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How to write word boundary inside character class in python without losing its meaning? I wish to add underscore(_) in definition of word boundary(\b)
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Closed 2 years ago.
I have to handle a long filename in a specific format that contains two dates and someone's full name. Here is a template that describes this format:
firstname_middlename_lastname_yyyy-mm-dd_text1_text2_yyyy-mm-dd.xls
How to extract the fullname, first date, and second date from that filename using regular expression?
I've tried to extract the first date like:
string1 = 'CHEN_MOU_MOU_1999-04-11_Scientific_Report_2020-03-14.xlsx'
ptn = re.compile('\b(\d{4}-\d{2}-\d{2})\b')
print(ptn.match(string1))
But it doesn't seem to work. The output I get is None.
Any help will be appreciated.
The reason your solution does not work is because _ is considered an alphanumeric character in Python.
From docs:
\w
Matches any alphanumeric character; this is equivalent to the class [a-zA-Z0-9_].
So \b does not match _ in your string. But it'll match -.
From docs:
\b
This is a zero-width assertion that matches only at the beginning or end of a word. A word is defined as a sequence of alphanumeric characters, so the end of a word is indicated by whitespace or a non-alphanumeric character.
But if you replace _ around your dates with a - (hyphen), then your solution works just fine.
>>> import re
>>> string1 = 'CHEN_MOU_MOU-1999-04-11-Scientific Report-2020-03-14.xlsx'
>>> ptn = re.compile(r'\b(\d{4}-\d{2}-\d{2})\b')
>>> ptn.findall(string1)
['1999-04-11', '2020-03-14']
Following is a solution that should work for your task:
$ python
Python 3.7.3 (v3.7.3:ef4ec6ed12, Mar 25 2019, 21:26:53) [MSC v.1916 32 bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import re
>>> string1 = 'CHEN_MOU_MOU_1999-04-11_Scientific_Report_2020-03-14.xlsx'
>>> fullnamepattern = r'[a-zA-Z]+_[a-zA-Z]+_[a-zA-Z]+'
>>> datepattern = r'\d{4}-\d{2}-\d{2}'
>>> re.search(fullnamepattern, string1).group()
'CHEN_MOU_MOU'
>>> re.findall(datepattern, string1)
['1999-04-11', '2020-03-14']
My two cents...
import re
pattern = re.compile(r'^(.*?)_(?=\d)(.*?)_(.*)_(.*?)\.(.*)$')
string = 'CHEN_MOU_MOU_1999-04-11_Scientific_Report_2020-03-14.xlsx'
split_filename = pattern.findall(string)
split_filename
Output:
[('CHEN_MOU_MOU', '1999-04-11', 'Scientific_Report', '2020-03-14', 'xlsx')]
Running...
split_filename[0][3]
shows...
'2020-03-14'
At least this way you can pick what you would like to read/check/record. Just change the [3] above to another number to get a different part...
split_filename[0][0]
'CHEN_MOU_MOU'
split_filename[0][4]
'xlsx'
This will work for you:
\d[^_\.]+
\d match a digit
[^_\.]+ match all except "." or "_"
Try this to extract all dates with the format yyyy-mm-dd in the string :
string1 = 'CHEN_MOU_MOU_1999-04-11_Scientific_Report_2020-03-14.xlsx'
ptn = re.compile("\d{4}-\d{2}-\d{2}")
all_dates = ptn.findall(string1)
# ['1999-04-11', '2020-03-14']
full_name = " ".join(string1.split('_')[:3])
# 'CHEN MOU MOU'
You can extract dates using reguler expression but you can extract full name easily using split function try following code to get more idea
string1 = 'CHEN_MOU_MOU_1999-04-11_Scientific_Report_2020-03-14.xlsx'
pattern = re.compile("\d{4}-\d{2}-\d{2}")
dates=pattern.findall(string1) # it will return first date and last date in dates's array
fullname = string1.split("_") # split data using _ character and it stored in array
fullname = " ".join(fullname[:3]) #join with fullname blank space (first three data)
Or you can join last two line in one
fullname = " ".join(string1.split("_")[:3])
Please let me know what you think about it
Related
This is in python
Input string:
Str = 'Y=DAT,X=ZANG,FU=_COG-GAB-CANE-,FU=FARE,T=TART,RO=TOP,FU=#-_MAP.com-,Z=TRY'
Expected output
'FU=_COG-GAB-CANE_,FU=FARE,FU=#-_MAP.com_'
here 'FU=' is the occurence we are looking for and the value which follows FU=
return all occurrences of FU=(with the associated value for FU=) in a comma-separated string, they can occur anywhere within the string and special characters are allowed.
