Below is a piece of code I have been working on and not getting the desired result. I would like to only use groups to split
elements into group 1 and group 2. For the last two elements I would like to only match 0567 and not 567 for group 2. I get the desired result for '+1 234 567' but not for '0567' or '567'. Please help with this.
regex_str = "^(?:\+1)?\s?([123456789]\d{2})\s?([123456789]\d{2})"
PATTERN = re.compile(regex_str)
num = ['+1 234 567','0567', '567']
for i in num:
m = PATTERN.match(i)
if m != None:
print (i," and ",m.group(1),m.group(2))
else:
print (i, " has no match")
output:
+1 234 567 and 234 567
0789 has no match
789 has no match
Use a ? after group 1 to make it optional.
regex_str = "^(?:\+1)?\s?([123456789]\d{2})?\s?([123456789]\d{2})"
Thanks for the suggestion using ? works to make group 1 optional. In addition, I had to use ?(1) to make sure group 1 has a correct match before proceeding to group 2 that gave me the answer I desired. Thanks for the help.
Related
I have a list of names with different notations:
for example:
myList = [ab2000, abc2000_2000, AB2000, ab2000_1, ABC2000_01, AB2000_2, ABC2000_02, AB2000_A1]
the standarized version for those different notations are, for example:
'ab2000' is 'ABC2000'
'ab2000_1' is 'ABC2000_01'
'AB2000_A1' is 'ABC2000_A1'
What I tried is to separate the different characters of the string using compile.
input:
compiled = re.compile(r'[A-Za-z]+|\d+|\W+')
compiled.findall("AB2000_2000_A1")
output:
characters = ['AB', '2000', '2000', 'A', '1']
Then applying:
characters = list(set(characters))
To finally try to match the values of that list with the main components of the string: an alpha format followed by a digit format followed by an alphanumeric format.
But as you can see in the previous output I can't match 'A1' into a single character using \W+. My desired output is:
characters = ['AB', '2000', '2000', 'A1']
any idea to fix that?
o any better idea to solve my problem in general. Thank you, in advance.
Use the following pattern with optional groups and capturing groups:
r'([A-Z]+)(\d+)(?:_([A-Z\d]+))?(?:_([A-Z\d]+))?'
and re.I flag.
Note that (?:_([A-Z\d]+))? must be repeated in order to match both
third and fourth group. If you attempted to "repeat" this group, putting
it once with "*" it would match only the last group, skipping the third
group.
To test it, I ran the following test:
myList = ['ab2000', 'abc2000_2000', 'AB2000', 'ab2000_1', 'ABC2000_01',
'AB2000_2', 'ABC2000_02', 'AB2000_A1', 'AB2000_2000_A1']
pat = re.compile(r'([A-Z]+)(\d+)(?:_([A-Z\d]+))?(?:_([A-Z\d]+))?', re.I)
for tt in myList:
print(f'{tt:16} ', end=' ')
mtch = pat.match(tt)
if mtch:
for it in mtch.groups():
if it is not None:
print(f'{it:5}', end=' ')
print()
getting:
ab2000 ab 2000
abc2000_2000 abc 2000 2000
AB2000 AB 2000
ab2000_1 ab 2000 1
ABC2000_01 ABC 2000 01
AB2000_2 AB 2000 2
ABC2000_02 ABC 2000 02
AB2000_A1 AB 2000 A1
AB2000_2000_A1 AB 2000 2000 A1
im working on a project where i have to use speech to text as an input to determine who to call, however using the speech to text can give some unexpected results so i wanted to have a little dynamic matching of the strings, i'm starting small and try to match 1 single name, my name is Nick Vaes, and i try to match my name to the spoken text, but i also want it to match when for example some text would be Nik or something, idealy i would like to have something that would match everything if only 1 letter is wrong so
Nick
ick
nik
nic
nck
would all match my name, the current simple code i have is:
def user_to_call(s):
if "NICK" or "NIK" in s.upper(): redirect = "Nick"
if redirect: return redirect
for a 4 letter name its possible to put all possibilities in the filter, but for names with 12 letters it is a little bit of overkill since i'm pretty sure it can be done way more efficient.
You need to use Levenshtein_distance
A python implementation is nltk
import nltk
nltk.edit_distance("humpty", "dumpty")
What you basically need is fuzzy string matching, see:
https://en.wikipedia.org/wiki/Approximate_string_matching
https://www.datacamp.com/community/tutorials/fuzzy-string-python
Based on that you can check how similar is the input compared your dictionary:
from fuzzywuzzy import fuzz
name = "nick"
tomatch = ["Nick", "ick", "nik", "nic", "nck", "nickey", "njick", "nickk", "nickn"]
for str in tomatch:
ratio = fuzz.ratio(str.lower(), name.lower())
print(ratio)
This code will produce the following output:
100
86
86
86
86
80
89
89
89
You have to experiment with different ratios and check which will suit your requirements to miss only one letter
From what I understand, you are not looking at any fuzzy matching. (Because you did not upvote other responses).
