original string having been converted with str2num(str2mat(str)'):
function res=findshortestrepel(str);
[~,ii] = max(fft(str-mean(str)));
res = str(1:round(numel(str)/(ii-1)));
I performed a small test, comparing this to the regexp solution and found it to be faster overall (blue squares), although somewhat inconsistently, and only if you don't consider the time required to convert the string into a vector of floats (green squares). However I did not pursue this further (not breaking records with this):
Times in sec.
How can I find maximum possible pattern in a string in Matlab, which matches some expression. Example will clarify what I mean:
str = 'tan(sin*cos)';
str = 'tan(sin(exp)*cos(exp))';
I want to find the patterns, which look like tan(\w*). But I want brackets in tan to be balanced. Is there any approach to do it?
It's not possible without recusrsive regular expressions. For example, this string:
str = 'tan(tan(tan(x) + 4) + cos(x))'
would have to be regex'ed "from the inside out", something only recursion can do.
Instead, I'd just use a more practical solution:
regexprep(str, 'tan', '')
and/or split further when necessary. Or, as Ruud already suggested, just use a loop:
str{1} = 'tan(x)';
str{2} = 'tan(sin(exp)*cos(exp)) + tan(tan(x) + 4)';
S = regexp(str, 'tan\(');
match = cell(size(str));
[match{:}] = deal({});
for ii = 1:numel(str)
if ~isempty(S{ii})
for jj = 1:numel(S{ii})
open = false;
start = S{ii}(jj)+4;
for kk = start : numel(str{ii})
switch str{ii}(kk)
case '('
open = true;
case ')'
if open
open = false;
else
match{ii}{end+1} = str{ii}(start:kk-1);
break;
end
end
end
end
end
end