如何在不使用计数器或python中的字典的情况下查找项目出现在字符串列表中的次数？

Question

If I have

animal = [['cat', 'cat', 'dog'], ['cat', 'cat', 'dog'], ['cat', 'cat', 'dog']] 

and

item_to_find = ['cat', 'dog']

how do I calculate the number of times each string inside item_to_find can be found in animal ? for example, for 'cat' , the answer would be 6, and for 'dog' it would be 3. I need to do this without using collectiouns.Counter or a dictionary. I tried to use count , but it only works if it's a list of strings, and not a list of list of strings.

I tried to do it this way:

for i in item_to_find:
    print(animal.count(i))

but like I was saying this only works for a list of strings and not a list of list of strings.

My result would look like something like this: [6, 3] (as a list).

Answer 1

Use sum along with count to sum up the counts in all the sub-lists:

>>> animal = [['cat', 'cat', 'dog'], ['cat', 'cat', 'dog'], ['cat', 'cat', 'dog']]
>>> item_to_find = ['cat', 'dog']
>>> [(item, sum(a.count(item) for a in animal)) for item in item_to_find]
[('cat', 6), ('dog', 3)]

If you wanted to avoid using count you can do it almost as easily by using a nested generator with sum :

>>> [(item, sum(i == item for a in animal for i in a)) for item in item_to_find]
[('cat', 6), ('dog', 3)]

If you don't want to know which item each count goes with, make it a list of just the counts instead of a tuple that includes the item:

>>> [sum(a.count(item) for a in animal) for item in item_to_find]
[6, 3]

Answer 2

You can use iteration:

occurrences = [0] * len(item_to_find)
for sublist in animal:
    for idx, item in enumerate(item_to_find):
        occurrences[idx] += sublist.count(item)

Now occurrences will be [6, 3] for the example given.