Python: regex condition to find lower case/digit before capital letter

Question

I would like to split a string in python and make it into a dictionary such that a key is any chunk of characters between two capital letters and the value should be the number of occurrences of these chunk in the string.

As an example: string = 'ABbACc1Dd2E' should return this: {'A': 2, 'Bb': 1, 'Cc1': 1, 'Dd2': 1, 'E': 1}

I have found two working solution so far (see below), but I am looking for a more general/elegant solution to this, possibly a one-line regex condition.

Thank you

Answer 1

Solution 1

string = 'ABbACc1Dd2E'
string = ' '.join(string)

for ii in re.findall("([A-Z] [a-z])",string) + \
          re.findall("([A-Z] [0-9])",string) + \
          re.findall("([a-x] [0-9])",string):
            new_ii = ii.replace(' ','')
            string = string.replace(ii, new_ii)

string = string.split()
all_dict = {}
for elem in string:
    all_dict[elem] = all_dict[elem] + 1 if elem in all_dict.keys() else 1 

print(all_dict)

{'A': 2, 'Bb': 1, 'Cc1': 1, 'Dd2': 1, 'E': 1}

Solution 2

string = 'ABbACc1Dd2E'
all_upper = [ (pos,char) for (pos,char) in enumerate(string) if char.isupper() ]

all_dict = {}
for (pos,char) in enumerate(string):
    if (pos,char) in all_upper:
        new_elem = char
    else:
        new_elem += char

    if pos < len(string) -1 :
        if  string[pos+1].isupper():
            all_dict[new_elem] = all_dict[new_elem] + 1 if new_elem in all_dict.keys() else 1 
        else:
            pass
    else:
        all_dict[new_elem] = all_dict[new_elem] + 1 if new_elem in all_dict.keys() else 1 

print(all_dict)

{'A': 2, 'Bb': 1, 'Cc1': 1, 'Dd2': 1, 'E': 1}

Answer 2

Thanks to usr2564301 for this suggestion:

The right regex is '[AZ][az]*\\d*'

import re

string = 'ABbACc1Dd2E'
print(re.findall(r'[A-Z][a-z]*\d*', string))

['A', 'Bb', 'A', 'Cc1', 'Dd2', 'E']

One can then use itertools.groupby to make an iterator that returns consecutive keys and groups from the iterable.

from itertools import groupby

all_dict = {}
for i,j in groupby(re.findall(r'[A-Z][a-z]*\d*', string)):
    all_dict[i] = all_dict[i] + 1 if i in all_dict.keys() else 1 
print(all_dict)

{'A': 2, 'Bb': 1, 'Cc1': 1, 'Dd2': 1, 'E': 1}

Ultimately, one could use sorted() to get this in one line with the correct counting:

print({i:len(list(j)) for i,j in groupby(sorted(re.findall(r'[A-Z][a-z]*\d*', string))) })

{'A': 2, 'Bb': 1, 'Cc1': 1, 'Dd2': 1, 'E': 1}