How To Remove Duplicate Item In List?

If I have a list for example : courses = [{name: a, course: math, count:1}] and if I input again name: a course: math the list will be courses = {name: a, course: math, count:

Solution 1:

May I suggest you a different approach?

Instead of using a list of dictionaries wich may be complicated to manage in your case, write your own class to store "name" and "course".

class NC:
    def __init__(self, n, c):
        self.name = n
        self.course = c

    def __hash__(self):
        return hash((self.name, self.course))

    def __eq__(self, other):
        if self.__hash__() == other.__hash__():
            return True
        else:
            return False

    def __repr__(self):
        return "[{}: {}]".format(self.name, self.course)

By defining the special method __hash__ and __eq__ you make your objects hashable, so they can be counted by Counter. If you write something like this:

from collections import Counter

a = NC("a", "math")
b = NC("b", "physics")
c = NC("a", "math")

l = [a, b, c]

lc = Counter(l)
print lc

the print will gives you Counter({[a: math]: 2, [b: physics]: 1})

With this approach, you may just append all the NC objects to your list and at the end use Counter to get the repetitions.

EDIT after request in the comment.

To do the same thing in "real time" you can create an empty counter an then update it.

from collections import Counter

lc = Counter()

lc.update([NC("a", "math")])
print lc #this prints: Counter({[a: math]: 1})

lc.update([NC("b", "physics")])
print lc #this prints: Counter({[a: math]: 1, [b: physics]: 1})

lc.update([NC("a", "math")])
print lc #this prints: Counter({[a: math]: 2, [b: physics]: 1})

Just remember that Counter.update wants an iterable, so if you want to add one element to the Counter, you have to give in input a list with that one element. Of course you may also add more elements togheter, for example: lc.update([NC("b", "physics"), NC("c", "chemistry")]) is valid and both objects are added to the counter.

Solution 2:

You can use a for else clause. The else part will be called only if break is not reached, here is an example for you

courses = []
courses.append({'name': 'a', 'course': 'math', 'count': 1})
def add_course(d):
    for course in courses:
        if course['course'] == d['course'] and course['name'] == d['name']:
            course['count'] += 1
            break
    else:
        d['count'] = 1
        courses.append(d)


add_course({'name': 'a', 'course': 'math'})
add_course({'name': 'a', 'course': 'english'})
print(courses)

As an output you have [{'name': 'a', 'course': 'math', 'count': 2}, {'name': 'a', 'course': 'english', 'count': 1}]