I have a pandas dataframe containing addresses. Some are formatted correctly like 481 Rogers Rd York ON. Others have a space missing between the city quandrant and the city name, for example: 101 9 Ave SWCalgary AB or even possibly: 101 9 Ave SCalgary AB, where SW refers to south west and S to south.
I'm trying to find a regex that will add a space between second and third capital letters if they are followed by lowercase letters, or if there are only 2 capitals followed by lower case, add a space between the first and second.
So far, I've found that ([A-Z]{2,3}[a-z]) will match the situation correctly, but I can't figure out how to look back into it and sub at position 2 or 3. Ideally, I'd like to use an index to split the match at [-2:] but I can't figure out how to do this.
I found that re.findall('(?<=[A-Z][A-Z])[A-Z][a-z].+', '101 9 Ave SWCalgary AB')
will return the last part of the string and I could use a look forward regex to find the start and then join them but this seems very inefficient.
Thanks
You may use
df['Test'] = df['Test'].str.replace(r'\b([A-Z]{1,2})([A-Z][a-z])', r'\1 \2')
See this regex demo
Details
\b - a word boundary
([A-Z]{1,2}) - Capturing group 1 (later referred with \1 from the replacement pattern): one or two uppercase letters
([A-Z][a-z]) - Capturing group 2 (later referred with \2 from the replacement pattern): an uppercase letter + a lowercase one.
If you want to specifically match city quadrants, you may use a bit more specific regex:
df['Test'] = df['Test'].str.replace(r'\b([NS][EW]|[NESW])([A-Z][a-z])', r'\1 \2')
See this regex demo. Here, [NS][EW]|[NESW] matches N or S that are followed with E or W, or a single N, E, S or W.
Pandas demo:
import pandas as pd
df = pd.DataFrame({'Test':['481 Rogers Rd York ON',
'101 9 Ave SWCalgary AB',
'101 9 Ave SCalgary AB']})
>>> df['Test'].str.replace(r'\b([A-Z]{1,2})([A-Z][a-z])', r'\1 \2')
0 481 Rogers Rd York ON
1 101 9 Ave SW Calgary AB
2 101 9 Ave S Calgary AB
Name: Test, dtype: object
You can use
([A-Z]{1,2})(?=[A-Z][a-z])
to capture the first (or first and second) capital letters, and then use lookahead for a capital letter followed by a lowercase letter. Then, replace with the first group and a space:
re.sub(r'([A-Z]{1,2})(?=[A-Z][a-z])', r'\1 ', str)
https://regex101.com/r/TcB4Ph/1
Related
I am trying to remove the middle initial at the end of a name string. An example of how the data looks:
df = pd.DataFrame({'Name': ['Smith, Jake K',
'Howard, Rob',
'Smith-Howard, Emily R',
'McDonald, Jim T',
'McCormick, Erica']})
I am currently using the following code, which works for all names except for McCormick, Erica. I first use regex to identify all capital letters. Then any rows with 3 or more capital letters, I remove [:-1] from the string (in an attempt to remove the middle initial and extra space).
df['Cap_Letters'] = df['Name'].str.findall(r'[A-Z]')
df.loc[df['Cap_Letters'].str.len() >= 3, 'Name'] = df['Name'].str[:-1]
This outputs the following:
As you can see, this properly removes the middle initial for all names except for McCormick, Erica. Reason being she has 3 capital letters but no middle initial, which incorrectly removes the 'a' in Erica.
You can use Series.str.replace directly:
df['Name'] = df['Name'].str.replace(r'\s+[A-Z]$', '', regex=True)
Output:
0 Smith, Jake
1 Howard, Rob
2 Smith-Howard, Emily
3 McDonald, Jim
4 McCormick, Erica
Name: Name, dtype: object
See the regex demo. Regex details:
\s+ - one or more whitespaces
[A-Z] - an uppercase letter
$ - end of string.
Another solution(not so pretty) would be to split then take 2 elements then join again
df['Name'] = df['Name'].str.split().str[0:2].str.join(' ')
# 0 Smith, Jake
# 1 Howard, Rob
# 2 Smith-Howard, Emily
# 3 McDonald, Jim
# 4 McCormick, Erica
# Name: Name, dtype: object
I would use something like that :
def removeMaj(string):
tab=string.split(',')
tab[1]=lower(tab[1])
string=",".join(tab)
return(string)
I have a dataframe with multiple forms of names:
JOSEPH W. JASON
Ralph Landau
RAYMOND C ADAMS
ABD, SAMIR
ABDOU TCHOUSNOU, BOUBACAR
ABDL-ALI, OMAR R
For first 3, the rule is last word. For the last three, or anything with comma, the first word is the last name. However, for name like Abdou Tchousnou, I only took the last word, which is Tchousnou.
