Count letters in a word in python debugging

Question

I am trying to count the number of times “e” appears in a word.

def has_no_e(word):     #counts 'e in a word
    letters = len(word)
    count = 0
    while letters >= 0:
        if word[letters-1] == 'e':
            count = count + 1
        letters = letters - 1
    print count

It seems to work fine, except that the word ends with "e". He will count this "e" twice. I have no idea why. Any help?

I know that my code may be messy, I'm a beginner! I'm just trying to understand the logic of what is happening.

Answer 1

As already mentioned, you can implement the test with a simple one word.count('e'). If you do not do this as a simple exercise, it is much better than trying to invent a wheel.

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def count_letter(word, char='e'):
    count = 0
    for c in word:
        if c == char:
            count += 1
    return count

:

>>> count_letter('tee')
2
>>> count_letter('tee', 't')
1
>>> count_letter('tee', 'f')
0
>>> count_letter('wh' + 'e'*100)
100

Answer 2

>>> word = 'eeeooooohoooooeee'
>>> word.count('e')
6

?

Answer 3

def has_no_e(word):
    return sum(1 for letter in word if letter=="e")

Answer 4

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def has_no_e(word):
    count = 0
    for letter in word:
        if letter == "e":
            count += 1
    print count

- :

def has_no_e(word):
    return sum(1 for letter in word if letter=="e")

Answer 5

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Answer 6

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Answer 7

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Answer 8

@marcog ;

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def has_no_e(word):
    letters = len(word)
    count = 0
    while letters >= 0:
        ch = word[letters-1]         # what is it looking at?
        if ch == 'e':
            count = count + 1
            print('{0} <-'.format(ch))
        else:
            print('{0}'.format(ch))
        letters = letters - 1
    print count

has_no_e('tease')

e <-
s
a
e <-
t
e <-
3

,

• e
• you "wrap" to the end of the line - therefore, an extra e if your line ends on one

Answer 9

If you really need "has_no_e", then the following might be more appropriate than counting "e" and then checking for zero,

def has_no_e(word):
  return 'e' not in word

>>> has_no_e('Adrian')
True
>>> has_no_e('test')
False
>>> has_no_e('NYSE')
True

If you want to check if there is an "E",

def has_no_e(word):
  return 'e' not in word.lower()

>>> has_no_e('NYSE')
False