CODEWITHSUNDEEP
pythonpython-3.x
misheekoh
misheekoh

Reputation: 460

Counting a set of characters in a string

I'm trying to count a set of characters, 'qed' in a string. My idea is to iterate through each character in a given string and if N(i), N(i-1), N(i-2) matches 'qed', update the count but failed so far. Any suggestions? Thanks!

def test(N):
    s = ('qed')
    count = 0
        for i in range(len(N)):
            if N[i] + N[i-1] + N[i-2] == s:
                count = count + 1
        return print(count)
test('qedmlqedlolqed')

Upvotes: 0

Views: 87

Answers (5)

Stefan Pochmann
Stefan Pochmann

Reputation: 28596

>>> 'qedmlqedlolqed'.count('qed')
3

Edit: Why the downvote? The question asks "Any suggestions?" and I think this is a good one.

Upvotes: 2

Ashish Kumar
Ashish Kumar

Reputation: 347

though Stefan's answer is simplest and lucid, here's another way to do it using list comprehension

s = 'qedmlqedlolqed'

result = len([1 for i in range(len(s)) if s[i:i+3] == 'qed']) (thanks Stefan)

Upvotes: 2

Tamas Hegedus
Tamas Hegedus

Reputation: 29896

Fixing your code:

def test(N):
  s = 'qed'
  count = 0
  for i in range(2, len(N)):
    if N[i-2] + N[i-1] + N[i] == s:
      count = count + 1
  return count
print(test('qedmlqedlolqed'))

Or you could count how many suffixes of the string starts with qed:

def test2(word, sub):
  return sum(1 for i,_ in enumerate(word) if word[i:].startswith(sub))
print(test2('qedmlqedlolqed', 'qed'))

Or you could just literally count all the substrings, and check the number of yours:

import collections
def all_substrings(s):
  for i in range(len(s)):
    for j in range(i, len(s)):
      yield s[i:j+1]

def test3(word, sub):
  return collections.Counter(all_substrings(word))[sub]
print(test3('qedmlqedlolqed', 'qed'))

Upvotes: 1

Stefan Pochmann
Stefan Pochmann

Reputation: 28596

Using find repeatedly:

>>> text = 'qedmlqedlolqed'
>>> count = index = 0
>>> while True:
        index = text.find('qed', index) + 1
        if not index:
            break
        count += 1

>>> count
3

I went with + 1 instead of + 3 to also support overlapping occurrences (can't happen with 'qed' or with any other "set of characters", assuming what you meant with that is a string without duplicates).

Upvotes: 0

mhawke
mhawke

Reputation: 87054

Fixing your code:

def test(N):
    s = 'qed'
    count = 0
    for i in range(len(N)-2):
        if N[i:i+3] == s:
            count += 1
    return count

>>> test('qedmlqedlolqed')
3

Or more generally:

def test(N, s):
    count = 0
    if s:
        for i in range(len(N)-len(s)+1):
            if N[i:i+len(s)] == s:
                count += 1
    return count

>>> test('qedmlqedlolqed', 'qed')
3
>>> test('qedmlqedlolqed', 'ed')
3
>>> test('qedmlqedlolqed', 'd')
3
>>> test('qedmlqedlolqed', '')
0
>>> test('qedmlqedlolqed', 'lol')
1
>>> test('qedmlqedlolqed', 'rofl')
0

Or, much easier, using str.count():

>>> 'qedmlqedlolqed'.count('qed')
3

'

Upvotes: 2

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