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Python matching various keyword from dictionary issues

I have a complex text where I am categorizing different keywords stored in a dictionary:
text = 'data-ls-static="1">Making Bio Implants, Drug Delivery and 3D Printing in Medicine,MEDICINE</h3>'
sector = {"med tech": ['Drug Delivery' '3D printing', 'medicine', 'medical technology', 'bio cell']}
this can successfully find my keywords and categorize them with some limitations:
pattern = r'[a-zA-Z0-9]+'
[cat for cat in sector if any(x in re.findall(pattern,text) for x in sector[cat])]
The limitations that I cannot solve are:
For example, keywords like "Drug Delivery" that are separated by a space are not recognized and therefore categorized.
I was not able to make the pattern case insensitive, as words like MEDICINE are not recognized. I tried to add (?i) to the pattern but it doesn't work.
The categorized keywords go into a pandas df, but they are printed into []. I tried to loop again the script to take them out but they are still there.
Data to pandas df:
ind_list = []
for site in url_list:
ind = [cat for cat in indication if any(x in re.findall(pattern,soup_string) for x in indication[cat])]
ind_list.append(ind)
websites['Indication'] = ind_list
Current output:
Website Sector Sub-sector Therapeutical Area Focus URL status
0 url3.com [med tech] [] [] [] []
1 www.url1.com [med tech, services] [] [oncology, gastroenterology] [] []
2 www.url2.com [med tech, services] [] [orthopedy] [] []
In the output I get [] that I'd like to avoid.
Can you help me with these points?
Thanks!

Give you some hints here the problem that can readily be spot:
Why can't match keywords like "Drug Delivery" that are separated by a space ? This is because the regex pattern r'[a-zA-Z0-9]+' does not match for a space. You can change it to r'[a-zA-Z0-9 ]+' (added a space after 9) if you want to match also for a space. However, if you want to support other types of white spaces (e.g. \t, \n), you need to further change this regex pattern.
Why don't support case insensitive match ? Your code fragment any(x in re.findall(pattern,text) for x in sector[cat]) requires x to have the same upper/lower case for BOTH being in result of re.findall and being in sector[cat]. This constrain even cannot be bypassed by setting flags=re.I in the re.findall() call. Suggest you to convert them all to the same case before checking. That is, for example change them all to lower cases before matching: any(x in re.findall(pattern,text.lower()) for x.lower() in sector[cat]) Here we added .lower() to both text and x.lower().
With the above 2 changes, it should allow you to capture some categorized keywords.
Actually, for this particular case, you may not need to use regular expression and re.findall at all. You may just check e.g. sector[cat][i].lower()) in text.lower(). That is, change the list comprehension as follows:
[cat for cat in sector if any(x in text.lower() for x in [y.lower() for y in sector[cat]])]
Edit
Test Run with 2-word phrase:
text = 'drug delivery'
sector = {"med tech": ['Drug Delivery', '3D printing', 'medicine', 'medical technology', 'bio cell']}
[cat for cat in sector if any(x in text.lower() for x in [y.lower() for y in sector[cat]])]
Output: # Successfully got the categorizing keyword even with dictionary values of different upper/lower cases
['med tech']
text = 'Drug Store fast delivery'
[cat for cat in sector if any(x in text.lower() for x in [y.lower() for y in sector[cat]])]
Ouptput: # Correctly doesn't match with extra words in between
[]

Can you try a different approach other than regex,
I would suggest difflib when you have two similar matching words.

findall is pretty wasteful here since you are repeatedly breaking up the string for each keyword.
If you want to test whether the keyword is in the string:
[cat for cat in sector if any(re.search(word, text, re.I) for word in sector[cat])]
# Output: med tech