If you are just trying to evaluate what you specified in your request, here is the code. I have put some additional conditions where I printed the appropriate message. Feel free to remove them.
def wordmatch(baseword, wordtoMatch, lengthOfMatch):
lis_of_baseword = list(baseword.lower())
lis_of_wordtoMatch = list(wordtoMatch.lower())
sum = 0
for index_i, i in enumerate(lis_of_wordtoMatch):
for index_j, j in enumerate(lis_of_baseword):
if i in lis_of_baseword:
if i == j and index_i <= index_j:
sum = sum + 1
break
else:
pass
else:
print("word to match has characters which are not in baseword")
return 0
if sum >= lengthOfMatch and len(wordtoMatch) <= len(baseword):
return 1
elif sum >= lengthOfMatch and len(wordtoMatch) > len(baseword):
print("word to match has no of characters more than that of baseword")
return 0
else:
return 0
base = "Nick"
tomatch = ["Nick", "ick", "nik", "nic", "nck", "nickey","njick","nickk","nickn"]
wordlength_match = 3 # this says how many words to match in the base word. In your case, its 3
for t_word in tomatch:
print(wordmatch(base,t_word,wordlength_match))
the output looks like this
1
1
1
1
1
word to match has characters which are not in baseword
0
word to match has characters which are not in baseword
0
word to match has no of characters more than that of baseword
0
word to match has no of characters more than that of baseword
0
Let me know if this served your purpose.
Input: 1 10 avenue
Desired Output: 1 10th avenue
As you can see above I have given an example of an input, as well as the desired output that I would like. Essentially I need to look for instances where there is a number followed by a certain pattern (avenue, street, etc). I have a list which contains all of the patterns and it's called patterns.
If that number does not have "th" after it, I would like to add "th". Simply adding "th" is fine, because other portions of my code will correct it to either "st", "nd", "rd" if necessary.
Examples:
1 10th avenue OK
1 10 avenue NOT OK, TH SHOULD BE ADDED!
I have implemented a working solution, which is this:
def Add_Th(address):
try:
address = address.split(' ')
except AttributeError:
pass
for pattern in patterns:
try:
location = address.index(pattern) - 1
number_location = address[location]
except (ValueError, IndexError):
continue
if 'th' not in number_location:
new = number_location + 'th'
address[location] = new
address = ' '.join(address)
return address
I would like to convert this implementation to regex, as this solution seems a bit messy to me, and occasionally causes some issues. I am not the best with regex, so if anyone could steer me in the right direction that would be greatly appreciated!
Here is my current attempt at the regex implementation:
def add_th(address):
find_num = re.compile(r'(?P<number>[\d]{1,2}(' + "|".join(patterns + ')(?P<following>.*)')
check_th = find_num.search(address)
if check_th is not None:
if re.match(r'(th)', check_th.group('following')):
return address
else:
# this is where I would add th. I know I should use re.sub, i'm just not too sure
# how I would do it
else:
return address
I do not have a lot of experience with regex, so please let me know if any of the work I've done is incorrect, as well as what would be the best way to add "th" to the appropriate spot.
Thanks.
Just one way, finding the positions behind a digit and ahead of one of those pattern words and placing 'th' into them:
>>> address = '1 10 avenue 3 33 street'
>>> patterns = ['avenue', 'street']
>>>
>>> import re
>>> pattern = re.compile(r'(?<=\d)(?= ({}))'.format('|'.join(patterns)))
>>> pattern.sub('th', address)
'1 10th avenue 3 33th street'
Hello all…I want to pick up the texts ‘DesingerXXX’ from a text file which contains below contents:
C DesignerTEE edBore 1 1/42006
Cylinder SingleVerticalB DesignerHHJ e 1 1/8Cooling 1
EngineBore 11/16 DesignerTDT 8Length 3Width 3
EngineCy DesignerHEE Inline2008Bore 1
Height 4TheChallen DesignerTET e 1Stroke 1P 305
Height 8C 606Wall15ccG DesignerQBG ccGasEngineJ 142
Height DesignerEQE C 60150ccGas2007
Anidea is to use the ‘Designer’ as a key, to consider each line into 2 parts, before the key, and after the key.
file_object = open('C:\\file.txt')
lines = file_object.readlines()
for line in lines:
if 'Designer' in line:
where = line.find('Designer')
before = line[0:where]
after = line[where:len(line)]
file_object.close()
In the ‘before the key’ part, I need to find the LAST space (‘ ’), and replace to another symbol/character.
In the ‘after the key’ part, I need to find the FIRST space (‘ ’), and replace to another symbol/character.
Then, I can slice it and pick up the wanted according to the new symbols/characters.
is there a better way to pick up the wanted texts? Or not, how can I replace the appointed key spaces?
