Develop Reference

Is there a way to find (potentially) multiple results with re.search?

While parsing file names of TV shows, I would like to extract information about them to use for renaming. I have a working model, but it currently uses 28 if/elif statements for every iteration of filename I've seen over the last few years. I'd love to be able to condense this to something that I'm not ashamed of, so any help would be appreciated.
Phase one of this code repentance is to hopefully grab multiple episode numbers. I've gotten as far as the code below, but in the first entry it only displays the first episode number and not all three.
import re
def main():
pattern = '(.*)\.S(\d+)[E(\d+)]+'
strings = ['blah.s01e01e02e03', 'foo.s09e09', 'bar.s05e05']
#print(strings)
for string in strings:
print(string)
result = re.search("(.*)\.S(\d+)[E(\d+)]+", string, re.IGNORECASE)
print(result.group(2))
if __name__== "__main__":
main()
This outputs:
blah.s01e01e02e03
01
foo.s09e09
09
bar.s05e05
05
It's probably trivial, but regular expressions might as well be Cuneiform most days. Thanks in advance!

No. You can use findall to find all e\d+, but it cannot find overlapping matches, which makes it impossible to use s\d+ together with it (i.e. you can't distinguish e02 in "foo.s01e006e007" from that of "age007.s01e001"), and Python doesn't let you use variable-length lookbehind (to make sure s\d+ is before it without overlapping).
The way to do this is to find \.s\d+((?:e\d+)+)$ then split the resultant group 1 in another step (whether by using findall with e\d+, or by splitting with (?<!^)(?=e)).
text = 'blah.s01e01e02e03'
match = re.search(r'\.(s\d+)((?:e\d+)+)$', text, re.I)
season = match.group(1)
episodes = re.findall(r'e\d+', match.group(2), re.I)
print(season, episodes)
# => s01 ['e01', 'e02', 'e03']

re.findall instead of re.search will return a list of all matches

If you can make use of the PyPi regex module you could make use of repeating capture groups in the pattern, and then use .captures()
For example:
import regex
s = "blah.s01e01e02e03"
pattern = r"\.(s\d+)(e\d+)+"
m = regex.search(pattern, s, regex.IGNORECASE)
if m:
print(m.captures(1)[0], m.captures(2))
Output:
s01 ['e01', 'e02', 'e03']
See a Python demo and a regex101 demo.
Or using .capturesdict () with named capture groups.
For example:
import regex
s = "blah.s01e01e02e03"
pattern = r"\.(?P<season>s\d+)(?P<episodes>e\d+)+"
m = regex.search(pattern, s, regex.IGNORECASE)
if m:
print(m.capturesdict())
Output:
{'season': ['s01'], 'episodes': ['e01', 'e02', 'e03']}
See a Python demo.
Note that the notation [E(\d+)] that you used is a character class, that matches 1 or the listed characters like E ( a digit + )

Remove Characters From A String Until A Specific Format is Reached

So I have the following strings and I have been trying to figure out how to manipulate them in such a way that I get a specific format.
string1-itd_jan2021-internal
string2itd_mar2021-space
string3itd_feb2021-internal
string4-itd_mar2021-moon
string5itd_jun2021-internal
string6-itd_feb2021-apollo
I want to be able to get rid of any of the last string so I am just left with the month and year, like below:
string1-itd_jan2021
string2itd_mar2021
string3itd_feb2021
string4-itd_mar2021
string5itd_jun2021
string6-itd_feb2021
I thought about using string.split on the - but then realized that for some strings this wouldn't work. I also thought about getting rid of a set amount of characters by putting it into a list and slicing but the end is varying characters length?
Is there anything I can do it with regex or any other python module?

Use str.rsplit with the appropriate maxsplit parameter:
s = s.rsplit("-", 1)[0]
You could also use str.split (even though this is clearly the worse choice):
s = "-".join(s.split("-")[:-1])
Or using regular expressions:
s = re.sub(r'-[^-]*$', '', s)
# "-[^-]*" a "-" followed by any number of non-"-"

With a regex:
import re
re.sub(r'([0-9]{4}).*$', r'\1', s)

Use re.sub like so:
import re
lines = '''string1-itd_jan2021-internal
string2itd_mar2021-space
string3itd_feb2021-internal
string4-itd_mar2021-moon
string5itd_jun2021-internal
string6-itd_feb2021-apollo'''
for old in lines.split('\n'):
new = re.sub(r'[-][^-]+$', '', old)
print('\t'.join([old, new]))
Prints:
string1-itd_jan2021-internal string1-itd_jan2021
string2itd_mar2021-space string2itd_mar2021
string3itd_feb2021-internal string3itd_feb2021
string4-itd_mar2021-moon string4-itd_mar2021
string5itd_jun2021-internal string5itd_jun2021
string6-itd_feb2021-apollo string6-itd_feb2021
Explanation:
r'[-][^-]+$' : Literal dash (-), followed by any character other than a dash ([^-]) repeated 1 or more times, followed by the end of the string ($).