Python replace strings using regex on large dataset

I have recently started using the re package in order to clean up transaction descriptions.
Example of original transaction descriptions:
['bread','payment to facebook.com', 'milk', 'savings', 'amazon.com $xx ased lux', 'holiday_amazon']
For a list of expressions I would like to replace the current description with a better one, e.g. if one of the list entries contains 'facebook' or 'amazon' preceded by a space (or at the beginning of the string), I want to replace the entire list entry by the word 'facebook' or 'amazon' respectively, i.e.:
['bread', 'facebook', 'milk', 'savings', 'amazon', 'holiday_amazon']
As I only want to pick it up if the word facebook is preceded by a space or if it is at the beginning of a word, I have created regex that represent this, e.g. (^|\s)facebook. Note that this is only an example, in reality I want to filter out more complex expressions as well.
In total I have a dataframe with 90 such expressions that I want to replace.
My current code (with minimum workable example) is:
import pandas as pd
import re
def specialCases(list_of_narratives, replacement_dataframe):
# Create output array
new_narratives = []
special_cases_identifiers = replacement_dataframe["REGEX_TEST"]
# For each string element of the list
for memo in list_of_narratives:
index_count = 0
found_count = 0
for i in special_cases_identifiers:
regex = re.compile(i)
if re.search(regex, memo.lower()) is not None:
new_narratives.append(replacement_dataframe["NARRATIVE_TO"].values[index_count].lower())
index_count += 1
found_count += 1
break
else:
index_count += 1
if found_count == 0:
new_narratives.append(memo.lower())
return new_narratives
# Minimum example creation
list_of_narratives = ['bread','payment to facebook.com', 'milk', 'savings', 'amazon.com $xx ased lux', 'holiday_amazon']
list_of_regex_expressions = ['(^|\s)facebook', '(^|\s)amazon']
list_of_regex_replacements = ['facebook', 'amazon']
replacement_dataframe = pd.DataFrame({'REGEX_TEST': list_of_regex_expressions, 'NARRATIVE_TO': list_of_regex_replacements})
# run code
new_narratives = specialCases(list_of_narratives, replacement_dataframe)
However, with over 1 million list entries and 90 different regex expressions to be replaced (i.e. len(list_of_regex_expressions) is 90) this is extremely slow, presumably due to the double for loop.
Could someone help me improve the performance of this code?

Python function to find similarity between differently formatted strings

I have 2 excel files with names of items. I want to compare the items but the only remotely similar column is the name column which too has different formatting of the names like
KIDS-Piano as kids piano
Butter Gel 100mg as Butter-Gel-100MG
I know it can't be 100% accurate so I would instead ask the human operating the code to make the final verification but how do I show the closest matching names?

The proper way of doing this is writing a regular expression.
But the vanilla code below might do the trick as well:
column_a = ["KIDS-Piano", "Butter Gel 100mg"]
column_b = ["kids piano", "Butter-Gel-100MG"]
new_column_a = []
for i in column_a:
# convert strings into lowercase
a = i.lower()
# replace dashes with spaces
a = a.replace('-', ' ')
new_column_a.append(a)
# do the same for column b
new_column_b = []
for i in column_b:
# convert strings into lowercase
a = i.lower()
# replace dashes with spaces
a = a.replace('-', ' ')
new_column_b.append(a)
as_not_found_in_b = []
for i in new_column_a:
if i not in new_column_b:
as_not_found_in_b.append(i)
bs_not_found_in_a = []
for i in new_column_b:
if i not in new_column_a:
bs_not_found_in_a.append(i)
# find the problematic ones and manually fix them
print(as_not_found_in_b)
print(bs_not_found_in_a)