Here is one approach.
>>> import re
>>> str_ = 'Y=DAT,X=ZANG,FU=FAT,T=TART,FU=GEM,RO=TOP,FU=MAP,Z=TRY'
>>> re.findall.__doc__[:58]
'Return a list of all non-overlapping matches in the string'
>>> re.findall(r'FU=\w+', str_)
['FU=FAT', 'FU=GEM', 'FU=MAP']
>>> ','.join(re.findall(r'FU=\w+', str_))
'FU=FAT,FU=GEM,FU=MAP'
Got it working
Python Code
import re
str_ = 'Y=DAT,X=ZANG,FU=_COG-GAB-CANE-,FU=FARE,T=TART,RO=TOP,FU=#-_MAP.com-,Z=TRY'
str2='FU='+',FU='.join(re.findall(r'FU=(.*?),', str_))
print(str2)
Gives the desired output:
'FU=_COG-GAB-CANE-,FU=FARE,FU=#-_MAP.com-'
Successfully gives me all the occurrences of FU= followed by values, irrespective of order and number of special characters.
Although a bit unclean way as I am manually adding FU= for the first occurrence.
Please suggest if there is a cleaner way of doing it ? , but yes it gets the work done.
A columns in data frame contains the keywords I want to match with.
I want to check if each column contains any of the keywords. If yes, print them.
Tried below:
import pandas as pd
import re
Keywords = [
"Caden(S, A)",
"Caden(a",
"Caden(.A))",
"Caden.Q",
"Caden.K",
"Caden"
]
data = {'People' : ["Caden(S, A) Charlotte.A, Caden.K;", "Emily.P Ethan.B; Caden(a", "Grayson.Q, Lily; Caden(.A))", "Mason, Emily.Q Noah.B; Caden.Q - Riley.P"]}
df = pd.DataFrame(data)
pat = '|'.join(r"\b{}\b".format(x) for x in Keywords)
df["found"] = df['People'].str.findall(pat).str.join('; ')
print df["found"]
It returns Nan. I guess the challenge lies in the spaces and brackets in the keywords.
What's the right way to get the ideal outputs? Thank you.
Caden(S, A); Caden.K
Caden(a
Caden(.A))
Caden.Q
Since you do not need to find every keyword, but the longest ones if they are overlapping you may use a regex with findall approach.
The point here is that you need to sort the keywords by length in the descending order first (because there are whitespaces in them), then you need to escape these values as they contain special characters, then you must amend the word boundaries to use unambiguous word boundaries, (?<!\w) and (?!\w) (note that \b is context dependent).
Use
pat = r'(?<!\w)(?:{})(?!\w)'.format('|'.join(map(re.escape, sorted(Keywords,key=len,reverse=True))))
See an online Python test:
import re
Keywords = ["Caden(S, A)", "Caden(a","Caden(.A))", "Caden.Q", "Caden.K", "Caden"]
rx = r'(?<!\w)(?:{})(?!\w)'.format('|'.join(map(re.escape, sorted(Keywords,key=len,reverse=True))))
# => (?<!\w)(?:Caden\(S,\ A\)|Caden\(\.A\)\)|Caden\(a|Caden\.Q|Caden\.K|Caden)(?!\w)
strs = ["Caden(S, A) Charlotte.A, Caden.K;", "Emily.P Ethan.B; Caden(a", "Grayson.Q, Lily; Caden(.A))", "Mason, Emily.Q Noah.B; Caden.Q - Riley.P"]
for s in strs:
print(re.findall(rx, s))
Output
['Caden(S, A)', 'Caden.K']
['Caden(a']
['Caden(.A))']
['Caden.Q']
Hey don't know if this solution is optimal but it works. I just replaced dot by 8 and '(' by 6 and ')' by 9 don't know why those character are ignored by str.findall ?
