Count the number of occurrences of characters in a string?-CodePudding

Given a string "a1a3b5a2c4b1". The characters are followed by numbers representing the number of occurrences of the character in the string. The correct solution would return "a6b6c4" (a1 a3 a2 = a6, b5 b1 = b6, c4)

My original idea was to convert the string into a list, then to a dict of key value pairs.

data="a1a3b5a2c4b1"
lst = list(data)
{lst[i]: lst[i  1] for i in range(0, len(lst), 2)}

Returns:

{'a': '2', 'b': '1', 'c': '4'}

The issue is that it does not increase the number of occurrences (values) but takes the last seen value.

How can I create a dictionary that increases the values rather than replaces them?

CodePudding user response：

You keep overriding the key with the latest count in that comprehension. You would have to rather update them by addition:

data = "a1a3b5a2c4b1"

counts = {}
i = iter(data)
for char, count in zip(i, i):
    counts[char] = counts.get(char, 0)   int(count)

# {'a': 6, 'b': 6, 'c': 4}

The other natural util to handle counts is, well, a collections.Counter:

from collections import Counter

counts = Counter()

i = iter(data)
for char, count in zip(i, i):
    counts.update(**{char: int(count)})

This also uses the "zip the same iterator" trick to produce chunks of 2. AS for turning these dictionaries into the desired string output:

"".join(f"{k}{v}" for k, v in counts.items())

CodePudding user response：

If the string is based on sequences of pairs where the first character is your key and the second character is a single digit then:

mystring = 'a1a3b5a2c4b1'
mydict = dict()
for i in range(0, len(mystring), 2):
    pair = mystring[i:i 2]
    key = pair[0]
    value = int(pair[1:])
    if key in mydict:
        mydict[key]  = value
    else:
        mydict[key] = value
print(mydict)

You could play around with defaultdict and/or dictionary comprehension but I think this is clearer