What Regex to use in this example

I am parsing a string that I know will definitely only contain the following distinct phrases that I want to parse:
'Man of the Match'
'Goal'
'Assist'
'Yellow Card'
'Red Card'
The string that I am parsing could contain everything from none of the elements above to all of them (i.e. the string being parsed could be anything from None to 'Man of the Match Goal Assist Yellow Card Red Card'.
For those of you that understand football, you will also realise that the elements 'Goal' and 'Assist' could in theory be repeated an infinite number of times. The element 'Yellow Card' could be repeated 0, 1 or 2 times also.
I have built the following Regex (where 'incident1' is the string being parsed), which I believed would return an unlimited number of all preceding Regexes, however all I am getting is single instances:
regex1 = re.compile("Man of the Match*", re.S)
regex2 = re.compile("Goal*", re.S)
regex3 = re.compile("Assist*", re.S)
regex4 = re.compile("Red Card*", re.S)
regex5 = re.compile("Yellow Card*", re.S)
mysearch1 = re.search(regex1, incident1)
mysearch2 = re.search(regex2, incident1)
mysearch3 = re.search(regex3, incident1)
mysearch4 = re.search(regex4, incident1)
mysearch5 = re.search(regex5, incident1)
#print mystring
print "incident1 = ", incident1
if mysearch1 is not None:
print "Man of the match = ", mysearch1.group()
if mysearch2 is not None:
print "Goal = ", mysearch2.group()
if mysearch3 is not None:
print "Assist = ", mysearch3.group()
if mysearch4 is not None:
print "Red Card = ", mysearch4.group()
if mysearch5 is not None:
print "Yellow Card = ", mysearch5.group()
This works as long as there is only one instance of every element encountered in a string, however if a player was for example to score more than one goal, this code only returns one instance of 'Goal'.
Can anyone see what I am doing wrong?

You can try something like this:
import re
s = "here's an example Man of the Match match and a Red Card match, and another Red Card match"
patterns = [
'Man of the Match',
'Goal',
'Assist',
'Yellow Card',
'Red Card',
]
repattern = '|'.join(patterns)
matches = re.findall(repattern, s, re.IGNORECASE)
print matches # ['Man of the Match', 'Red Card', 'Red Card']