The expected output is
JASON
LANDAU
ADAMS
ABD
TCHOUNOU
ABDL-ALI
The left is the name, and the right is what I want to return.
str.extract(r'(^(?=[^,]*,?$)[\w-]+|(?<=, )[\w-]+)', expand=False)
Is there anyway to solve this? The current code only returns the first name instead of surname which is the one that I want.
Something like this would work:
(.+(?=,)|\S+$)
( - start capture group #1
.+(?=,) - get everything before a comma
| - or
\S+$ - get everything which is not a whitespace before the end of the line
) - end capture group #1
https://regex101.com/r/myvyS0/1
Python:
str.extract(r'(.+(?=,)|\S+$)', expand=False)
You may use this regex to extract:
>>> print (df)
name
0 JOSEPH W. JASON
1 Ralph Landau
2 RAYMOND C ADAMS
3 ABD, SAMIR
4 ABDOU TCHOUSNOU, BOUBACA
5 ABDL-ALI, OMAR R
>>> df['name'].str.extract(r'([^,]+(?=,)|\w+(?:-\w+)*(?=$))', expand=False)
0 JASON
1 Landau
2 ADAMS
3 ABD
4 ABDOU TCHOUSNOU
5 ABDL-ALI
RegEx Details:
(: Start capture group
[^,]+(?=,): Match 1+ non-comma characters tha
|: OR
\w+: Match 1+ word charcters
(?:-\w+)*: Match - followed 1+ word characters. Match 0 or more of this group
): End capture group
(?=,|$): Lookahead to assert that we have comma or end of line ahead
I have a string from a NWS bulletin:
LTUS41 KCAR 141558 AAD TMLB Forecast for the National Parks
KHNX 141001 RECHNX Weather Service San Joaquin Valley
My aim is to extract a couple fields with regular expressions. In the first string I want "AAD" and from the second string I want "RECHNX". I have tried:
( )\w{3} #for the first string
and
\w{6} #for the 2nd string
But these find all 3 and 6 character strings leading up to the string I want.
Assuming the fields you want to extract are always in capital letters and preceded by 6 digits and a space, this regular expression would do the trick:
(?<=\d{6}\s)[A-Z]+
Demo: https://regex101.com/r/dsDHTs/1
Edit: if you want to match up to two alpha-numeric uppercase words preceded by 6 digits, you can use:
(?<=\d{6}\s)([A-Z0-9]+\b)\s(?:([A-Z0-9]+\b))*
Demo: https://regex101.com/r/dsDHTs/5
If you have a specific list of valid fields, you could also simply use:
(AAD|TMLB|RECHNX|RR4HNX)
https://regex101.com/r/dsDHTs/3
Since the substring you want to extract is a word that follows a number, separated by a space, you can use re.search with the following regex (given your input stored in s):
re.search(r'\b\d+ (\w+)', s).group(1)
To read first groups of word chars from each line, you can use a pattern like
(\w+) (\w+) (\w+) (\w+).
Then, from the first line read group No 4 and from the second line read group No 3.
Look at the following program. It prints four groups from each source line:
import re
txt = """LTUS41 KCAR 141558 AAD TMLB Forecast for the National Parks
KHNX 141001 RECHNX Weather Service San Joaquin Valley"""
n = 0
pat = re.compile(r'(\w+) (\w+) (\w+) (\w+)')
for line in txt.splitlines():
n += 1
print(f'{n:2}: {line}')
mtch = pat.search(line)
if mtch:
gr = [ mtch.group(i) for i in range(1, 5) ]
print(f' {gr}')
The result is:
1: LTUS41 KCAR 141558 AAD TMLB Forecast for the National Parks
['LTUS41', 'KCAR', '141558', 'AAD']
2: KHNX 141001 RECHNX Weather Service San Joaquin Valley
['KHNX', '141001', 'RECHNX', 'Weather']
I have following requirements in date which can be any of the following format.
