Python | Return the number of times the item occurs in the list

Question

def count(squence,item):
    count=0
    i=0
    for item in squence:
        if item == squence:
            count+=1

    print count

count([1,7,8,7,7],7)

I don't understand why the If statment doesn't work :( Thanks,

Answer 1

list already has a function count() : [1,7,8,7,7].count(7) returns 3

But what you are trying to do is:

def count(squence, item):
    cnt = 0
    for i in squence:
        if i == item:
            cnt += 1

    print cnt

In your code, you overwrite item : item is the value you want to count but it is also the values you check, so it does not work...

Answer 2

You are overwriting your the variables your trying to check against. Also you need to compare the target item to the item that is in the sequence. The code below works.

def count(squence,target):
    count=0
    i=0
    for item in squence:
        if target == item:
            count+=1

    print count

count([1,7,8,7,7],7)

Answer 3

why not use the count method in python? Unless you are doing this as an exercise it makes sense to use pythons standard functions, right?

>>> mylist=[1,7,8,7,7]
>>> mylist.count(7)
3