Product code looks like abcd2343, what to split by letters and numbers

Question

I have a list of product codes in a text file, on each like is the product code that looks like:

abcd2343 abw34324 abc3243-23A

Answer 1

In [32]: import re

In [33]: s='abcd2343 abw34324 abc3243-23A'

In [34]: re.split('(\d+)',s)
Out[34]: ['abcd', '2343', ' abw', '34324', ' abc', '3243', '-', '23', 'A']

Or, if you want to split on the first occurrence of a digit:

In [43]: re.findall('\d*\D+',s)
Out[43]: ['abcd', '2343 abw', '34324 abc', '3243-', '23A']

---

• \d+ matches 1-or-more digits.
• \d*\D+ matches 0-or-more digits followed by 1-or-more non-digits.
• \d+|\D+ matches 1-or-more digits or 1-or-more non-digits.

Consult the docs for more about Python's regex syntax.

---

re.split(pat, s) will split the string s using pat as the delimiter. If pat begins and ends with parentheses (so as to be a "capturing group"), then re.split will return the substrings matched by pat as well. For instance, compare:

In [113]: re.split('\d+', s)
Out[113]: ['abcd', ' abw', ' abc', '-', 'A']   # <-- just the non-matching parts

In [114]: re.split('(\d+)', s)
Out[114]: ['abcd', '2343', ' abw', '34324', ' abc', '3243', '-', '23', 'A']  # <-- both the non-matching parts and the captured groups

In contrast, re.findall(pat, s) returns only the parts of s that match pat:

In [115]: re.findall('\d+', s)
Out[115]: ['2343', '34324', '3243', '23']

Thus, if s ends with a digit, you could avoid ending with an empty string by using re.findall('\d+|\D+', s) instead of re.split('(\d+)', s):

In [118]: s='abcd2343 abw34324 abc3243-23A 123'

In [119]: re.split('(\d+)', s)
Out[119]: ['abcd', '2343', ' abw', '34324', ' abc', '3243', '-', '23', 'A ', '123', '']

In [120]: re.findall('\d+|\D+', s)
Out[120]: ['abcd', '2343', ' abw', '34324', ' abc', '3243', '-', '23', 'A ', '123']

Answer 2

Try this code it will work fine

import re
text = "MARIA APARECIDA 99223-2000 / 98450-8026"
parts = re.split(r' (?=\d)',text, 1)
print(parts)

Output:

['MARIA APARECIDA', '99223-2000 / 98450-8026']

Answer 3

import re

m = re.match(r"(?P<letters>[a-zA-Z]+)(?P<the_rest>.+)$",input)

m.group('letters')
m.group('the_rest')

This covers your corner case of abc3243-23A and will output abc for the letters group and 3243-23A for the_rest

Since you said they are all on individual lines you'll obviously need to put a line at a time in input

Answer 4

This function handles float and negative numbers as well.

def separate_number_chars(s):
    res = re.split('([-+]?\d+\.\d+)|([-+]?\d+)', s.strip())
    res_f = [r.strip() for r in res if r is not None and r.strip() != '']
    return res_f

For example:

utils.separate_number_chars('-12.1grams')
> ['-12.1', 'grams']

Answer 5

def firstIntIndex(string):
    result = -1
    for k in range(0, len(string)):
        if (bool(re.match('\d', string[k]))):
            result = k
            break
    return result

Answer 6

To partition on the first digit

parts = re.split('(\d.*)','abcd2343')      # => ['abcd', '2343', '']
parts = re.split('(\d.*)','abc3243-23A')   # => ['abc', '3243-23A', '']

So the two parts are always parts[0] and parts[1].

Of course, you can apply this to multiple codes:

>>> s = "abcd2343 abw34324 abc3243-23A"
>>> results = [re.split('(\d.*)', pcode) for pcode in s.split(' ')]
>>> results
[['abcd', '2343', ''], ['abw', '34324', ''], ['abc', '3243-23A', '']]

If each code is in an individual line then instead of s.split( ) use s.splitlines().