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.
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
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.
Use filter
function like this
len(filter(lambda x: x==letter, myString))
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)))