I want to convert strings in the format
The European Union - A Very Short Introduction - Pinder, John
to
John Pinder - The European Union - A Very Short Introduction
I am having trouble matching on "Pinder" and "John" to reformat in the desired way.
You can use:
^(.*?)(?:-\s+(\w+),\s+(\w+))$
Demo
If you may have authors with multiple names (such as 'von Clausewitz, Carl') this won't work. Instead, maybe:
^(.*)(?:-\s+([^,]+?),\s+(\w+))$
Demo 2
There are many ways to approach the problem, all requiring some assumptions not specified in your question. Here is one solution...
^.+-\s+(.+),\s+(.+)$
regexper diagram
It is working by consuming as many characters as possible (up to first capture group, using hyphen and whitespace as delimiter) then it assumes there is a comma followed by whitespace separating first name from last name, which it assumes is the end of the string.
Depending on what you know about the uniformity of the data, this may or may not work for you, but I thought it would be nice to have a solution which does not try to restrict characters in name, but rather the rest of the format.
Use this code:
$code = preg_match_all('/(?:.*?) - (?:.*?) -(.*?),(.*)/', $string,$matches);
This will give you an array and $matches[1] will give you the last name (in this case "Pinder") and $matches[2] will give you the first name ("John"). You can then turn it back into a string if you want to using $lastname = implode('',$matches[1]);.
Related
I don't have a great grasp on Regex; but I am attempting to grab names following the word "sortname", but only after the nth time that word appears.
I have (thanks to Wikipedia's API) a list of governors in the United States, listed in order of their states name alphabetically. (https://en.wikipedia.org/w/api.php?action=parse&prop=wikitext&page=List_of_current_United_States_governors§ion=1&format=json)
If you do ctrl+f you will see that each name follows the word "sortname" and there are 50 of them. So if I wanted to see who the Governor of Texas is, I would get the name that follows the 43rd instance of the word "sortname". furthermore the first and last name of each governor is formatted as "sortname|Kay|Ivey" or "sortname|Michelle|Lujan Grisham".
Thanks for the help!
After some more testing I have ended up with the following pattern sortname([^;]*)[^}|]}
It collects more than necessary but its going in the right direction. I can use python to sort it out from there.
Assuming a string str contains the whole text, would you please try:
m = re.findall(r'sortname\|[^|]+\|[^}]+', str, re.DOTALL)
print(m[42])
Output:
sortname|Greg|Abbott
So I had this code working for a few months already, lets say I have a table called Categories, which has a string column called name, so I receive a string and I want to know if any category was mentioned (a mention occur when the string contains the substring: #name_of_a_category), the approach I follow for this was something like below:
categories.select { |category_i| content_received.downcase.match(/##{category_i.downcase}/)}
That worked pretty well until today suddenly started to receive an exception unmatched close parenthesis, I realized that the categories names can contain special chars so I decided to not consider special chars or spaces anymore (don't want to add restrictions to the user and at the same time don't want to deal with those cases so the policy is just to ignore it).
So the question is there a clean way of removing these special chars (maintaining the #) and matching the string (don't want to modify the data just ignore it while looking for mentions)?
You can also use
prep_content_received = content_received.gsub(/[^\w\s]|_/,'')
p categories.select { |c|
prep_content_received.match?(/\b#{c.gsub(/[^\w\s]|_/, '').strip()}\b/i)
}
See the Ruby demo
Details:
The prep_content_received = content_received.gsub(/[^\w\s]|_/,'') creates a copy of content_received with no special chars and _. Using it once reduced overhead if there are a lot of categories
Then, you iterate over the categories list, and each time check if the prep_content_received matches \b (word boundary) + category with all special chars, _ and leading/trailing whitespace stripped from it + \b in a case insensitive way (see the /i flag, no need to .downcase).
So after looking around I found some answers on the platform but nothing with my specific requirements (maybe I missed something, if so please let me know), and this is how I fix it for my case:
content_received = 'pepe is watching a #comedy :)'
categories = ['comedy :)', 'terror']
temp_content = content_received.downcase
categories.select { |category_i| temp_content.gsub(/[^\sa-zA-Z0-9]/, '#' => '#').match?(/##{category_i.downcase.
gsub(/[^\sa-zA-Z0-9]/, '')}/) }
For the sake of the example, I reduced the categories to a simple array of strings, basically the first gsub, remove any character that is not a letter or a number (any special character) and replace each # with an #, the second gsub is a simpler version of the first one.