mm/dd/yyyy or dd Mon YYYY
Few examples are shown below
04/20/2009 and 24 Jan 2001
To handle this I have written regular expression as below
Few text scenarios are metnioned below
txt1 = 'Lithium 0.25 (7/11/77). LFTS wnl. Urine tox neg. Serum tox
+ fluoxetine 500; otherwise neg. TSH 3.28. BUN/Cr: 16/0.83. Lipids unremarkable. B12 363, Folate >20. CBC: 4.9/36/308 Pertinent Medical
Review of Systems Constitutional:'
txt2 = "s The patient is a 44 year old married Caucasian woman,
unemployed Decorator, living with husband and caring for two young
children, who is referred by Capitol Hill Hospital PCP, Dr. Heather
Zubia, for urgent evaluation/treatment till first visit with Dr. Toney
Winkler IN EIGHT WEEKS on 24 Jan 2001."
date = re.findall(r'(?:\b(?<!\.)[\d{0,2}]+)'
'(?:[/-]\d{0,}[/-]\d{2,4}) | (?:\b(?<!\.)[\d{1,2}]+)[th|st|nd]*'
' (?:[Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec][a-z]*) \d{2,4}', txtData)
I am not getting 24 Jan 2001 where as if I run individually (?:\b(?<!\.)[\d{1,2}]+)[th|st|nd]* (?:[Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec][a-z]*) \d{2,4}' I am able to get output.
Question 1: What is bug in above expression?
Question 2: I want to combine both to make more readable as I have to parse any other formats so I used join as shown below
RE1 = '(?:\b(?<!\.)[\d{0,2}]+) (?:[/-]\d{0,}[/-]\d{2,4})'
RE2 = '(?:\b(?<!\.)[\d{1,2}]+)[th|st|nd]* (?:[Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec][a-z]*) \d{2,4}'
regex_all = '|'.join([RE1, RE2])
regex_all = re.compile(regex_all)
date = regex_all.findall(txtData) // notice here txtData can be any one of the above string.
I am getting output as NaN in case of above for date.
Please suggest what is the mistake if I join.
Thanks for your help.
Note that it is a very bad idea to join such long patterns that also match at the same location within the string. That would cause the regex engine to backtrack too much, and possibly lead to crashes and slowdown. If there is a way to re-write the alternations so that they could only match at different locations, or even get rid of them completely, do it.
Besides, you should use grouping constructs (...) to groups sequences of patterns, and only use [...] character classes when you need to matches specific chars.
Also, your alternatives are overlapping, you may combine them easily. See the fixed regex:
\b(?<!\.)\d{1,2}(?:[/-]\d+[/-]|(?:th|st|[nr]d)?\s*(?:(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]*))\s*(?:\d{4}|\d{2})\b
See the regex demo.
Details
\b - a word boundary
(?<!\.) - no . immediately to the left of the current location
\d{1,2} - 1 or 2 digits
(?: - start of a non-capturing alternation group:
[/-]\d+[/-] - / or -, 1+ digits, - or /
| - or
(?:th|st|[nr]d)?\s*(?:
(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]*)) - th, st, nd or rd (optionally), followed with 0+ whitespaces, and then month names
\s* - 0+ whitespaces
(?:\d{4}|\d{2}) - 2 or 4 digits
\b - trailing word boundary.
Another note: if you want to match the date-like strings with two matching delimiters, you will need to capture the first one, and use a backreference to match the second one, see this regex demo. In Python, you would need a re.finditer to get those matches.
See this Python demo:
import re
rx = r"\b(?<!\.)\d{1,2}(?:([/-])\d+\1|(?:th|st|[nr]d)?\s*(?:(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]*))\s*(?:\d{4}|\d{4})\b"
s = "Lithium 0.25 (7/11/77). LFTS wnl. Urine tox neg. Serum tox\nfluoxetine 500; otherwise neg. TSH 3.28. BUN/Cr: 16/0.83. Lipids unremarkable. B12 363, Folate >20. CBC: 4.9/36/308 Pertinent Medical\nReview of Systems Constitutional:\n\nThe patient is a 44 year old married Caucasian woman, unemployed Decorator, living with husband and caring for two young children, who is referred by Capitol Hill Hospital PCP, Dr. Heather Zubia, for urgent evaluation/treatment till first visit with Dr. Toney Winkler IN EIGHT WEEKS on 24 Jan 2001"
print([x.group(0) for x in re.finditer(rx, s, re.I)])
# => ['7/11/77', '24 Jan 2001']
I think your approach is too complicated. I suggest using a combination of a simple regex and strptime().