Match characters and digits of fixed length and one occurance in Python

I have a list in Python with values
['JUL_2018', 'AUG_2018', 'SEP_2018', 'OCT_2018', 'NOV_2018', 'DEC_2018', 'JAN_2019', 'FEB_2019', 'MAR_2019', 'APR_2019', 'MAY_2019', 'JUN_2019', 'MAT_YA_1', 'MAT_TY_1', 'YTD_YA_1', 'YTD_TY_1', 'L3M_YA_1', 'L1M_YA_1']
I want to match only strings where length is 8 and there are 3 characters before underscore and 4 digits after underscore so I eliminate values not required. I am interested only in the MMM_YYYY values from above list.
Tried below and I am not able to filter values like YTD_TY_1 which has multiple underscores.
for c in col_headers:
d= (re.match('^(?=.*\d)(?=.*[A-Z0-9])[A-Z_0-9\d]{8}$',c))
if d:
data_period.append(d[0])

Update: based on #WiktorStribiżew observation that re.match does not require a full string match in Python
The regex I am using is based upon the one that #dvo provided in a comment:
import re
REGEX = '^[A-Z]{3}_[0-9]{4}$'
col_headers = ['JUL_2018', 'AUG_2018', 'SEP_2018', 'OCT_2018', 'NOV_2018', 'DEC_2018', 'JAN_2019', 'FEB_2019', 'MAR_2019', 'APR_2019', 'MAY_2019', 'JUN_2019', 'MAT_YA_1', 'MAT_TY_1', 'YTD_YA_1', 'YTD_TY_1', 'L3M_YA_1', 'L1M_YA_1']
regex = re.compile(REGEX)
data_period = list(filter(regex.search, col_headers))
Once again, based on a comment made by #WiktorStribiżew, if you do not want to match something as "SXX_0012" or "XYZ_0000", you should use the regex he has provided in a comment:
REGEX = r'^(?:JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC)-[0-9]{4}$'

Rather than use regex for this, you should just try to parse it as a date in the first place:
from datetime import datetime
date_fmt = "%b_%Y"
for c in col_headers:
try:
d = datetime.strptime(c, date_fmt)
data_period.append(c) # Or just save the datetime object directly
except ValueError:
pass

The part of this code that is actually doing the matching in your solution is this
[A-Z_0-9\d]{8}
The problem with this is that you're asking to find exactly 8 characters that include A-Z, _, 0-9, and \d. Now, \d is equivalent to 0-9, so you can eliminate that, but that doesn't solve the whole problem, the issue here is that you've encased the entire solution in brackets []. Basically, your string will match anything that is 8 characters long and includes the above characters, ie: A_19_KJ9
What you need to do is specify that you want exactly 3 A-Z characters, then a single _, then 4 \d, see below:
[A-Z]{3}_\d{4}
This will match anything with exactly 3 A-Z characters, then a single _, then 4 \d(any numeric digit)
For a better understanding of regex, I'd encourage you to use an online tool, like regex101

How do I extract some string from a long string in Python?

I have a lot of long strings - not all of them have the same length and content, so that's why I can't use indices - and I want to extract a string from all of them. This is what I want to extract:
http://www.someDomainName.com/anyNumber
SomeDomainName doesn't contain any numbers and and anyNumber is different in each long string. The code should extract the desired string from any string possible and should take into account spaces and any other weird thing that might appear in the long string - should be possible with regex right? -. Could anybody help me with this? Thank you.
Update: I should have said that www. and .com are always the same. Also someDomainName! But there's another http://www. in the string

import re
results = re.findall(r'\bhttp://www\.someDomainName\.com/\d+\b', long_string)