In the string replace function, I can limit the times of replacing but not exactly which I can replace. How can I do that?
thanks
Using regular expressions, its a trivial task:
>>> s = '''C DesignerTEE edBore 1 1/42006
... Cylinder SingleVerticalB DesignerHHJ e 1 1/8Cooling 1
... EngineBore 11/16 DesignerTDT 8Length 3Width 3
... EngineCy DesignerHEE Inline2008Bore 1
... Height 4TheChallen DesignerTET e 1Stroke 1P 305
... Height 8C 606Wall15ccG DesignerQBG ccGasEngineJ 142
... Height DesignerEQE C 60150ccGas2007'''
>>> import re
>>> exp = 'Designer[A-Z]{3}'
>>> re.findall(exp, s)
['DesignerTEE', 'DesignerHHJ', 'DesignerTDT', 'DesignerHEE', 'DesignerTET', 'DesignerQBG', 'DesignerEQE']
The regular expression is Designer[A-Z]{3} which means the letters Designer, followed by any letter from capital A to capital Z that appears 3 times, and only three times.
So, it won't match DesignerABCD (4 letters), it also wont match Desginer123 (123 is not valid letters).
It also won't match Designerabc (abc are small letters). To make it ignore the case, you can pass an optional flag re.I as a third argument; but this will also match designerabc (you have to be very specific with regular expressions).
So, to make it so that it matches Designer followed by exactly 3 upper or lower case letters, you'd have to change the expression to Designer[Aa-zZ]{3}.
If you want to search and replace, then you can use re.sub for substituting matches; so if I want to replace all matches with the word 'hello':
>>> x = re.sub(exp, 'hello', s)
>>> print(x)
C hello edBore 1 1/42006
Cylinder SingleVerticalB hello e 1 1/8Cooling 1
EngineBore 11/16 hello 8Length 3Width 3
EngineCy hello Inline2008Bore 1
Height 4TheChallen hello e 1Stroke 1P 305
Height 8C 606Wall15ccG hello ccGasEngineJ 142
Height hello C 60150ccGas2007
and what if both before and after 'Designer', there are characters,
and the length of character is not fixed. I tried
'[Aa-zZ]Designer[Aa-zZ]{0~9}', but it doesn't work..
For these things, there are special characters in regular expressions. Briefly summarized below:
When you want to say "1 or more, but at least 1", use +
When you want to say "0 or any number, but there maybe none", use *
When you want to say "none but if it exists, only repeats once" use ?
You use this after the expression you want to be modified with the "repetition" modifiers.
For more on this, have a read through the documentation.
Now your requirements is "there are characters but the length is not fixed", based on this, we have to use +.
Try with re.sub. The regular expression match with your keyword surrounded by spaces. The second parameter of sub, replace the surrounder spaces by your_special_char (in my script a hyphen)
>>> import re
>>> with open('file.txt') as file_object:
... your_special_char = '-'
... for line in file_object:
... formated_line = re.sub(r'(\s)(Designer[A-Z]{3})(\s)', r'%s\2%s' % (your_special_char,your_special_char), line)
... print formated_line
...
C -DesignerTEE-edBore 1 1/42006
Cylinder SingleVerticalB-DesignerHHJ-e 1 1/8Cooling 1
EngineBore 11/16-DesignerTDT-8Length 3Width 3
EngineCy-DesignerHEE-Inline2008Bore 1
Height 4TheChallen-DesignerTET-e 1Stroke 1P 305
Height 8C 606Wall15ccG-DesignerQBG-ccGasEngineJ 142
Height-DesignerEQE-C 60150ccGas2007
Maroun Maroun mentioned 'Why not simply split the string'. so guessing one of the working way is:
import re
file_object = open('C:\\file.txt')
lines = file_object.readlines()
b = []
for line in lines:
a = line.split()
for aa in a:
b.append(aa)
for bb in b:
if 'Designer' in bb:
print bb
file_object.close()
I have one thing and need suggest. Please help me on this.
my scenario is:
i have one model like ordercode
in this table i have prefix like
1 | US
12 | Canada
13 | UK
134 | Australia
and more.
then, i have string like 12345678, and i need to get best match for this string .
if user enter 12345678 best match is 12|Canada , if user enter 135678975 best match is 13|Uk, and if user enter 1345676788 best match is 134.
How can i do it in django query?
Thanks,
it will require multiple request to check every char in the given order code...
def get_matching_country(order_num):
i = 1
matching_req = Country.objects.none()
while true:
req = Country.objects.filter(country_code=order_num[:i])
if res.exists():
matching_req = req
i += 1
else:
break
return matching_req
def match_country(request, string):
qs = Country.objects.filter(country_code__startswith=string) #or int(string) but you might want to write an exception if person provides a letters in that numbers.
return ...
Have you saved your country codes as strings or integers?