Some general overview on regex methods in python:
re.search | re.match
In your previous attempt, you tried to use re.search. This only returned one result, and as you'll see this isn't unusual. These two functions are used to identify if a line contains a certain regex. You'd use these for something like:
s = subprocess.check_output('ipconfig') # calls ipconfig and sends output to s
for line in s.splitlines():
if re.search("\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}", str(line)):
# if line contains an IP address...
print(line)
You use re.match to specifically check if the regex matches at the BEGINNING of the string. This is usually used with a regex that matches the WHOLE string. For example:
lines = ['Adam Smith, Age: 24, Male, Favorite Thing: Reading page: 16',
'Adam Smith, Age: 16, Male, Favorite Thing: Being a regex example']
# two Adams, but we only want the one who is 16 years old.
repattern = re.compile(r'''Adam \w+, Age: 16, (?:Male|Female), Favorite Thing: [^,]*?''')
for line in lines:
if repattern.match(line):
print(line)
# Adam Smith, Age: 16, Male, Favorite Thing: Being a regex example
# note if we'd used re.search for Age: 16, it would have found both lines!
The take away is that you use these two functions to select lines in a longer document (or any iterable)
re.findall | re.finditer
It seems in this case, you aren't trying to match a line, you're trying to pull some specifically-formatted information from the string. Let's see some examples of that.
s = """Phone book:
Adam: (555)123-4567
Joe: (555)987-6543
Alice:(555)135-7924"""
pat = r'''(?:\(\d{3}\))?\d{3}-?\d{4}'''
phone_numbers = re.findall(pat, s)
print(phone_numbers)
# ['(555)123-4567','(555)987-6543','(555)135-7924']
re.finditer returns a generator instead of a list. You'd use this the same way you'd use xrange instead of range in Python2. re.findall(some_pattern, some_string) can make a GIANT list if there are a TON of matches. re.finditer will not.
other methods: re.split | re.sub
re.split is great if you have a number of things you need to split by. Imagine you had the string:
s = '''Hello, world! It's great that you're talking to me, and everything, but I'd really rather you just split me on punctuation marks. Okay?'''
There's no great way to do that with str.split like you're used to, so instead do:
separators = [".", "!", "?", ","]
splitpattern = '|'.join(map(re.escape, separators))
# re.escape takes a string and escapes out any characters that regex considers
# special, for instance that . would otherwise be "any character"!
split_s = re.split(splitpattern, s)
print(split_s)
# ['Hello', ' world', " It's great that you're talking to me", ' and everything', " but I'd really rather you just split me on punctuation marks", ' Okay', '']
re.sub is great in cases where you know something will be formatted regularly, but you're not sure exactly how. However, you REALLY want to make sure they're all formatted the same! This will be a little advanced and use several methods, but stick with me....
dates = ['08/08/2014', '09-13-2014', '10.10.1997', '9_29_09']
separators = list()
new_sep = "/"
match_pat = re.compile(r'''
\d{1,2} # two digits
(.) # followed by a separator (capture)
\d{1,2} # two more digits
\1 # a backreference to that separator
\d{2}(?:\d{2})? # two digits and optionally four digits''', re.X)
for idx,date in enumerate(dates):
match = match_pat.match(date)
if match:
sep = match.group(1) # the separator
separators.append(sep)
else:
dates.pop(idx) # this isn't really a date, is it?
repl_pat = '|'.join(map(re.escape, separators))
final_dates = re.sub(repl_pat, new_sep, '\n'.join(dates))
print(final_dates)
# 08/08/2014
# 09/13/2014
# 10/10/1997
# 9/29/09
A slightly less advanced example, you can use re.sub with any sort of formatted expression and pass it a function to return! For instance:
def get_department(dept_num):
departments = {'1': 'I.T.',
'2': 'Administration',
'3': 'Human Resources',
'4': 'Maintenance'}
if hasattr(dept_num, 'group'): # then it's a match, not a number
dept_num = dept_num.group(0)
return departments.get(dept_num, "Unknown Dept")
file = r"""Name,Performance Review,Department
Adam,3,1
Joe,5,2
Alice,1,3
Eve,12,4""" # this looks like a csv file
dept_names = re.sub(r'''\d+$''', get_department, file, flags=re.M)
print(dept_names)
# Name,Performance Review,Department
# Adam,3,I.T.
# Joe,5,Administration
# Alice,1,Human Resources
# Eve,12,Maintenance
Without using regex here you could do:
replaced_lines = []
departments = {'1': 'I.T.',
'2': 'Administration',
'3': 'Human Resources',
'4': 'Maintenance'}
for line in file.splitlines():
the_split_line = line.split(',')
replaced_lines.append(','.join(the_split_line[:-1]+ \
departments.get(the_split_line[-1], "Unknown Dept")))
new_file = '\n'.join(replaced_lines)
# LOTS OF STRING MANIPULATION, YUCK!
Instead we replace all that for loop and string splitting, list slicing, and string manipulation with a function and a re.sub call. In fact, if you use a lambda it's even easier!
departments = {'1': 'I.T.',
'2': 'Administration',
'3': 'Human Resources',
'4': 'Maintenance'}
re.sub(r'''\d+$''', lambda x: departments.get(x, "Unknown Dept"), file, flags=re.M)
# DONE!

How to strip variable spaces in each line of a text file based on special condition - one-liner in Python?

I have some data (text files) that is formatted in the most uneven manner one could think of. I am trying to minimize the amount of manual work on parsing this data.
Sample Data :
Name Degree CLASS CODE EDU Scores
--------------------------------------------------------------------------------------
John Marshall CSC 78659944 89989 BE 900
Think Code DB I10 MSC 87782 1231 MS 878
Mary 200 Jones CIVIL 98993483 32985 BE 898
John G. S Mech 7653 54 MS 65
Silent Ghost Python Ninja 788505 88448 MS Comp 887
Conditions :
More than one spaces should be compressed to a delimiter (pipe better? End goal is to store these files in the database).
Except for the first column, the other columns won't have any spaces in them, so all those spaces can be compressed to a pipe.
Only the first column can have multiple words with spaces (Mary K Jones). The rest of the columns are mostly numbers and some alphabets.
First and second columns are both strings. They almost always have more than one spaces between them, so that is how we can differentiate between the 2 columns. (If there is a single space, that is a risk I am willing to take given the horrible formatting!).
The number of columns varies, so we don't have to worry about column names. All we want is to extract each column's data.
Hope I made sense! I have a feeling that this task can be done in a oneliner. I don't want to loop, loop, loop :(
Muchos gracias "Pythonistas" for reading all the way and not quitting before this sentence!