>>> import re
>>> pattern = re.compile("(http://www\\.)(\\w*)(\\.com/)(\\d+)")
>>> matches = pattern.search("http://www.someDomainName.com/2134")
>>> if matches:
print matches.group(0)
print matches.group(1)
print matches.group(2)
print matches.group(3)
print matches.group(4)
http://www.someDomainName.com/2134
http://www.
someDomainName
.com/
2134
In the above pattern, we have captured 5 groups -
One is the complete string that is matched
Rest are in the order of the brackets you see.. (So, you are looking for the second one..) - (\\w*)
If you want, you can capture only the part of the string you are interested in.. So, you can remove the brackets from rest of the pattern that you don't want and just keep (\w*)
>>> pattern = re.compile("http://www\\.(\\w*)\\.com/\\d+")
>>> matches = patter.search("http://www.someDomainName.com/2134")
>>> if matches:
print matches.group(1)
someDomainName
In the above example, you won't have groups - 2, 3 and 4, as in the previous example, as we have captured only 1 group.. And yes group 0 is always captured.. That is the complete string that matches..

Yeah, your simplest bet is regex. Here's something that will probably get the job done:
import re
matcher = re.compile(r'www.(.+).com\/(.+)
matches = matcher.search(yourstring)
if matches:
str1,str2 = matches.groups()

If you are sure that there are no dots in SomeDomainName you can just take the first occurence of the string ".com/" and take everything from that index on
this will avoid you the use of regex which are harder to maintain
exp = 'http://www.aejlidjaelidjl.com/alieilael'
print exp[exp.find('.com/')+5:]

Parsing and reformatting CSV/text data using Python

sorry if this a bit of a beginner's question, but I haven't had much experience with python, and could really use some help in figuring this out. If there is a better programming language for tackling this, I'd be more than open to hearing it
I'm working on a small project, and I have two blocks of data, formatted differently from each other. They're all spreadsheets saved as CSV files, and I'd really like to make one group match the other without having to manually edit all the data.
What I need to do is go through a CSV, and format any data saved like this:
10W
20E
15-16N
17-18S
To a format like this (respective line to respective format):
10,W
20,E
,,15,16,N
,,17,18,S
So that they can just be copied over when opened as spreadsheets
I'm able to get the files into a string in python, but I'm unsure of how to properly write something to search for a number-hyphen-number-letter format.
I'd be immensely grateful for any help I can get. Thanks

This sounds like a good use-case for regular expressions. Once you've split the lines up into individual strings and stripped the whitespace (using s.strip()) these should work (I'm assuming those are cardinal directions; you'll need to change [NESW] to something else if that assumption is incorrect.):
>>> import re
>>> re.findall('\A(\d+)([NESW])', '16N')
[('16', 'N')]
>>> re.findall('\A(\d+)([NESW])', '15-16N')
[]
>>> re.findall('\A(\d+)-(\d+)([NESW])', '15-16N')
[('15', '16', 'N')]
>>> re.findall('\A(\d+)-(\d+)([NESW])', '16N')
[]
The first regex '\A(\d+)([NESW])' matches only a string that begins with a sequence of digits followed by a capital letter N, E, S, or W. The second matches only a string that begins with a sequence of digits followed by a hyphen, followed by another sequence of digits, followed by a capital letter N, E, S, or W. Forcing it to match at the beginning ensures that these regexes don't match a suffix of a longer string.
Then you can do something like this:
>>> vals = re.findall('\A(\d+)([NESW])', '16N')[0]
>>> ','.join(vals)
'16,N'
>>> vals = re.findall('(\d+)-(\d+)([NESW])', '15-16N')[0]
>>> ',,' + ','.join(vals)
',,15,16,N'

This is a whole solution that uses regexs. #senderle has beat me to the answer, so feel free to tick his response. This is just added here as I know how difficult it was to wrap my head around re in my code at first.
import re
dash = re.compile('(\d{2})-(\d{2})([WENS])')
no_dash = re.compile( '(\d{2})([WENS])' )
raw = '''10W
20E
15-16N
17-18S'''
lines = raw.split('\n')
data = []
for l in lines:
if '-' in l:
match = re.search(dash, l).groups()
data.append( ',,%s,%s,%s' % (match[0], match[1], match[2] ) )
else:
match = re.search(no_dash, l).groups()
data.append( '%s,%s' % (match[0], match[1] ) )
print '\n'.join(data)

In your case, I think the quick solution would involve regexps
You can either use the match method to extract your different tokens when they match a given regular expression, or the split method to split your string into tokens given a separator.
However, in your case, the separator would be a single character, so you can use the split method from the str class.