You can test the snippet above here
I am using string replace method to clean-up column names.
df.columns=df.columns.str.replace("#$%./- ","").str.replace(' ', '_').str.replace('.', '_').str.replace('(','').str.replace(')','').str.replace('.','').str.lower()
Though it works, certainly does not look pythonic. Any suggestion?
I need only A-Za-z and underscore _ if required as column names.
Update:
I tried using Regular expression in the first replace method, but I still need to chain the string like this...
terms.columns=terms.columns.str.replace(r"^[^a-zA-Z1-9]*", '').str.replace(' ', '_').str.replace('(','').str.replace(')','').str.replace('.', '').str.replace(',', '')
Update showing test data:
Original string (Tab separated):
[Sr.No. Course Terms Besic of Education Degree Course Course Approving Authority (i.e Medical Council, etc.) Full form of Course 1 year Duration 2nd year 3rd year Duration 4 th year Duration]
Change column names:
terms.columns=terms.columns.str.replace(r"^[^a-zA-Z1-9]*", '').str.replace(' ', '_').str.replace('(','').str.replace(')','').str.replace('.', '').str.replace(',', '').str.lower()
Output:
['srno', 'course', 'terms', 'besic_of_education', 'degree_course',
'course_approving_authority_ie_medical_council_etc',
'full_form_of_course', '1_year_duration', '2nd_year_',
'3rd_year_duration', '4_th_year_duration']
Above output is correct. The question: Is there any way to achive the same other than the way I have used?
You can use a smaller number of .replace operations by replacing non-word strings with an empty string and subsequently removing the whitespace characters with an underscore.
df.columns.str.replace("[^\w\s]+","").str.replace("\s+","_").str.lower()
I hope this helps.
Looking for a regex that can split expressions like:
A-6-b 10/%XYZ
into:
A-6-b
10%
/XYZ
Note that the first group can also contain spaces and numbers:
AQDF 100 56%/ABC
and percentage can be a float:
SFSDF 0.1%/ABC
I've come up with (^[A-Z\s\d-]*)(?!%)(\d+%)(.*$) but this doe snot match any percentages that are floats and more importantly even simple examples like ABC 10%/XYZ fail because the first digit of the percentage is assigned to the first capturing group.
Any idea how I can achieve what I want? I'm not a regex expert...
EDIT: fixed errors in example
EDIT2:
The examples are not complete. Here one more:
ABC Dwsd 0.01%/XYZ QST
First part can contain spaces
Last Part can contain spaces
number can be a float
Super simple:
/^(.*) ([1-9][0-9]*(?:\.[0-9]+)?%)(.*)$/
The most easily identifiable item is your percentage, so the ([1-9][0-9]*(?:\.[0-9]+)?%) part deals with finding that.
Then it's simply a case of getting everything before (excluding the final space) to get the name, and everything after to get the solvent.
Done.
Don't overcomplicate this by using one unreadable regex.
Based on what you've said, your separators are well defined (the last space and the last %). In JavaScript, for example, you could use:
var str = "A-6-b 10/%XYZ";
var firstSeparator = str.lastIndexOf(' ');
var secondSeparator = str.lastIndexOf('%');
var name = str.substring(0, firstSeparator);
var percentage = str.substring(firstSeparator + 1, secondSeparator + 1); // we want to include the % separator in this one
var solvent = str.substring(secondSeparator + 1);
console.log(name, percentage, solvent);
Working JSFiddle: http://jsfiddle.net/rL5uymhm/
(There may be a typo in your question, as your examples differ on where the / symbol appears. So the code may need tweaking. My point still stands – don't use a regex for the sake of it when there is a more readable alternative.)
IF you really want to use a regex, /^(.+ )([^%]+%)(.*)$/ should work.
I try this Let me know if you have any problem in comment.
((?:(?!\s*[0-9]*\/%).)*)\s*([\d\/%]*)\s*(.*)
SEE DEMO : http://regex101.com/r/lL8oN4/1
This one works for me (using PCRE):
/^(.+) ([0-9.]+)[\/%]+([^\/]+)$/
Is it possible to write a regular expression that matches all strings that does not only contain numbers? If we have these strings:
abc
a4c
4bc
ab4
123
It should match the four first, but not the last one. I have tried fiddling around in RegexBuddy with lookaheads and stuff, but I can't seem to figure it out.
(?!^\d+$)^.+$
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