I've got the following exercise: Write the function countA(word) that takes in a word as argument and returns the number of 'a' in that word. Examples
>>> countA("apple")
1
>>> countA("Apple")
0
>>> countA("Banana")
3
My solution is:
def countA(word):
return len([1 for x in word if x is 'a'])
And it's ok. But I'm not sure if it's the best pythonic way to solve this, since I create a list of 1 values and check its length. Can anyone suggest something more pythonic?
Assuming you're not just allowed to use word.count() - here's another way - possibly faster than summing a generator and enables you to do multiple letters if you wanted...
def CountA(word):
return len(word) - len(word.translate(None, 'A'))
Or:
def CountA(word):
return len(word) - len(word.replace('A', ''))
use this
def countA(word):
return word.count('a')
As another example, myString.count('c') will return the number of occurrences of c in the string
In python, strings are sequences (just like a list).
Then you could use count :
>>> 'Apple'.count('p')
2
Using this, the answer to your exercise would be:
def countA(word):
return word.count('a')
I know this has already been answered, but I can't believe nobody mentioned using a Counter for this, since it's very well suited to this problem. It might have a little more overhead, but if you want to find multiple letters it's probably the best way to go. Here it is in a simple, readable lambda function.
from collections import Counter
count = lambda word, letter: Counter(word)[letter]
count("test", "t")
>>> 2
Counter(word).get(letter, 0) can be replaced with Counter(word)[letter] as Counter returns 0 for missing items
Use sum:
>>> def countA(word):
return sum(1 for x in word if x == 'a')
>>> countA("apple")
1
>>> countA("Apple")
0
>>> countA("Banana")
3
sum and length return the same, as long as 1 is returned inside the list comprehension.
[1 for x in word if x is 'a'] and then calculate its length. Sum on the other hand just goes trough iterator over the original word.
is 'a' use == ais is incorrect. It returned correct results only accidentally (because in Python small equal strings point to the same objects, but for longer equal strings objects will be different, so is will return False).
word.count('a')?