A kind of bijection between {8,6,9} and {'.','(',')'}
for i in range(len(Keywords)):
Keywords[i] = Keywords[i].replace('(','6').replace(')','9').replace('.','8')
for i in range(len(df['People'])):
df['People'][i] = df['People'][i].replace('(','6').replace(')','9').replace('.','8')
And then you apply your function
pat = '|'.join(r"\b{}\b".format(x) for x in Keywords)
df["found"] = df['People'].str.findall(pat).str.join('; ')
Final step get back the {'.','(',')'}
for i in range(len(df['found'])):
df['found'][i] = df['found'][i].replace('6','(').replace('9',')').replace('8','.')
df['People'][i] = df['People'][i].replace('6','(').replace('9',')').replace('8','.')
Voilà
I have a input text like this (actual text file contains tons of garbage characters surrounding these 2 string too.)
(random_garbage_char_here)**value=xxx**;(random_garbage_char_here)**value=yyy**;(random_garbage_char_here)
I am trying to parse the text to store something like this:
value1="xxx" and value2="yyy".
I wrote python code as follows:
value1_start = content.find('value')
value1_end = content.find(';', value1_start)
value2_start = content.find('value')
value2_end = content.find(';', value2_start)
print "%s" %(content[value1_start:value1_end])
print "%s" %(content[value2_start:value2_end])
But it always returns:
value=xxx
value=xxx
Could anyone tell me how can I parse the text so that the output is:
value=xxx
value=yyy
Use a regex approach:
re.findall(r'\bvalue=[^;]*', s)
Or - if value can be any 1+ word (letter/digit/underscore) chars:
re.findall(r'\b\w+=[^;]*', s)
See the regex demo
Details:
\b - word boundary
value= - a literal char sequence value=
[^;]* - zero or more chars other than ;.
See the Python demo:
import re
rx = re.compile(r"\bvalue=[^;]*")
s = "$%$%&^(&value=xxx;$%^$%^$&^%^*value=yyy;%$#^%"
res = rx.findall(s)
print(res)
Use regex to filter the data you want from the "junk characters":
>>> import re
>>> _input = '#4#5%value=xxx38u952035983049;3^&^*(^%$3value=yyy#%$#^&*^%;$#%$#^'
>>> matches = re.findall(r'[a-zA-Z0-9]+=[a-zA-Z0-9]+', _input)
>>> matches
['value=xxx', 'value=yyy']
>>> for match in matches:
print(match)
value=xxx
value=yyy
>>>
Summary or the regular expression:
[a-zA-Z0-9]+: One or more alphanumeric characters
=: literal equal sign
[a-zA-Z0-9]+: One or more alphanumeric characters
For this input:
content = '(random_garbage_char_here)**value=xxx**;(random_garbage_char_here)**value=yyy**;(random_garbage_char_here)'
use a simple regex and manually strip off the first and last two characters:
import re
values = [x[2:-2] for x in re.findall(r'\*\*value=.*?\*\*', content)]
for value in values:
print(value)
Output:
value=xxx
value=yyy
Here the assumption is that there are always two leading and two trailing * as in **value=xxx**.
You already have good answers based on the re module. That would certainly be the simplest way.
If for any reason (perfs?) you prefere to use str methods, it is indeed possible. But you must search the second string past the end of the first one :
value2_start = content.find('value', value1_end)
value2_end = content.find(';', value2_start)
I'm trying to delete some things from a block of text using regex. I have all of my patterns ready, but I can't seem to be able to remove two (or more) that overlap.
For example:
import re
r1 = r'I am'
r2 = r'am foo'
text = 'I am foo'
re.sub(r1, '', text) # Returns ' foo'
re.sub(r2, '', text) # Returns 'I '
How do I replace both of the occurrences simultaneously and end up with an empty string?
I ended up using a slightly modified version of Ned Batchelder's answer:
def clean(self, text):
mask = bytearray(len(text))
for pattern in patterns:
for match in re.finditer(pattern, text):
r = range(match.start(), match.end())
mask[r] = 'x' * len(r)
return ''.join(character for character, bit in zip(text, mask) if not bit)
You can't do it with consecutive re.sub calls as you have shown. You can use re.finditer to find them all. Each match will provide you with a match object, which has .start and .end attributes indicating their positions. You can collect all those together, and then remove characters at the end.
