Counting in sequences - function

Question

Say I want to count values in sequence xs of how many times it appears in v and return the integers in a list in the same order. Including dups. This the code I have so far and I'm kind of stuck on what to do. Trying to keep it simple without .count funcs and what not.
def count_each(xs,v):
count = []
for i in range(len(xs)):
if xs(i) == v:
return count.append(i)
return count

Answer 1

You could use the list method count().
>>> keys = [10, 20, 30]
>>> search = [10, 20, 50, 20, 40, 20]
>>> print [search.count(key) for key in keys]
[1, 3, 0]
alternatively O(n),
>>> from collections import Counter
>>> c = Counter(search)
>>> print [c[key] for key in keys]
[1, 3, 0]

Answer 2

Below is the sample function to achieve this using collections.defaultdict:
from collections import defaultdict
def count_each(xs, v):
count = defaultdict(int)
for item in v:
if item in xs:
count[item] += 1
return [count[item] for item in xs]
OR, using simple dict as:
def count_each(xs, v):
count = {}
for item in v:
if item in xs:
if item not in count:
count[item] = 0
count[item] += 1
return [count.get(item, 0) for item in xs]
Sample call:
>>> count_each([10,20,30],[10,20,50,20,40,20])
[1, 3, 0]

Answer 3

You can use function list.count(element) to count element in list
So your code can look like this
def count_each(xs, v):
result = []
for element in xs:
result.append( v.count(element) )
return result
count_each([10,20,30],[10,20,50,20,40,20])
or you can write it shorter using list comprehension - see #mrdomoboto answer.
EDIT: the same without count()
def count_each(xs, v):
result = []
for element in xs:
count = 0
for x in v:
if x == element:
count += 1
result.append( count )
return result
count_each([10,20,30],[10,20,50,20,40,20])
BTW: you can't do this with dict because dict don't have to keep order. One time you can get result [1,3,0] but another time [3,1,0] or [0,1,3], etc.