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

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

abcd2343 abw34324 abc3243-23A

So it is letters followed by numbers and other characters.

I want to split on the first occurrence of a number.

Answers:

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Method 1

import re
s='abcd2343 abw34324 abc3243-23A'
re.split('(d+)',s)

> ['abcd', '2343', ' abw', '34324', ' abc', '3243', '-', '23', 'A']

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

re.findall('d*D+',s)
> ['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:

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

re.split('(d+)', s)
> ['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:

re.findall('d+', s)
> ['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):

s='abcd2343 abw34324 abc3243-23A 123'

re.split('(d+)', s)
> ['abcd', '2343', ' abw', '34324', ' abc', '3243', '-', '23', 'A ', '123', '']

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

Method 2

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']

Method 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

Method 4

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().

Method 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

Method 6

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’]

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0