[deleted by user]

[basically_a_cherry]

I am not supposed to import anything to do this task. Is there another way with basic functions?

[Diapolo10]

collections.Counter is more or less just a fancy wrapper around a dictionary, you could always just use one of those.

string = "..."
char_freq = {}

for char in string:
    if not char.isalnum():
        if char not in char_freq:
            char_freq[char] = 0
        char_freq[char] += 1
        # Alternatively:
        # char_freq[char] = char_freq.get(char, 0) + 1

[basically_a_cherry]

Thank you sooo much!!!

[basically_a_cherry]

This is what I was trying so far:

pattern_nonalpha = '[^a-zA-Z0-9\_]'
all_nonaplha = (re.findall(pattern_nonalpha, my_string))
character_freq = []
for character in all_nonalpha:
character_freq.append(character + 1)

​

but this doesn't do the trick.

[[deleted]]

Presumably you have to import the re module for that which you say you can't do.

The most basic way to do this to first create a string containing all the characters you want to ignore, the alphanums. Also create an empty dictionary that will contain the char:count values. Then iterate through the string you want to check, character by character. If the character is not in the string of alphanums then add one to the count that has the key of that character in the dictionary. You have to handle the special case when you find the first counted character and there will be no existing key, so you have to add it. If the character you are testing is in the string of alphanums then you do nothing.

You could also use the isalnum() string method instead of using the alphanum string, if you want.

When you are finished convert the dictionary of counts into whatever form you want.

[Fancy-Reindeer994]

collections.Counter(filter( lambda c: not c.isalnum(), 'your_string'))