I have a '3 x 1' cell array the contents of which appear like the following:
'ASDF_LE_NEWYORK Fixedafdfgd_ML'
'Majo_LE_WASHINGTON FixedMonuts_ML'
'Array_LE_dfgrt_fdhyuj_BERLIN Potato Price'
I want to be able to elegantly extract and create another '3x1' cell array with contents as:
'NEWYORK'
'WASHINGTON'
'BERLIN'
If you notice in above the NAME's are after the last underscore and before the first SPACE or '_ML'. How do I write such code in a concise manner.
Thanks
Edit:
Sorry guys I should have used a better example. I have it corrected now.
You can use lookbehind for _ and lookahead for space:
names = regexp(A, '(?<=_)[^\s_]*(?=\s)', 'match', 'once');
Where A is the cell array containing the strings:
A = {...
'ASDF_LE_NEWYORK Fixedafdfgd_ML'
'Majo_LE_WASHINGTON FixedMonuts_ML'
'Array_LE_dfgrt_fdhyuj_BERLIN Potato Price'};
>> names = regexp(A, '(?<=_)[^\s_]*(?=\s)', 'match', 'once')
names =
'NEWYORK'
'WASHINGTON'
'BERLIN'
NOTE: The question was changed, so the answer is no longer complete, but hopefully the regexp example is still useful.
Try regexp like this:
names = regexp(fullNamesCell,'_(NAME\d?)\s','tokens');
names = cellfun(#(x)(x{1}),names)
In the pattern _(NAME\d?)\s, the parenthesis define a subexpression, which will be returned as a token (a portion of matched text). The \d? specifies zero or one digits, but you could use \d{1} for exactly one digit or \d{1,3} if you expect between 1 and 3 digits. The \s specified whitespace.
The reorganization of names is a little convoluted, but when you use regexp with a cell input and tokens you get a cell of cells that needs some reformatting for your purposes.
Related
I have a string, and I want to extract, using regular expressions, groups of characters that are between the character : and the other character /.
typically, here is a string example I'm getting:
'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh'
and so, I want to retrieved, 45.72643,4.91203 and also hereanotherdata
As they are both between characters : and /.
I tried with this syntax in a easier string where there is only 1 time the pattern,
[tt]=regexp(str,':(\w.*)/','match')
tt = ':45.72643,4.91203/'
but it works only if the pattern happens once. If I use it in string containing multiples times the pattern, I get all the string between the first : and the last /.
How can I mention that the pattern will occur multiple time, and how can I retrieve it?
Use lookaround and a lazy quantifier:
regexp(str, '(?<=:).+?(?=/)', 'match')
Example (Matlab R2016b):
>> str = 'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
>> result = regexp(str, '(?<=:).+?(?=/)', 'match')
result =
1×2 cell array
'45.72643,4.91203' 'hereanotherdata'
In most languages this is hard to do with a single regexp. Ultimately you'll only ever get back the one string, and you want to get back multiple strings.
I've never used Matlab, so it may be possible in that language, but based on other languages, this is how I'd approach it...
I can't give you the exact code, but a search indicates that in Matlab there is a function called strsplit, example...
C = strsplit(data,':')
That should will break your original string up into an array of strings, using the ":" as the break point. You can then ignore the first array index (as it contains text before a ":"), loop the rest of the array and regexp to extract everything that comes before a "/".
So for instance...
'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh'
Breaks down into an array with parts...
1 - 'abcd'
2 - '45.72643,4.91203/Rou'
3 - 'hereanotherdata/defgh'
Then Ignore 1, and extract everything before the "/" in 2 and 3.
As John Mawer and Adriaan mentioned, strsplit is a good place to start with. You can use it for both ':' and '/', but then you will not be able to determine where each of them started. If you do it with strsplit twice, you can know where the ':' starts :
A='abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
B=cellfun(#(x) strsplit(x,'/'),strsplit(A,':'),'uniformoutput',0);
Now B has cells that start with ':', and has two cells in each cell that contain '/' also. You can extract it with checking where B has more than one cell, and take the first of each of them:
C=cellfun(#(x) x{1},B(cellfun('length',B)>1),'uniformoutput',0)
C =
1×2 cell array
'45.72643,4.91203' 'hereanotherdata'
Starting in 16b you can use extractBetween:
>> str = 'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
>> result = extractBetween(str,':','/')
result =
2×1 cell array
{'45.72643,4.91203'}
{'hereanotherdata' }
If all your text elements have the same number of delimiters this can be vectorized too.
I am using string replace method to clean-up column names.
df.columns=df.columns.str.replace("#$%./- ","").str.replace(' ', '_').str.replace('.', '_').str.replace('(','').str.replace(')','').str.replace('.','').str.lower()
Though it works, certainly does not look pythonic. Any suggestion?
