Count of each unique element in a list [duplicate]

Question

Say I have a list of countries

l = ['India', 'China', 'China', 'Japan', 'USA', 'India', 'USA']  

and then I have a list of unique countries

ul = ['India', 'China', 'Japan', 'USA']

I want to have a count of each unique country in the list in ascending order. So output should be as follows:

Japan 1
China 2
India 2
USA   2

Answer 1

You can use Counter from collections:

from collections import Counter

l = ["India", "China", "China", "Japan", "USA", "India", "USA"]

new_vals = Counter(l).most_common()
new_vals = new_vals[::-1] #this sorts the list in ascending order

for a, b in new_vals:
    print a, b

Answer 2

If you don't want to use a Counter you can count yourself (you already know the unique elements because you have ul) using a dictionary:

l = ['India', 'China', 'China', 'Japan', 'USA', 'India', 'USA'] 
ul = ['India', 'China', 'Japan', 'USA']

cnts = dict.fromkeys(ul, 0)  # initialize with 0

# count them
for item in l:
    cnts[item] += 1

# print them in ascending order
for name, cnt in sorted(cnts.items(), key=lambda x: x[1]):  # sort by the count in ascending order
    print(name, cnt)   
    # or in case you need the correct formatting (right padding for the name):
    # print('{:<5}'.format(name), cnt)  

which prints:

Japan 1
China 2
India 2
USA   2