complex string match with regexprep
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What should be the expression in the following script to generate out from str?
str='diff([v0(k-1,2)-v0(k-1,1) v(2)-v(1)])/diff(t0(k-1:k))+((v(2)-0)*diff(t0(k-1:k))+sum((v0(2:k-1,2)-v0(2:k-1,10)).*diff(t0(1:k-1)).'',1))';
expression = ???;
replace = '0';
out=regexprep(str,expression,replace);
out='diff([v0(k-1,2)-v0(k-1,1) v(2)-v(1)])/diff(t0(k-1:k))+((v(2)-0)*diff(t0(k-1:k))+sum((v0(2:k-1,2)-0).*diff(t0(1:k-1)).'',1))'
7 个评论
Adam Danz
2019-5-14
It would be helpful for you to explicitly show us the the differences between str and out so we can be sure there weren't typos and so we can be certain of the pattern you want to extract.
For the string 'v0(2:k-1,10))' what parts of it might vary and what parts will always be the same. For example, will it always have any of these structures?
- v0(#:k-#,#)
- v0(#:_-_,#)
- v0(#:_,#)
- v0(#:_,_)
- v0(_:_,_)
- _(_:_,_)
- _(_,_)
The more specific (higher up on my list), the better.
I'm assuming the string won't always be identical to 'v0(2:k-1,10)'
Adam Danz
2019-5-14
Got it. Thoroughly test my answer and if there are any problems, you can leave a comment under my answer and we can continue the discussion.
S H
2019-5-14
Adam Danz
2019-5-14
There are so many options with regular expressions that it's hard to capture them all in one document. I usually just google awkward phrases like "regular expressions match any character until" to remind myself of the options. The website I suggested in my answer is another great tool.
采纳的回答
Here you are.
str='diff([v0(k-1,2)-v0(k-1,1) v(2)-v(1)])/diff(t0(k-1:k))+((v(2)-0)*diff(t0(k-1:k))+sum((v0(2:k-1,2)-v0(2:k-1,10)).*diff(t0(1:k-1)).'',1))';
expression = 'v0\([^,]+,10\)';
replace = '0';
out = regexprep(str, expression, replace)
Compare input/output (I added space in output for comparison)
diff([v0(k-1,2)-v0(k-1,1) v(2)-v(1)])/diff(t0(k-1:k))+((v(2)-0)*diff(t0(k-1:k))+sum((v0(2:k-1,2)-v0(2:k-1,10)).*diff(t0(1:k-1)).',1)) %input
diff([v0(k-1,2)-v0(k-1,1) v(2)-v(1)])/diff(t0(k-1:k))+((v(2)-0)*diff(t0(k-1:k))+sum((v0(2:k-1,2)-0 ).*diff(t0(1:k-1)).',1)) %output
2 个评论
S H
2019-5-14
Great! Just so you know...
- v0\( Start the expression at v0(
- [^,]+ Stop matching just before the next comma
- ,10\) make sure the expression ends with ,10)
A nice workstation to develop regular expressions: https://regex101.com/
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