I need only A-Za-z and underscore _ if required as column names.
Update:
I tried using Regular expression in the first replace method, but I still need to chain the string like this...
terms.columns=terms.columns.str.replace(r"^[^a-zA-Z1-9]*", '').str.replace(' ', '_').str.replace('(','').str.replace(')','').str.replace('.', '').str.replace(',', '')
Update showing test data:
Original string (Tab separated):
[Sr.No. Course Terms Besic of Education Degree Course Course Approving Authority (i.e Medical Council, etc.) Full form of Course 1 year Duration 2nd year 3rd year Duration 4 th year Duration]
Change column names:
terms.columns=terms.columns.str.replace(r"^[^a-zA-Z1-9]*", '').str.replace(' ', '_').str.replace('(','').str.replace(')','').str.replace('.', '').str.replace(',', '').str.lower()
Output:
['srno', 'course', 'terms', 'besic_of_education', 'degree_course',
'course_approving_authority_ie_medical_council_etc',
'full_form_of_course', '1_year_duration', '2nd_year_',
'3rd_year_duration', '4_th_year_duration']
Above output is correct. The question: Is there any way to achive the same other than the way I have used?
You can use a smaller number of .replace operations by replacing non-word strings with an empty string and subsequently removing the whitespace characters with an underscore.
df.columns.str.replace("[^\w\s]+","").str.replace("\s+","_").str.lower()
I hope this helps.
I am trying to add a new calculated column that counts the number of semi colons in a string and adds one to it. So the column i have contains a bunch of aliases and I need to know how many for each row.
For example,
A; B; C; D
So basically this means there are 4 aliases (3 semi colons + 1)
Need to do this for over 2 million rows. Help please!
Basic idea is to subtract length of your string without ; characters from it's original length:
len([columnName])-len(Substitute([columnName],";",""))+1
Here it is with a regular expression:
Len(RXReplace([Column 1], "(?!;).", "", "gis"))+1
RXReplace takes as arguments:
The string you are wanting to work on (in this case it is on Column 1)
The regular expression you want to use (here it is (?!;). )
What you want to replace matches with (blank in this situation so
that everything that matches the regex is removed)
Finally a parameter saying how you want it to work (we are passing
in gis which means replace all matches not just the first, ignore case, replace newlines)
We wrap this in a Len which gives us the amount of semicolons since that is all that is left and finally we add 1 to it to get the final result.
You can read more about the regular expression here: https://msdn.microsoft.com/en-us/library/az24scfc(v=vs.110).aspx but in a nutshell it says match everything that isn't a semi colon.
You can read more about RXReplace and Len here: https://docs.tibco.com/pub/spotfire/6.0.0-november-2013/userguide-webhelp/ncfe/ncfe_text_functions.htm
A Bloomberg futures ticker usually looks like:
MCDZ3 Curcny
where the root is MCD, the month letter and year is Z3 and the 'yellow key' is Curcny.
Note that the root can be of variable length, 2-4 letters or 1 letter and 1 whitespace (e.g. S H4 Comdty).
The letter-year allows only the letter listed below in expr and can have two digit years.
Finally the yellow key can be one of several security type strings but I am interested in (Curncy|Equity|Index|Comdty) only.
In Matlab I have the following regular expression
expr = '[FGHJKMNQUVXZ]\d{1,2} ';
[rootyk, monthyear] = regexpi(bbergtickers, expr,'split','match','once');
where
rootyk{:}
ans =
'mcd' 'curncy'
and
monthyear =
'z3 '
I don't want to match the ' ' (space) in the monthyear. How can I do?
Assuming there are no leading or trailing whitespaces and only upcase letters in the root, this should work:
^([A-Z]{2,4}|[A-Z]\s)([FGHJKMNQUVXZ]\d{1,2}) (Curncy|Equity|Index|Comdty)$
You've got root in the first group, letter-year in the second, yellow key in the third.
I don't know Matlab nor whether it covers Perl Compatible Regex. If it fails, try e.g. with instead of \s. Also, drop the ^...$ if you'd like to extract from a bigger source text.
The expression you're feeding regexpi with contains a space and is used as a pattern for 'match'. This is why the matched monthyear string also has a space1.
If you want to keep it simple and let regexpi do the work for you (instead of postprocessing its output), try a different approach and capture tokens instead of matching, and ignore the intermediate space:
%// <$1><----------$2---------> <$3>
expr = '(.+)([FGHJKMNQUVXZ]\d{1,2}) (.+)';
tickinfo = regexpi(bbergtickers, expr, 'tokens', 'once');
You can also simplify the expression to a more genereic '(.+)(\w{1}\d{1,2})\s+(.+)', if you wish.
