How to remove alphabets and extract numbers using regex in python?
import re
l=["098765432123 M","123456789012"]
s = re.findall(r"(?<!\d)\d{12}", l)
print(s)
Expected Output:
123456789012
If all you want is to have filtered list, consisting elements with pure digits, use filter with str.isdigit:
list(filter(str.isdigit, l))
Or as #tobias_k suggested, list comprehension is always your friend:
[s for s in l if s.isdigit()]
Output:
['123456789012']
I would suggest to use a negative lookahead assertion, if as stated you want to use regex only.
l=["098765432123 M","123456789012"]
res=[]
for a in l:
s = re.search(r"(?<!\d)\d{12}(?! [a-zA-Z])", a)
if s is not None:
res.append(s.group(0))
The result would then be:
['123456789012']
To keep only digits you can do re.findall('\d',s), but you'll get a list:
s = re.findall('\d', "098765432123 M")
print(s)
> ['0', '9', '8', '7', '6', '5', '4', '3', '2', '1', '2', '3']
So to be clear, you want to ignore the whole string if there is a alphabetic character in it? Or do you still want to extract the numbers of a string with both numbers and alphabetic characters in it?
If you want to find all numbers, and always find the longest number use this:
regex = r"\d+"
matches = re.finditer(regex, test_str, re.MULTILINE)
\d will search for digits, + will find one or more of the defined characters, and will always find the longest consecutive line of these characters.
If you only want to find strings without alphabets:
import re
regex = r"[a-zA-Z]"
test_str = ("098765432123 M", "123456789012")
for x in test_str:
if not re.search(regex, x):
print(x)
Related
Hello I'm trying to split a string without removing the delimiter and it can have multiple delimiters.
The delimiters can be 'D', 'M' or 'Y'
For example:
>>>string = '1D5Y4D2M'
>>>re.split(someregex, string) #should ideally return
['1D', '5Y', '4D', '2M']
To keep the delimiter I use Python split() without removing the delimiter
>>> re.split('([^D]+D)', '1D5Y4D2M')
['', '1D', '', '5Y4D', '2M']
For multiple delimiters I use In Python, how do I split a string and keep the separators?
>>> re.split('(D|M|Y)', '1D5Y4D2M')
['1', 'D', '5', 'Y', '4', 'D', '2', 'M', '']
Combining both doesn't quite make it.
>>> re.split('([^D]+D|[^M]+M|[^Y]+Y)', string)
['', '1D', '', '5Y4D', '', '2M', '']
Any ideas?
I'd use findall() in your case. How about:
re.findall(r'\d+[DYM]', string
Which will result in:
['1D', '5Y', '4D', '2M']
(?<=(?:D|Y|M))
You need 0 width assertion split.Can be done using regex module python.
See demo.
https://regex101.com/r/aKV13g/1
You can split at the locations right after D, Y or M but not at the end of the string with
re.split(r'(?<=[DYM])(?!$)', text)
See the regex demo. Details:
(?<=[DYM]) - a positive lookbehind that matches a location that is immediately preceded with D or Y or M
(?!$) - a negative lookahead that fails the match if the current position is the string end position.
Note
In the current scenario, (?<=[DYM]) can be used instead of a more verbose (?<=D|Y|M) since all alternatives are single characters. If you have multichar delimiters, you would have to use a non-capturing group, (?:...), with lookbehind alternatives inside it. For example, to separate right after Y, DX and MZB you would use (?:(?<=Y)|(?<=DX)|(?<=MZB)). See Python Regex Engine - "look-behind requires fixed-width pattern" Error
I think it will work fine without regex or split
time complexity O(n)
string = '1D5Y4D2M'
temp=''
res = []
for x in string:
if x=='D':
temp+='D'
res.append(temp)
temp=''
elif x=='M':
temp+='M'
res.append(temp)
temp=''
elif x=='Y':
temp+='Y'
res.append(temp)
temp=''
else:
temp+=x
print(res)
using translate
string = '1D5Y4D2M'
delimiters = ['D', 'Y', 'M']
result = string.translate({ord(c): f'{c}*' for c in delimiters}).strip('.*').split('*')
print(result)
>>> ['1D', '5Y', '4D', '2M']
Using python 3.8
Given str = A11B11C32D34,....
I want to split it into [11, 11, 32, 34 ...]. Meaning split using alphabets. How could I do this?
Thanks in advance!
