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I am trying to write a program to count the occurrences of a specific letter in a string without the count function. I made the string into a list and set a loop to count but the count is never changing and i cant figure out why. This is what I have right now:

letter = 'a'
myString = 'aardvark'
myList = []

for i in myString:
    myList.append(i)

count = 1

for i in myList:
    if i == letter:
        count == count + 1

    else:
        continue

print (count)

Any help is greatly appreciated.

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4
  • 1
    count == count + 1 is not an assignment. Simply remove one =. Commented Mar 10, 2015 at 14:37
  • Why inventing things badly if 'aardvark'.count('a') does the job better and faster? (At least it isn't as errorprone) Commented Mar 10, 2015 at 14:38
  • Your for loop features an if...else... where the else features a continue. The continue is not necessary because there is nothing to avoid computing with the for loop. A solitary if would suffice. Commented Mar 10, 2015 at 14:39
  • 1
    Creating a list out of a string is also completely superfluous: for char in myString: ... does the job and is very Pythonic at that. Commented Mar 10, 2015 at 14:40

8 Answers 8

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2

Be careful, you are using count == count + 1, and you must use count = count + 1

The operator to attribute a new value is =, the operator == is for compare two values

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1 Comment

Thank you! That was just my stupidity. I cant believe I made that error.
1

Instead of 

count == count + 1

you need to have

count = count + 1
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Comments

1

Although someone else has solved your problem, the simplest solution to do what you want to do is to use the Counter data type:

>>> from collections import Counter
>>> letter = 'a'
>>> myString = 'aardvark'
>>> counts = Counter(myString)
>>> print(counts)
Counter({'a': 3, 'r': 2, 'v': 1, 'k': 1, 'd': 1})
>>> count = counts[letter]
>>> print(count)
3

Or, more succinctly (if you don't want to check multiple letters):

>>> from collections import Counter
>>> letter = 'a'
>>> myString = 'aardvark'
>>> count = Counter(myString)[letter]
>>> print(count)
3

The simplest way to do your implementation would be:

count = sum(i == letter for i in myString)

or:

count = sum(1 for i in myString if i == letter)

This works because strings can be iterated just like lists, and False is counted as a 0 and True is counted as a 1 for arithmetic.

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6 Comments

how would you go about storing each count of each letter? for example if we had aaabbccbb, and we wanted it as a3b42c??
You can just loop over the counter using items(), then combine them with a ''.join. This is a one-liner: ''.join('{}{}'.format(x, y) for x, y in Counter('aaabbccbb').items()). Output is 'a3b4c2'
Thanks! another question, how would you go about it without any built in functions?
This is sounding like a homework question. In real life you would never do this sort of thing without built in functions. You should be able to adapt my sum based approach above to do that, but I won't give you the answer to a homework question.
i was more interested in the approach rather than an answer, thanks anyways.
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1

Apart from the above methods, another easiest way is to solve the problem by using the python dictionary

 word="purple"
 dic={}
 for letter in word:
     if letter in dic:
        dic[letter]+=1
     else:
        dic[letter]=1
 print(dic)
 {'p': 2, 'u': 1, 'r': 1, 'l': 1, 'e': 1}

In case if you want to count the occurences of a particular character in the word.we can get that it by following the below mentioned way,

        dic['p']
        2
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1 Comment

This is what I needed... Thank you
0

Use filter function like this

len(filter(lambda x: x==letter, myString))
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Comments

0

Your count is never changing because you are using == which is equality testing, where you should be using = to reassign count. Even better, you can increment with 

count += 1

Also note that else: continue doesn't really do anything as you will continue with the next iteration of the loop anyways. If I were to have to come up with an alternative way to count without using the count function, I would lean towards regex:

import re
stringy = "aardvark"
print(len(re.findall("a", stringy)))
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0

Your code logic is right except for considering count == 1 while using count == 1 you are comparing if count == 1 and count = 1 is for assigning and count += 1 is for incrementing.
Probably you know this, you might have got confused also, you have to initialize count = 0

letter = 'a'
myString  = 'aardvark'
myList = []

for i in myString:
    myList.append(i)
count = 0
for i in myList:
    if i == letter:
        count +=1
    else:
        continue
print(count)
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Comments

0

For the code to function properly, two alterations need to be made:

  1. Modify the initialization of count
count = 0
  1. Use = instead of ==
count = count + 1

or

count += 1
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1 Comment

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