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Python Python Collections (Retired) Dictionaries Membership

Stefan Vaziri
Stefan Vaziri
17,453 Points

Help

Unsure how to have two arguments and count how many of the items in the list are in the other list. Used the count function but that didn't work either.

counts.py
# You can check for dictionary membership using the
# "key in dict" syntax from lists.

### Example
# my_dict = {'apples': 1, 'bananas': 2, 'coconuts': 3}
# my_list = ['apples', 'coconuts', 'grapes', 'strawberries']
# members(my_dict, my_list) => 2
def members(d, l):
  d = {'apples':1,'bananas':2,'coconuts':3}
  l = ['apples','coconuts','grapes','strawberries']
  return members(d,l))

3 Answers

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,454 Points

The code needs to iterate over the list and check if item is in the dict.

# members(my_dict, my_list) => 2
def members(d, l):
    count = 0
    # for each item in list
    for item in l:
        # check if otem is a dict key
        if item in d:  # same as 'if kitem in d.keys()'
            # add to count
            count += 1
    return count
Stefan Vaziri
Stefan Vaziri
17,453 Points

Can you explain to me why you used "count = 0" and "count +=1"?

The goal of the function is to return a count of how many times we can find a key from the dictionary that is also listed in the list. The count variable is first set to zero then each time a key is found in the dictionary and list, the count variable is increased by one. The for loop does the search and the if condition checks each loop/cycle to see if there is a match.

Chris Freeman
Chris Freeman
Treehouse Moderator 68,454 Points

A standard way to accumulate a "count" is to add one and assign back to count:

count = count + 1

The Python shorthand for this is to use the "+=" notation to mean the same thing:

count += 1  # Same as count = count + 1

The final part is setting an initial value for count. This is done so the right-side of the equation "count + 1" does not raise an NameError error due to count not yet being set on the first iteration. The simplest way to set count is to give it a default starting value before the loop:

count = 0
def members(d, l):
    count = 0
    for key in d:
        if key in l:
            count += 1
    return count
Chris Freeman
Chris Freeman
Treehouse Moderator 68,454 Points

I suggest reversing your use of l and d. If the dictionary is very large relative to the list, it will run quicker looking for each list item in the dict keys.

Thanks, I see what you are saying.

Bart Bruneel
Bart Bruneel
27,212 Points

Hello Stefan,

First of all, there is no need to the define the list and dictionary inside your function. These arguments will be passed to the function if it gets called in the evaluation. I would approach the problem as follows:

  • first you'll need to extract the keys from the library. This can be done with the .keys() method. This method returns a list of the keys.
  • Then you set a count to zero. -Subsequently you loop over the list-items in my_list.
  • If the item is also in the keys-list, you add 1 to the count.
  • return the count.

Happy coding.