Check with
s= 'A11B11C32D34'
s
Out[388]: 'A11B11C32D34'
import re
re.findall(r'\d+', s)
Out[390]: ['11', '11', '32', '34']
I might also suggest using a regex split approach here:
inp = "A11B11C32D34"
nums = [x for x in re.split(r'\D+', inp) if x]
print(nums) # ['11', '11', '32', '34']
The idea here is to split the string on any one or more collection of non digit characters. I also use a list comprehension to remove any leading/trailing empty entries in the output from re.split which might arise due to the string starting/ending with a non digit character.
note that the final two numbers of this pattern for example FBXASC048 are ment to be ascii code for numbers (0-9)
input example list ['FBXASC048009Car', 'FBXASC053002Toy', 'FBXASC050004Human']
result example ['1009Car', '5002Toy', '2004Human']
what is the proper way to searches for any of these pattern in an input list
num_ascii = ['FBXASC048', 'FBXASC049', 'FBXASC050', 'FBXASC051', 'FBXASC052', 'FBXASC053', 'FBXASC054', 'FBXASC055', 'FBXASC056', 'FBXASC057']
and then replaces the pattern found with one of the items in the conv list but not randomally
because each element in the pattern list equals only one element in the conv_list
conv_list = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
this is the solution in mind:
it has two part
1st part--> is to find for ascii pattern[48, 49, 50, 51, 52, 53, 54, 55, 56,57]
and then replace those with the proper decimal matching (0-9)
so we will get new input list will be called input_modi_list that has ascii replaced with decimal
2nd part-->another process to use fixed pattern to replace using replace function which is this 'FBXASC0'
new_list3
for x in input_modi_list:
y = x.replace('FBXASC0', '')
new_list3.append(new_string)
so new_list3 will have the combined result of the two parts mentioned above.
i don't know if there would be a simplar solution or a better one maybe using regex
also note i don't have any idea on how to replace ascii with decimal for a list of items
I think this should do the trick:
import re
input_list = ['FBXASC048009Car', 'FBXASC053002Toy', 'FBXASC050004Human']
pattern = re.compile('FBXASC(\d{3,3})')
def decode(match):
return chr(int(match.group(1)))
result = [re.sub(pattern, decode, item) for item in input_list]
print(result)
Now, there is some explanation due:
1- the pattern object is a regular expression that will match any part of a string that starts with 'FBXASC' and ends with 3 digits (0-9). (the \d means digit, and {3,3} means that it should occur at least 3, and at most 3 times, i.e. exactly 3 times). Also, the parenthesis around \d{3,3} means that the three digits matched will be stored for later use (explained in the next part).
2- The decode function receives a match object, uses .group(1) to extract the first matched group (which in our case are the three digits matched by \d{3,3}), then uses the int function to parse the string into an integer (for example, convert '048' to 48), and finally uses the chr function to find which character has that ASCII-code. (for example chr(48) will return '0', and chr(65) will return 'A')
3- The final part applies the re.sub function to all elements of list which will replace each occurrence of the pattern you described (FBXASC048[3-digits]) with it's corresponding ASCII character.
You can see that this solution is not limited only to your specific examples. Any number can be used as long as it has a corresponding ASCII character recognized by the chr function.
But, if you do want to limit it just to the 48-57 range, you can simply modify the decode function:
def decode(match):
ascii_code = int(match.group(1))
if ascii_code >= 48 and ascii_code <= 57:
return chr(ascii_code)
else:
return match.group(0) # returns the entire string - no modification
This is how I would do it.
make the regex pattern by simply joining the strings with |:
>>> num_ascii = ['FBXASC048', 'FBXASC049', 'FBXASC050', 'FBXASC051', 'FBXASC052', 'FBXASC053', 'FBXASC054', 'FBXASC055', 'FBXASC056', 'FBXASC057']
>>> conv_list = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
>>> regex_pattern = '|'.join(num_ascii)
>>> regex_pattern
'FBXASC048|FBXASC049|FBXASC050|FBXASC051|FBXASC052|FBXASC053|FBXASC054|FBXASC055
|FBXASC056|FBXASC057'
make a look-up dictionary by simply zipping the two lists:
>>> conv_table = dict(zip(num_ascii, conv_list))
>>> conv_table
{'FBXASC048': '0', 'FBXASC049': '1', 'FBXASC050': '2', 'FBXASC051': '3', 'FBXASC
052': '4', 'FBXASC053': '5', 'FBXASC054': '6', 'FBXASC055': '7', 'FBXASC056': '8
', 'FBXASC057': '9'}
iterate over the data and replace the matched string with the corresponding digit:
>>> import re
>>> result = []
>>> for item in ['FBXASC048009Car', 'FBXASC053002Toy', 'FBXASC050004Human']:
... m = re.match(regex_pattern, item)
... matched_string = m[0]
... digit = (conv_table[matched_string])
... print(f'replacing {matched_string} with {digit}')
... result.append(item.replace(matched_string, digit))
...