Here I use a bytearray as a mutable string, used as a mask. It's initialized to zero bytes, and I mark with an 'x' all the bytes that match any regex. Then I use the bit mask to select the characters to keep in the original string, and build a new string with only the unmatched characters:
bits = bytearray(len(text))
for pat in patterns:
for m in re.finditer(pat, text):
bits[m.start():m.end()] = 'x' * (m.end()-m.start())
new_string = ''.join(c for c,bit in zip(text, bits) if not bit)
Not to be a downer, but the short answer is that I'm pretty sure you can't. Can you change your regex so that it doesn't require overlapping?
If you still want to do this, I would try keeping track of the start and stop indices of each match made on the original string. Then go through the string and only keep characters not in any deletion range?
Quite efficient too is a solution coming from ... Perl combine the regexps in one:
# aptitude install regexp-assemble
$ regexp-assemble
I am
I am foo
Ctrl + D
I am(?: foo)?
regexp-assemble takes all the variants of regexps or string you want to match and then
combine them in one. And yes it changes the initial problem to another one since it is not about matching overlapping regexp anymore, but combining regexp for a match
And Then you can use it in your code:
$ python
Python 2.7.3 (default, Aug 1 2012, 05:14:39)
[GCC 4.6.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import re
>>> re.sub("I am foo","I am(?: foo)?","")
''
A port of Regexp::Assemble in python would be nice :)
Here is an alternative that filters the strings on the fly using itertools.compress on the text with a selector iterator. The selector returns True if the character should be kept. selector_for_patterns creates one selector for every pattern. The selector are combined with the all function (only when all pattern want to keep a character it should be in the resulting string).
import itertools
import re
def selector_for_pattern(text, pattern):
i = 0
for m in re.finditer(pattern, text):
for _ in xrange(i, m.start()):
yield True
for _ in xrange(m.start(), m.end()):
yield False
i = m.end()
for _ in xrange(i, len(text)):
yield True
def clean(text, patterns):
gen = [selector_for_pattern(text, pattern) for pattern in patterns]
selector = itertools.imap(all, itertools.izip(* gen))
return "".join(itertools.compress(text, selector))
How can I use regex in python to capture something between two strings or phrases, and removing everything else on the line?
For example, the following is a protein sequence preceded by a one-line header. How can I sift off "CG33289-PC" from the header below based on the stipulation that is occurs after the phrase "FlyBase_Annotation_IDs:" and before the next comma "," ?
I need to substitute the header with this simplified result "CG33289-PC" and not destroy the protein sequence (found below the header-line in all caps).
This is what each protein sequence entry looks like - a header followed by a sequence:
>FBpp0293870 type=protein;loc=3L:join(21527760..21527913,21527977..21528076,21528130..21528390,21528443..21528653,21528712..21529192,21529254..21529264); ID=FBpp0293870; name=CG33289-PC; parent=FBgn0053289,FBtr0305327; dbxref=FlyBase:FBpp0293870,FlyBase_Annotation_IDs:CG33289-PC; MD5=478485a27487608aa2b6c35d39a3295c; length=405; release=r5.45; species=Dmel;
