I am a beginner in python and I would like your opinion on a more sophisticated way to write the code that follows. I want the code to iterate through the whole list (x) of items, count how many times each item appears and store the output in a new list (z). Here is the code (it works as it is) :
x = ["apple", "orange", "cherry", "apple"]
def new_list(a):
z=[]
for i in x:
y = x.count(i)
(z.append(y))
return z
print(new_list(x))
I would suggest to do a list comprehension:
>>> x = ["apple", "orange", "cherry", "apple"]
>>> [x.count(val) for val in x]
[2, 1, 1, 2]
>>>
Or even better with map:
>>> [*map(x.count, x)]
[2, 1, 1, 2]
>>>
With using list.count() order of your programming is O(n^2) and with Counter and hashmap order of your programmer is O(n).
Try this:
x = ["apple", "orange", "cherry", "apple"]
[x.count(i) for i in x]
# [2,1,1,2]
Or use Counter:
from collections import Counter
dct = Counter(x)
# Counter({'apple': 2, 'orange': 1, 'cherry': 1})
[dct[i] for i in x]
# [2, 1, 1, 2]
Check Runtime: (Counter is faster than list.count())
Try using the Counter class:
from collections import Counter
x = ["foo", "bar", "foo", "foo"];
z = Counter(x)
print(z)
# Output: {"foo": 3, "bar": 1}
You can use the Counter class in collections. So iterate over input list and then find count of every element.
from collections import Counter
x = ["apple", "orange", "cherry", "apple"]
c = Counter(x)
print([c[item] for item in x])
#[2, 1, 1, 2]
You can use the Counter subclass of the collections module:
from collections import Counter
x = ["apple", "orange", "cherry", "apple"]
counter = Counter(x)
z = list(counter.values())
print(z) #[2,1,1]
I didn't use count because it may not be good for performance which is also pointed out by this answer.
