CODEWITHSUNDEEP
pythonstringfrequency
Wicaledon
Wicaledon

Reputation: 810

Python - Counting Letter Frequency in a String

I want to write my each string's letter frequencies. My inputs and expected outputs are like this.

"aaaa" -> "a4"
"abb" -> "a1b2"
"abbb cc a" -> "a1b3 c2 a1"
"bbbaaacddddee" -> "b3a3c1d4e2"
"a   b" -> "a1 b1"

I found this solution but it gives the frequencies in random order. How can I do this?

Upvotes: 0

Views: 332

Answers (3)

seldomspeechless
seldomspeechless

Reputation: 174

There is no need to iterate every character in every word.
This is an alternate solution. (If you don't want to use itertools, that looked pretty tidy.)

def word_stats(data: str=""):
    all = []
    for word in data.split(" "):
        res = []
        while len(word)>0:
            res.append(word[:1] + str(word.count(word[:1])))
            word = word.replace(word[:1],"")
        res.sort()
        all.append("".join(res))
    return " ".join(all)

print(word_stats("asjssjbjbbhsiaiic ifiaficjxzjooro qoprlllkskrmsnm mmvvllvlxjxj jfnnfcncnnccnncsllsdfi"))
print(word_stats("abbb cc a"))
print(word_stats("bbbaaacddddee"))

This would output:

c5d1f3i1j1l2n7s2  
a1b3 c2 a1  
a3b3c1d4e2

Upvotes: 0

nikeros
nikeros

Reputation: 3369

For me, it is a little bit trickier that it looks at first. For example, it does look that "bbbaaacddddee" -> "b3a3c1d4e2" needs the count results to be outputted in the order of appearance in the passed string:

import re

def unique_elements(t):
    l = []
    for w in t:
        if w not in l:
            l.append(w)
    return l

def splitter(s):
    res = []
    tokens = re.split("[ ]+", s)
    for token in tokens:
        s1 = unique_elements(token) # or s1 = sorted(set(token))
        this_count = "".join([k + str(v) for k, v in list(zip(s1, [token.count(x) for x in s1]))])
        res.append(this_count)
    return " ".join(res)

print(splitter("aaaa"))
print(splitter("abb")) 
print(splitter("abbb cc a"))
print(splitter("bbbaaacddddee")) 
print(splitter("a   b"))

OUTPUT

a4
a1b2
a1b3 c2 a1
b3a3c1d4e2
a1 b1

If the order of appearance is not a real deal, you can disregard the unique_elements function and simply substitute something like s1 = sorted(set(token)) within splitter, as indicated in the comment.

Upvotes: 1

mnist
mnist

Reputation: 6956

Does this satisfy your needs?

from itertools import groupby
s = "bbbaaac ddddee aa"

groups = groupby(s)
result = [(label, sum(1 for _ in group)) for label, group in groups]
res1 = "".join("{}{}".format(label, count) for label, count in result)
# 'b3a3c1 1d4e2 1a2'

# spaces just as spaces, do not include their count
import re
re.sub(' [0-9]+', ' ', res1)
'b3a3c1 d4e2 a2'

Upvotes: 1

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