使用for循环检查列表时如何追加到新列表项？

Question

I am having problems with this and can't figure out why my code is not working. Could someone help me out with this, please? I made this:

def f8(a_list, n):
    """
    The parameter a_list is a list of int's. The parameter n is an int.
    The function f8() should return a list that contains exactly one
    of each number in a_list that occurs less than n times in a_list.

    Example:
        f8([1, 1, 7, 7, 7, 3, 3, 3, 4, 4, 4, 5, 5], 3) should return
        [1, 5] (as 1 and 5 are the only numbers that occur less than 3
        times in a_list)
    """

    k = []
    for i in a_list:
        if i < a_list.count(n):
            k.append(i)
            return k


print f8([1, 7, 7, 3, 3, 3, 4, 4, 5], 2)

I would expect it should print [1,5], but it just gives me a None. Why is that? Could someone help me out, please? I am stuck here.

Answer 1

You have the counting the wrong way around! You need to count the occurrences of i and compare this against n . Also, you need to move the return outside the for-loop . Finally, to remove the repeats in the final list , we should iterate through set(a_list) so that we only iterate through unique elements . The neat thing about this is that since we are counting the occurrences in the original a_list , we don't need to create any copies or anything fiddly to deal with this.

This makes your function :

def f8(a_list, n):
    k = []
    for i in set(a_list):
        if a_list.count(i) < n:
            k.append(i)
    return k

which works if we give it a test:

>>> f8([1, 1, 7, 7, 7, 3, 3, 3, 4, 4, 4, 5, 5], 3)
[1, 5]
>>> f8([1, 1, 2, 2, 2, 2, 4, 5, 6, 7, 7, 7, 7], 3)
[1, 4, 5, 6]

---

Note, if you wanted to shorten the function , you could achieve the same result in a one line list-comprehension :

def f8(a_list, n):
    return [i for i in set(a_list) if a_list.count(i) < n]

which gives the same outputs to the tests as above.

Answer 2

Firstly, you should not call count repeatedly. Each call iterates the entire list. You can get all counts in one go using collections.Counter . Secondly, you need to check if you have added any element before to avoid duplicates, just calling set one a_list will not guarantee order of appearance:

from collections import Counter

def f8(a_list, n):
    c = Counter(a_list)
    k, seen = [], set()
    for x in a_list:
        if c[x] < n and x not in seen:
            seen.add(x)
            k.append(x)
    return k

Answer 3

The condition to append in your code is never met, thus, no value is being returned. Instead, return after the loop is finished. Also, your counting condition is reversed, as you should be finding the count of i occurring in the list, not n :

k = []
for i in a_list:
    if a_list.count(i) < n and i not in k:
        k.append(i)

return k

The reason why you are receiving None is that the return statement does not pass, so to satisfy the variable storage during the function call, None is returned.