How do I count letters occurring in a string without using a dictionary or list.count()?

Question

I am trying to count each letter up without using count() or dict().

I did write something but I am still having issues with my code.

myString = []
#countList = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
myString ="pynativepynvepynative"
countList = [len(myString)+1]
for i in range(len(myString)):
        #print("Here 0")
        for j in range(len(countList)):
            #print("Here 1")
            if i == countList[j]:
                #print("Here 1.1")
                countList[j+1] = (countList[j+1] + 1)
                break
            else:
                #print("Here 2")
                countList.append(myString[i])
                countList.append(1)
                break
print(countList)

Expected output:

['p', 3, 'y', 3, 'n', 3, 'a', 2, 't', 2, 'i', 2, 'v', 3, 'e', 3]

Actual output:

[22, 'p', 1, 'y', 1, 'n', 1, 'a', 1, 't', 1, 'i', 1, 'v', 1, 'e', 1, 'p', 1, 'y', 1, 'n', 1, 'v', 1, 'e', 1, 'p', 1, 'y', 1, 'n', 1, 'a', 1, 't', 1, 'i', 1, 'v', 1, 'e', 1]

Are you trying to count the occurrence of each character? What is your expected output for this example? — adamgy
– adamgy, Commented Jun 29, 2020 at 22:42
So you would like to output a dictionary without using a dictionary? — adamgy
– adamgy, Commented Jun 29, 2020 at 22:46
Your output is a dictionary @NoajmIsMyName you might want to change it to a list of tuples or something similar if you absolutely do not want to use a dict. — Ryan Schaefer
– Ryan Schaefer, Commented Jun 29, 2020 at 22:49
I don't see the point of manipulating a list if you would like to return a dictionary. Why not just use a dictionary from the beginning? — adamgy
– adamgy, Commented Jun 29, 2020 at 22:52

sahasrara62 · Accepted Answer · 2020-06-29 23:00:34Z

2

what can you do is get the unique letters from the string and the for each unique letter loop through the string to count its frequency.

def func_count(string):
    letter = []
    for char in string:
        if char not in letter:
            letter.append(char)
    res = []
    for let in letter:
        count = 0
        for char in string:
            if let == char:
                count+=1
        res.extend([let, count])
    # res = {a:b for a,b in zip(res[::2], res[1::2])} 
    return res

string = "pynativepynvepynative"

solution = func_count(string)
print(solution)

output

['p', 3, 'y', 3, 'n', 3, 'a', 2, 't', 2, 'i', 2, 'v', 3, 'e', 3]

edit, if you want solution in dict form add res = {a:b for a,b in zip(res[::2], res[1::2])} in function

edited Jun 29, 2020 at 23:00

answered Jun 29, 2020 at 22:52

sahasrara62

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Comments

Noajm IsMy Name · Accepted Answer · 2020-06-30 05:50:50Z

Using my question code, I was able to get the right answer modifying the question's code

My problem was that I did not know how to initiate countList properly

countList = []
myString ="pynativepynvepynative"
for i in range(len(myString)):
        #print("Here 0")
        for j in range(len(countList)):
            #print("Here 1")
            #print(myString[i])
            #print(j)
            #print(countList[j])
            if myString[i] == countList[j]:
                #print("Here 1.1")
                #print(myString[i])
                countList[j+1] = (countList[j+1] + 1)
                break
        else :
            #print("Here 2")
            countList.append(myString[i])
            countList.append(1)
print(countList)

Actual output:

['p', 3, 'y', 3, 'n', 3, 'a', 2, 't', 2, 'i', 2, 'v', 3, 'e', 3]

smci · Accepted Answer · 2020-07-15 19:15:43Z

0

Use collections.Counter, the dict subclass for counting objects, which makes this a one-liner:

from collections import Counter
c = Counter('pynativepynvepynative')

Counter({'p': 3, 'y': 3, 'n': 3, 'v': 3, 'e': 3, 'a': 2, 't': 2, 'i': 2})

(Technically this isn't a dict, it's a subclass of dict.)

You can get a list-of-tuple from it:

>>> c.most_common()
[('p', 3), ('y', 3), ('n', 3), ('v', 3), ('e', 3), ('a', 2), ('t', 2), ('i', 2)]

Lists or tuples are undesirable for counting things, because you want to be able to separately access/sort by the keys (objects that you're counting) and the values (counts). In theory you can do that on list-of-list/tuple, but it's a pain, and Counter alrady defines several of the methods you'll need.

edited Jul 15, 2020 at 19:15

answered Jun 29, 2020 at 22:50

smci

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1 Comment

Noajm IsMy Name Over a year ago

I truly do understand your point. I am only doing this example just to be better at looping and alternating elements inside a list... I am still learning!

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