I am working with a messy manually maintained "database" that has a column containing a string with name,value pairs. I am trying to parse the entire column with regexp to pull out the values. The column is huge (>100,000 entries). As a proxy for my actual data, let's use this code:
line1={'''thing1'': ''-583'', ''thing2'': ''245'', ''thing3'': ''246'', ''morestuff'':, '''''};
line2={'''thing1'': ''617'', ''thing2'': ''239'', ''morestuff'':, '''''};
line3={'''thing1'': ''unexpected_string(with)parens5'', ''thing2'': 245, ''thing3'':''246'', ''morestuff'':, '''''};
mycell=vertcat(line1,line2,line3);
This captures the general issues encountered in the database. I want to extract what thing1, thing2, and thing3 are in each line using cellfun to output a scalar cell array. They should normally be 3 digit numbers, but sometimes they have an unexpected form. Sometimes thing3 is completely missing, without the name even showing up in the line. Sometimes there are minor formatting inconsistencies, like single quotes missing around the value, spaces missing, or dashes showing up in front of the three digit value. I have managed to handle all of these, except for the case where thing3 is completely missing.
My general approach has been to use expressions like this:
expr1='(?<=thing1''):\s?''?-?([\w\d().]*?)''?,';
expr2='(?<=thing2''):\s?''?-?([\w\d().]*?)''?,';
expr3='(?<=thing3''):\s?''?-?([\w\d().]*?)''?,';
This looks behind for thingX' and then tries to match : followed by zero or one spaces, followed by 0 or 1 single quote, followed by zero or one dash, followed by any combination of letters, numbers, parentheses, or periods (this is defined as the token), using a lazy match, until zero or one single quote is encountered, followed by a comma. I call regexp as regexp(___,'tokens','once') to return the matching token.
The problem is that when there is no match, regexp returns an empty array. This prevents me from using, say,
out=cellfun(#(x) regexp(x,expr3,'tokens','once'),mycell);
unless I call it with 'UniformOutput',false. The problem with that is twofold. First, I need to then manually find the rows where there was no match. For example, I can do this:
emptyout=cellfun(#(x) isempty(x),out);
emptyID=find(emptyout);
backfill=cell(length(emptyID),1);
[backfill{:}]=deal('Unknown');
out(emptyID)=backfill;
In this example, emptyID has a length of 1 so this code is overkill. But I believe this is the correct way to generalize for when it is longer. This code will change every empty cell array in out with the string Unknown. But this leads to the second problem. I've now got a 'messy' cell array of non-scalar values. I cannot, for example, check unique(out) as a result.
Pardon the long-windedness but I wanted to give a clear example of the problem. Now my actual question is in a few parts:
Is there a way to accomplish what I'm trying to do without using 'UniformOutput',false? For example, is there a way to have regexp pass a custom string if there is no match (e.g. pass 'Unknown' if there is no match)? I can think of one 'cheat', which would be to use the | operator in the expression, and if the first token is not matched, look for something that is ALWAYS found. I would then still need to double back through the output and change every instance of that result to 'Unknown'.
If I take the 'UniformOutput',false approach, how can I recover a scalar cell array at the end to easily manipulate it (e.g. pass it through unique)? I will admit I'm not 100% clear on scalar vs nonscalar cell arrays.
If there is some overall different approach that I'm not thinking of, I'm also open to it.
Tangential to the main question, I also tried using a single expression to run regexp using 3 tokens to pull out the values of thing1, thing2, and thing3 in one pass. This seems to require 'UniformOutput',false even when there are no empty results from regexp. I'm not sure how to get a scalar cell array using this approach (e.g. an Nx1 cell array where each cell is a 3x1 cell).
At the end of the day, I want to build a table using these results:
mytable=table(out1,out2,out3);
Edit: Using celldisp sheds some light on the problem:
celldisp(out)
out{1}{1} =
246
out{2} =
Unknown
out{3}{1} =
246
I assume that I need to change the structure of out so that the contents of out{1}{1} and out{3}{1} are instead just out{1} and out{3}. But I'm not sure how to accomplish this if I need 'UniformOutput',false.
Note: I've not used MATLAB and this doesn't answer the "efficient" aspect, but...
How about forcing there to always be a match?
Just thinking about you really wanting a match to skip this problem, how about an empty match?
Looking on the MATLAB help page here I can see a 'emptymatch' option, perhaps this is something to try.