It still seems tome that there's some format in your files:
>>> regex = r'^(.+)\b\s{2,}\b(.+)\s+(\d+)\s+(\d+)\s+(.+)\s+(\d+)'
>>> for line in s.splitlines():
lst = [i.strip() for j in re.findall(regex, line) for i in j if j]
print(lst)
[]
[]
['John Marshall', 'CSC', '78659944', '89989', 'BE', '900']
['Think Code DB I10', 'MSC', '87782', '1231', 'MS', '878']
['Mary 200 Jones', 'CIVIL', '98993483', '32985', 'BE', '898']
['John G. S', 'Mech', '7653', '54', 'MS', '65']
['Silent Ghost', 'Python Ninja', '788505', '88448', 'MS Comp', '887']
Regex is quite straightforward, the only things you need to pay attention to are the delimiters (\s) and the word breaks (\b) in case of the first delimiter. Note that when the line wouldn't match you get an empty list as lst. That would be a read flag to bring up the user interaction described below. Also you could skip the header lines by doing:
>>> file = open(fname)
>>> [next(file) for _ in range(2)]
>>> for line in file:
... # here empty lst indicates issues with regex
Previous variants:
>>> import re
>>> for line in open(fname):
lst = re.split(r'\s{2,}', line)
l = len(lst)
if l in (2,3):
lst[l-1:] = lst[l-1].split()
print(lst)
['Name', 'Degree', 'CLASS', 'CODE', 'EDU', 'Scores']
['--------------------------------------------------------------------------------------']
['John Marshall', 'CSC', '78659944', '89989', 'BE', '900']
['Think Code DB I10', 'MSC', '87782', '1231', 'MS', '878']
['Mary 200 Jones', 'CIVIL', '98993483', '32985', 'BE', '898']
['John G. S', 'Mech', '7653', '54', 'MS', '65']
another thing to do is simply allow user to decide what to do with questionable entries:
if l < 3:
lst = line.split()
print(lst)
iname = input('enter indexes that for elements of name: ') # use raw_input in py2k
idegr = input('enter indexes that for elements of degree: ')
Uhm, I was all the time under the impression that the second element might contain spaces, since it's not the case you could just do:
>>> for line in open(fname):
name, _, rest = line.partition(' ')
lst = [name] + rest.split()
print(lst)

Variation on SilentGhost's answer, this time first splitting the name from the rest (separated by two or more spaces), then just splitting the rest, and finally making one list.
import re
for line in open(fname):
name, rest = re.split('\s{2,}', line, maxsplit=1)
print [name] + rest.split()

This answer was written after the OP confessed to changing every tab ("\t") in his data to 3 spaces (and not mentioning it in his question).
Looking at the first line, it seems that this is a fixed-column-width report. It is entirely possible that your data contains tabs that if expanded properly might result in a non-crazy result.
Instead of doing line.replace('\t', ' ' * 3) try line.expandtabs().
Docs for expandtabs are here.
If the result looks sensible (columns of data line up), you will need to determine how you can work out the column widths programatically (if that is possible) -- maybe from the heading line.
Are you sure that the second line is all "-", or are there spaces between the columns?
The reason for asking is that I once needed to parse many different files from a database query report mechanism which presented the results like this:
RecordType ID1 ID2 Description
----------- -------------------- ----------- ----------------------
1 12345678 123456 Widget
4 87654321 654321 Gizmoid
and it was possible to write a completely general reader that inspected the second line to determine where to slice the heading line and the data lines. Hint:
sizes = map(len, dash_line.split())
If expandtabs() doesn't work, edit your question to show exactly what you do have i.e. show the result of print repr(line) for the first 5 or so lines (including the heading line). It might also be useful if you could say what software produces these files.