replacing FBXASC048 with 0
replacing FBXASC053 with 5
replacing FBXASC050 with 2
>>> result
['0009Car', '5002Toy', '2004Human']
For example I get following input:
-9x+5x-2-4x+5
And I need to get following list:
['-9x', '5x', '-2', '-4x', '5']
Here is my code, but I don't know how to deal with minuses.
import re
text = '-3x-5x+2=9x-9'
text = re.split(r'\W', text)
print(text)
warning: I cannot use any libraries except re and math.
You could re.findall all groups of characters followed by + or - (or end-of-string $), then strip the + (which, like -, is still part of the following group) from the substrings.
>>> s = "-9x+5x-2-4x+5"
>>> [x.strip("+") for x in re.findall(r".+?(?=[+-]|$)", s)]
['-9x', '5x', '-2', '-4x', '5']
Similarly, for the second string with =, add that to the character group and also strip it off the substrings:
>>> s = '-3x-5x+2=9x-9'
>>> [x.strip("+=") for x in re.findall(r".+?(?=[+=-]|$)", s)]
>>> ['-3x', '-5x', '2', '9x', '-9']
Or apply the original comprehension to the substrings after splitting by =, depending on how the result should look like:
>>> [[x.strip("+") for x in re.findall(r".+?(?=[+-]|$)", s2)] for s2 in s.split("=")]
>>> [['-3x', '-5x', '2'], ['9x', '-9']]
In fact, now that I think of it, you can also just findall that match an optional minus, followed by some digits, and an optional x, with or without splitting by = first:
>>> [re.findall(r"-?\d+x?", s2) for s2 in s.split("=")]
[['-3x', '-5x', '2'], ['9x', '-9']]
One of many possible ways:
import re
term = "-9x+5x-2-4x+5"
rx = re.compile(r'-?\d+[a-z]?')
factors = rx.findall(term)
print(factors)
This yields
['-9x', '5x', '-2', '-4x', '5']
For your example data, you might split on either a plus or equals sign or split when asserting a minus sign on the right which is not at the start of the string.
[+=]|(?=(?<!^)-)
[+=] Match either + or =
| Or
(?=(?<!^)-) Positive lookahead, assert what is on the right is - but not at the start of the string
Regex demo | Python demo
Output for both example strings
['-9x', '5x', '-2', '-4x', '5']
['-3x', '-5x', '2', '9x', '-9']
I would like to split a string into sections of numbers and sections of text/symbols
my current code doesn't include negative numbers or decimals, and behaves weirdly, adding an empty list element on the end of the output
import re
mystring = 'AD%5(6ag 0.33--9.5'
newlist = re.split('([0-9]+)', mystring)
print (newlist)
current output:
['AD%', '5', '(', '6', 'ag ', '0', '.', '33', '--', '9', '.', '5', '']
desired output:
['AD%', '5', '(', '6', 'ag ', '0.33', '-', '-9.5']
Your issue is related to the fact that your regex captures one or more digits and adds them to the resulting list and digits are used as a delimiter, the parts before and after are considered. So if there are digits at the end, the split results in the empty string at the end to be added to the resulting list.
You may split with a regex that matches float or integer numbers with an optional minus sign and then remove empty values:
result = re.split(r'(-?\d*\.?\d+)', s)
result = filter(None, result)
To match negative/positive numbers with exponents, use
r'([+-]?\d*\.?\d+(?:[eE][-+]?\d+)?)'
The -?\d*\.?\d+ regex matches:
-? - an optional minus
\d* - 0+ digits
\.? - an optional literal dot
\d+ - one or more digits.
Unfortunately, re.split() does not offer an "ignore empty strings" option. However, to retrieve your numbers, you could easily use re.findall() with a different pattern:
import re
string = "AD%5(6ag0.33-9.5"
rx = re.compile(r'-?\d+(?:\.\d+)?')
numbers = rx.findall(string)
print(numbers)
# ['5', '6', '0.33', '-9.5']
As mentioned here before, there is no option to ignore the empty strings in re.split() but you can easily construct a new list the following way:
import re
mystring = "AD%5(6ag0.33--9.5"
newlist = [x for x in re.split('(-?\d+\.?\d*)', mystring) if x != '']
print newlist
output:
['AD%', '5', '(', '6', 'ag', '0.33', '-', '-9.5']