E.g.
the_thing_i_want_to_find|
Match "the_thing_i_want_to_find" or an empty match, note the | character.
In capture group it might look like this:
(the_thing_i_want_to_find|)
As a workaround, I have found that using regexprep can be used to find entries where thing3 is missing. For example:
replace='$1 ''thing3'': ''Unknown'', ''morestuff''';
missingexpr='(?<=thing2'':\s?)(''?-?[\w\d().]*?''?,) ''morestuff''';
regexprep(mycell{2},missingexpr,replace)
ans =
''thing1': '617', 'thing2': '239', 'thing3': 'Unknown', 'morestuff':, '''
Applying it to the entire array:
fixedcell=cellfun(#(x) regexprep(x,missingexpr,replace),mycell);
out=cellfun(#(x) regexp(x,expr3,'tokens','once'),fixedcell,'UniformOutput',false);
This feels a little roundabout, but it works.
cellfun can be replaced with a plain old for loop. Your code will either be equally fast, or maybe even faster. cellfun is implemented with a loop anyway, there is no advantage of using it other than fewer lines of code. In your explicit loop, you can then check the output of regexp, and build your output array any way you like.
I have recently been started to play around with Redis (and am extremely pleased with all the functions it has to offer).
In particular, I was looking for a way to find all elements within a set for which a regex matches.
For example:
>>smember WORDS
1) "person"
2) "saint church"
3) "saint house"
If I wanted to return only the elements where "saint" is present, how would I do so?
I have tried to use sscan as follows:
sscan WORDS match *saint*
for which I get an error.
My understanding was that sscan can return "array of elements is a list of Set members"
Please help! Thanks!
Assuming that you have no more than 1000 elements matching you can use
sscan WORDS 0 match *saint* count 1000
If you want know the exact numbers of the elements in the set you can use SMEMBERS command.
If you want know the exact numbers of the elements in the set that MATCH with your regex, in a single command no a specific command.
Just add a ZERO as the cursor parameter for the SCAN:
sscan WORDS 0 match *saint*
I'm comparing three lexical resources. I use entries from one of them to create queries — see first column — and see if the other two lexicons return the right answers. All wrong answers are written to a text file. Here's a sample out of 3000 lines:
réincarcérer<IND><FUT><REL><SG><1> réincarcèrerais réincarcérerais réincarcérerais
réinsérer<IND><FUT><ABS><PL><1> réinsèrerons réinsérerons réinsérerons
macérer<IND><FUT><ABS><PL><3> macèreront macéreront macéreront
répéter<IND><FUT><ABS><PL><1> répèterons répéterons répéterons
The first column is the query, the second is the reference. The third and fourth columns are the results returned by the lexicons. The values are tab-separated.
I'm trying to identify answers that only differ from the reference by their diacritics. That is, répèterons répéterons should match because the only difference between the two is that the second part has an acute accent on the e rather than a grave accent.
I'd like to match the entire line. I'd be grateful for a regex that would also identify answers that differ by their gemination — the following two lines should match because martellerait has two ls while martèlerait only has one.
modeler<IND><FUT><ABS><SG><2> modelleras modèleras modèleras
marteler<IND><FUT><REL><SG><3> martellerait martèlerait martèlerait
The last two values will always be identical. You can focus on values #2 and 3.
The first part can be achieved by doing a lossy conversion to ASCII and then doing a direct string comparison. Note, converting to ASCII effectively removes the diacritics.
To do the second part is not possible (as far as I know) with a regex pattern. You will need to do some research into things like the Levenshtein distance.
EDIT:
This regex will match duplicate consonants. It might be helpful for your gemination problem.
([b-df-hj-np-tv-xz])\\1+
Which means:
([b-df-hj-np-tv-xz]) # Match only consonants
\\1+ # Match one or times again what was captured in the first capture group
First, I'm using EditPadPro for my regex cleaning, so any answers given should work within that environment.
I get a large spreadsheet full of data that I have to clean every day. I've managed to get it down to a couple of different regexes that I run, and this works... but I'm curious to see if it's possible to reduce down to a single regex.