import re
from datetime import datetime
date_formats = ['%m/%d/%Y', '%d %b %Y']
pattern = re.compile(r'\b(\d\d?/\d\d?/\d{4}|\d\d? \w{3} \d{4})\b')
data = "... your string ..."
for match in re.findall(pattern, data):
print("Trying to parse '%s'" % match)
for fmt in date_formats:
try:
date = datetime.strptime(match, fmt)
print(" OK:", date)
break
except:
pass
The advantage of this approach is, besides a much more manageable regex, that it won't pick dates that look plausible but do not exist, like 2/29/2000 (whereas 2/29/2004 works).
r'(?:\b(?<!\.)[\d{0,2}]+)'
'(?:[/-]\d{0,}[/-]\d{2,4}) | (?:\b(?<!\.)[\d{1,2}]+)[th|st|nd]*'
' (?:[Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec][a-z]*) \d{2,4}'
you should use raw strings (r'foo') for each string, not only the first one. This way backslashes (\) will be considered as normal character and usable by the re library.
[abc|def] matches any character between the [], while (one|two|three) matches any expression (one, two, or three)
I've searched and searched, but can't find an any relief for my regex woes.
I wrote the following dummy sentence:
Watch Joe Smith Jr. and Saul "Canelo" Alvarez fight Oscar de la Hoya and Genaddy Triple-G Golovkin for the WBO belt GGG. Canelo Alvarez and Floyd 'Money' Mayweather fight in Atlantic City, New Jersey. Conor MacGregor will be there along with Adonis Superman Stevenson and Mr. Sugar Ray Robinson. "Here Goes a String". 'Money Mayweather'. "this is not a-string", "this is not A string", "This IS a" "Three Word String".
I'm looking for a regular expression that will return the following when used in Python 3.6:
Canelo, Money, Money Mayweather, Three Word String
The regex that has gotten me the closest is:
(["'])[A-Z](\\?.)*?\1
I want it to only match strings of 3 capitalized words or less immediately surrounded by single or double quotes. Unfortunately, so far it seem to match any string in quotes, no matter what the length, no matter what the content, as long is it begins with a capital letter.
I've put a lot of time into trying to hack through it myself, but I've hit a wall. Can anyone with stronger regex kung-fu give me an idea of where I'm going wrong here?
Try to use this one: (["'])((?:[A-Z][a-z]+ ?){1,3})\1
(["']) - opening quote
([A-Z][a-z]+ ?){1,3} - Capitalized word repeating 1 to 3 times separated by space
[A-Z] - capital char (word begining char)
[a-z]+ - non-capital chars (end of word)
_? - space separator of capitalized words (_ is a space), ? for single word w/o ending space
{1,3} - 1 to 3 times
\1 - closing quote, same as opening
Group 2 is what you want.
Match 1
Full match 29-37 `"Canelo"`
Group 1. 29-30 `"`
Group 2. 30-36 `Canelo`
Match 2
Full match 146-153 `'Money'`
Group 1. 146-147 `'`
Group 2. 147-152 `Money`
Match 3
Full match 318-336 `'Money Mayweather'`
Group 1. 318-319 `'`
Group 2. 319-335 `Money Mayweather`
Match 4
Full match 398-417 `"Three Word String"`
Group 1. 398-399 `"`
Group 2. 399-416 `Three Word String`
RegEx101 Demo: https://regex101.com/r/VMuVae/4
Working with the text you've provided, I would try to use regular expression lookaround to get the words surrounded by quotes and then apply some conditions on those matches to determine which ones meet your criterion. The following is what I would do:
[p for p in re.findall('(?<=[\'"])[\w ]{2,}(?=[\'"])', txt) if all(x.istitle() for x in p.split(' ')) and len(p.split(' ')) <= 3]
txt is the text you've provided here. The output is the following:
# ['Canelo', 'Money', 'Money Mayweather', 'Three Word String']
Cleaner:
matches = []
for m in re.findall('(?<=[\'"])[\w ]{2,}(?=[\'"])', txt):
if all(x.istitle() for x in m.split(' ')) and len(m.split(' ')) <= 3:
matches.append(m)
print(matches)
# ['Canelo', 'Money', 'Money Mayweather', 'Three Word String']
Here's my go at it: ([\"'])(([A-Z][^ ]*? ?){1,3})\1