I'm attempting to create a function that finds how many times an item occurs in a function. I have the code below, but it ends up counting all the items in the list(6), not just the one i want( in this case 1). I know i'm not returning anything, the print is mainly for me to see what answer I get.
def count(sequence, item):
found = 0
for i in sequence:
if item in sequence:
found = found + 1
print found
count([1,2,1,2,4,5], 1)
try this
def count(sequence, item):
found = 0
for i in sequence:
if i == item:
found = found + 1
print found
count([1,2,1,2,4,5], 1)
You have to compare element from sequence with item
if i == item:
Do you realize that this method already exists?
return sequence.count(item)
For instance, we get a value of 2 from
my_list = [1,2,1,2,4,5]
print my_list.count(1)
Okay, so you want something harder-headed. :-) How about this?
return len([i for i in sequence if i == item])
You could use sum and a comprehension
def count(haystack, needle):
return sum(el==needle for el in haystack)
This will work for any sequence you can iterate over.
>>> s = {"a": None, "b": None, "c": None}
>>> count(s, "a") # why are you counting a dict? You so crazy....
1
This works because True can be coerced to the integer 1. Summing a sequence of booleans results in the count of Trues in the sequence.
For future reference, python has an awesome Counter class that will do this for you.
from collections import Counter
someList = [1,2,1,2,4,5]
allCounts = Counter(someList)
print allCounts[1]
print allCounts
Python lists support this natively
>>> my_list = [1, 2, 1, 2, 4, 5]
>>> print my_list.count(1)
2
if item == i. printialso on each iteration. You probably will understand why you should check againstithat way.itemis always insequence. So for every iteration, your code gets intoifand incrementsfound.