MEMLKYVISDNNYSWWIKLYFAIIFALVLFVAVNLAVGIYNKWDSTPVII
GISSKMTPIDQIPFPTITVCNMNQAKKSKVEHLMPGSIRYAMLQKTCYKE
SNFSQYMDTQHRNETFSNFILDVSEKCADLIVSCIFHQQRIPCTDIFRET
FVDEGLCCIFNVLHPYYLYKFKSPYIRDFTSSDRFADIAVDWDPISGYPQ
RLPSSYYPRPGVGVGTSMGLQIVLNGHVDDYFCSSTNGQGFKILLYNPID
QPRMKESGLPVMIGHQTSFRIIARNVEATPSIRNIHRTKRQCIFSDEQEL
LFYRYYTRRNCEAECDSMFFLRLCSCIPYYLPLIYPNASVCDVFHFECLN
RAESQIFDLQSSQCKEFCLTSCHDLIFFPDAFSTPFSQKDVKAQTNYLTN
FSRAV
This is the desired output:
CG33289-PC
MEMLKYVISDNNYSWWIKLYFAIIFALVLFVAVNLAVGIYNKWDSTPVII
GISSKMTPIDQIPFPTITVCNMNQAKKSKVEHLMPGSIRYAMLQKTCYKE
SNFSQYMDTQHRNETFSNFILDVSEKCADLIVSCIFHQQRIPCTDIFRET
FVDEGLCCIFNVLHPYYLYKFKSPYIRDFTSSDRFADIAVDWDPISGYPQ
RLPSSYYPRPGVGVGTSMGLQIVLNGHVDDYFCSSTNGQGFKILLYNPID
QPRMKESGLPVMIGHQTSFRIIARNVEATPSIRNIHRTKRQCIFSDEQEL
LFYRYYTRRNCEAECDSMFFLRLCSCIPYYLPLIYPNASVCDVFHFECLN
RAESQIFDLQSSQCKEFCLTSCHDLIFFPDAFSTPFSQKDVKAQTNYLTN
FSRAV
Using regexps:
>>> s = """>FBpp0293870 type=protein;loc=3L:join(21527760..21527913,21527977..21528076,21528130..21528390,21528443..21528653,21528712..21529192,21529254..21529264); ID=FBpp0293870; name=CG33289-PC; parent=FBgn0053289,FBtr0305327; dbxref=FlyBase:FBpp0293870,FlyBase_Annotation_IDs:CG33289-PC; MD5=478485a27487608aa2b6c35d39a3295c; length=405; release=r5.45; species=Dmel; MEMLKYVISDNNYSWWIKLYFAIIFALVLFVAVNLAVGIYNKWDSTPVII
GISSKMTPIDQIPFPTITVCNMNQAKKSKVEHLMPGSIRYAMLQKTCYKE
SNFSQYMDTQHRNETFSNFILDVSEKCADLIVSCIFHQQRIPCTDIFRET
FVDEGLCCIFNVLHPYYLYKFKSPYIRDFTSSDRFADIAVDWDPISGYPQ
RLPSSYYPRPGVGVGTSMGLQIVLNGHVDDYFCSSTNGQGFKILLYNPID
QPRMKESGLPVMIGHQTSFRIIARNVEATPSIRNIHRTKRQCIFSDEQEL
LFYRYYTRRNCEAECDSMFFLRLCSCIPYYLPLIYPNASVCDVFHFECLN
RAESQIFDLQSSQCKEFCLTSCHDLIFFPDAFSTPFSQKDVKAQTNYLTN
FSRAV"""
>>> import re
>>> print re.sub(r'.*FlyBase_Annotation_IDs:([\w-]+).*;', r'\1\n', s)
CG33289-PC
MEMLKYVISDNNYSWWIKLYFAIIFALVLFVAVNLAVGIYNKWDSTPVII
GISSKMTPIDQIPFPTITVCNMNQAKKSKVEHLMPGSIRYAMLQKTCYKE
SNFSQYMDTQHRNETFSNFILDVSEKCADLIVSCIFHQQRIPCTDIFRET
FVDEGLCCIFNVLHPYYLYKFKSPYIRDFTSSDRFADIAVDWDPISGYPQ
RLPSSYYPRPGVGVGTSMGLQIVLNGHVDDYFCSSTNGQGFKILLYNPID
QPRMKESGLPVMIGHQTSFRIIARNVEATPSIRNIHRTKRQCIFSDEQEL
LFYRYYTRRNCEAECDSMFFLRLCSCIPYYLPLIYPNASVCDVFHFECLN
RAESQIFDLQSSQCKEFCLTSCHDLIFFPDAFSTPFSQKDVKAQTNYLTN
FSRAV
>>>
Not an elegant solution, but this should work for you:
>>> fly = 'FlyBase_Annotation_IDs'
>>> repl = 'CG33289-PC'
>>> part1, part2 = protein.split(fly)
>>> part2 = part2.replace(repl, "FooBar")
>>> protein = fly.join([part1, part2])
assuming FlyBase_Annotation_IDs can only appear once in the data.
I'm not sure about the format of the file, but this regex will capture the data in your example:
"FlyBase_Annotation_IDs:([A-Z0-9a-z-]*);"
Use findall function to get the match.
Assuming there is a newline after the header:
>>> import re
>>> protein = "..."
>>> r = re.compile(r"^.*FlyBase_Annotation_IDs:([A-Z0-9a-z-]*);.*$", re.MULTILINE)
>>> r.sub(r"\1", protein)
The group ([A-Z0-9a-z-]*) in the regular expression extracts any alphanumeric character and the dash. If ids can have other characters, just add them.