Here is some sample data:
3-CPC_114851_70095_70095_CAN-bre
3-CPC_114851_70095_70095_CAN
b11-ao1-113775-bre
b7-ao-114441
b7-ao-114441-bre
b7-ao1-114441
b7-ao1-114441-bre
http://go.nlvid.com/results1/?http://bo
go.nlv/results1/?click
b4-sm-1359
b6-sm-1356-bre
1359_195_1453814569-bre
1356_104_1456856729
b15-rad-8905
b15-rad-8905-bre
Here is how the above data needs to end up:
114851-bre
114851
113775-bre
114441
114441-bre
114441
114441-bre
http://go.nlvid.com/results1/
go.nlv/results1/
sm-1359
sm-1356-bre
sm-1359-bre
sm-1356
rad-8905
rad-8905-bre
So, there are numerous rules, such as:
In cases of more than 2 underscores, the result needs to contain only the value immediately after the first underscore, and everything from the dash onwards.
In cases where the string contains "-ao-", "-ao1-", everything prior to the final numeric string should be removed.
If a question mark is present, everything from the mark onwards should be removed.
If the string contains "-sm-" or "-rad-", everything prior to those alpha strings should be removed.
If the string contains 2 underscores, averything after the first numeric string up to a dash
(if present) should be removed, and the string "sm-" should be prepended.
Additionally there is other data that must be left untouched, including but not limited to:
113535|24905|24905
as well as many variations on this pattern of xxxxxx|yyyyy|zzzzz (and not always those string lengths)
This may be asking way too much of regex, I'm not sure as I'm not great with it. But I've seen some pretty impressive things done with it, so I thought I'd put this out to the community and see what you come back with.
Jonathan, I can wrap all of those into one regex, except the last one (where you prepend sm- to a string that does not contain sm). It is not possible in this context, because we cannot capture "sm" to reuse in the replacement, and because there is no "conditional replacement" syntax in EPP.
That being said, you can achieve what you want in EPP with two regexes and one macro to chain the two.
Here is how.
The solution below is tested in EPP.
Regex 1
Press Ctrl + Sh + F to enter Search / Replace mode
Enter the following Search and Replace in the appropriate boxes
At the top right of the Search bar, click the Favorite Searches pull-down, select "Add", give it a name, e.g. Regex 1
Search:
(?mx)^
(?=(?:[^_\r\n]*?_){3})[^_\r\n]+?_([^_\r\n]+)[^-\r\n]+(-[^\r\n]+)?
|
[^\r\n]*?-ao1?-\D*([^\r\n]+)
|
([^\r\n?]*)(?=\?)[^\r\n]+
|
[^\r\n]*?-((?:sm|rad)-[^\r\n]+)
Replace:
\1\2\3\4\5
Regex 2
Same 1-2-3 steps as above.
Search
^(?!(?:[^_\r\n]*?_){3})(?=(?:[^_\r\n]*?_){2})(\d+)(?:[^-\r\n]+(-[^\r\n]+)?)
Replace
sm-\1\2
Chaining Regex 1 and Regex 2
Top menu: Macros, Record Macro, give it a name.
Click the Favorite searches pulldown, select Regex 1
Hit Replace All.
Click the Favorite searches pulldown, select Regex 2
Hit Replace All.
Macros, Stop recording.
Whenever you want to do your sequence of replacements, pull it by name under the Macros menu.
Testing This
I have tested my "Jonathan macro" on your input. Here is the result:
114851-bre
114851
113775-bre
114441
114441-bre
114441
114441-bre
http://go.nlvid.com/results1/
go.nlv/results1/
sm-1359
sm-1356-bre
sm-1359-bre
sm-1356
rad-8905
rad-8905-bre
Try this:
Toggle the Search Panel : SHIFT+CTRL+F
SEARCH: .*?((?:sm-|rad-)?(?:(?:\d+|[\w\.]+\/.*?))(?:-\w+)?$)
REPLACE: $1
Check REGEX and WORDS
Click Replace All or Hit CTRL+ALT+F3
Check the image below:
Example:
let hits = []
:5s/regex-search/\=join(add(hits, submatch(0)))/g
This add all the matches in line 5 to a list.
However it does also a substitute in the text.
I tried to add the 'n' flag after the 'g'
but that doesn't add the matches to the list.
Is there any way to resolve my problem?
Almost there. First I don't think you need the join. Second, add returns the list with the match added. So you can just select the last element of the list to be the replaced element. (This makes it seem like nothing got replaced)
s/regex-search/\=add(hits,submatch(0))[-1]/g
With a recent enough Vim version, you can prevent that the actual substitution does take place (and messes up your undo-branches), while the expression on the right side of an :s command is still being evaluated.
You need at least Vim patch Vim patch 7.3.627 and then you can simply use
:s/foobar/\=add(hits, submatch(0))/gn