Example
bbergtickers = 'MCDZ3 Curncy';
expr = '(.+)([FGHJKMNQUVXZ]\d{1,2})\s+(.+)';
tickinfo = regexpi(bbergtickers, expr, 'tokens', 'once');
The result is:
tickinfo =
'MCD'
'Z3'
'Curncy'
1 This expression is also used as a delimiter for 'split'. Removing the trailing space from it won't help, as it will reappear in the rootyk output instead.
Assuming you just want to get rid of the leading and or trailing spaces at the edge, there is a very simple command for that:
monthyear = trim(monthyear)
For removing all spaces, you can do:
monthyear(isspace(monthyear))=[]
Here is a completely different approach, basically this searches the letter before your year number:
s = 'MCDZ3 Curcny'
p = regexp(s,'\d')
s(min(p)
s(min(p)-1:max(p))
Is it possible to write a regular expression that matches all strings that does not only contain numbers? If we have these strings:
abc
a4c
4bc
ab4
123
It should match the four first, but not the last one. I have tried fiddling around in RegexBuddy with lookaheads and stuff, but I can't seem to figure it out.
(?!^\d+$)^.+$
This says lookahead for lines that do not contain all digits and match the entire line.
Unless I am missing something, I think the most concise regex is...
/\D/
...or in other words, is there a not-digit in the string?
jjnguy had it correct (if slightly redundant) in an earlier revision.
.*?[^0-9].*
#Chad, your regex,
\b.*[a-zA-Z]+.*\b
should probably allow for non letters (eg, punctuation) even though Svish's examples didn't include one. Svish's primary requirement was: not all be digits.
\b.*[^0-9]+.*\b
Then, you don't need the + in there since all you need is to guarantee 1 non-digit is in there (more might be in there as covered by the .* on the ends).
\b.*[^0-9].*\b
Next, you can do away with the \b on either end since these are unnecessary constraints (invoking reference to alphanum and _).
.*[^0-9].*
Finally, note that this last regex shows that the problem can be solved with just the basics, those basics which have existed for decades (eg, no need for the look-ahead feature). In English, the question was logically equivalent to simply asking that 1 counter-example character be found within a string.
We can test this regex in a browser by copying the following into the location bar, replacing the string "6576576i7567" with whatever you want to test.
javascript:alert(new String("6576576i7567").match(".*[^0-9].*"));
/^\d*[a-z][a-z\d]*$/
Or, case insensitive version:
/^\d*[a-z][a-z\d]*$/i
May be a digit at the beginning, then at least one letter, then letters or digits
Try this:
/^.*\D+.*$/
It returns true if there is any simbol, that is not a number. Works fine with all languages.
Since you said "match", not just validate, the following regex will match correctly
\b.*[a-zA-Z]+.*\b
Passing Tests:
abc
a4c
4bc
ab4
1b1
11b
b11
Failing Tests:
123
if you are trying to match worlds that have at least one letter but they are formed by numbers and letters (or just letters), this is what I have used:
(\d*[a-zA-Z]+\d*)+
If we want to restrict valid characters so that string can be made from a limited set of characters, try this:
(?!^\d+$)^[a-zA-Z0-9_-]{3,}$
or
(?!^\d+$)^[\w-]{3,}$
/\w+/:
Matches any letter, number or underscore. any word character
.*[^0-9]{1,}.*
Works fine for us.
We want to use the used answer, but it's not working within YANG model.
And the one I provided here is easy to understand and it's clear:
start and end could be any chars, but, but there must be at least one NON NUMERICAL characters, which is greatest.
I am using /^[0-9]*$/gm in my JavaScript code to see if string is only numbers. If yes then it should fail otherwise it will return the string.
Below is working code snippet with test cases:
function isValidURL(string) {
var res = string.match(/^[0-9]*$/gm);
if (res == null)
return string;
else
return "fail";
};
var testCase1 = "abc";
console.log(isValidURL(testCase1)); // abc
var testCase2 = "a4c";
console.log(isValidURL(testCase2)); // a4c
var testCase3 = "4bc";
console.log(isValidURL(testCase3)); // 4bc
var testCase4 = "ab4";
console.log(isValidURL(testCase4)); // ab4
var testCase5 = "123"; // fail here
console.log(isValidURL(testCase5));
I had to do something similar in MySQL and the following whilst over simplified seems to have worked for me:
where fieldname regexp ^[a-zA-Z0-9]+$
and fieldname NOT REGEXP ^[0-9]+$
This shows all fields that are alphabetical and alphanumeric but any fields that are just numeric are hidden. This seems to work.
example:
name1 - Displayed
name - Displayed
name2 - Displayed
name3 - Displayed
name4 - Displayed
n4ame - Displayed
324234234 - Not Displayed