How do I create the right members function?

Question

I will admit for loops are a bit tricky for me, so I need a bail out when it comes to challenges like this

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
my_dict = {'red': 'f00', 'orange': 'f80', 'yellow': 'ff0', 'green': '0f0', 'blue': '00f', 'indigo': '40f', 'violet': '80f'}
my_list = ['red', 'green', 'blue']

def members(dic, lst):
  venn = []

  while True:
    for key in dic:
      for item in lst == key:
        venn.append(item)

    print (len(venn))

members(my_dict, my_list)

Answer 1 · 2016-02-17T05:24:32Z

February 17, 2016 5:24am

You are heading in the right direction. The for loop runs a limited number of times due to the collection iterated over. This makes the while unnecessary. Only one loop is needed since the other check can be an if statement:

# Write a function named members that takes two arguments,
# a dictionary and a list of keys.
# Return a count of how many of the items in the list are also keys in the dictionary.
def members(dic, lst):
  # initialize count
  count = 0  
  # for key in dict
  for key in dic:
      # check if key is in list
      if key in lst:
          # increase count
          count += 1
  # return final count
  return count

Additionally, It is more efficient to search for the relatively few list items within the dict keys, than to look for all of the dict keys in the small list.

def members(dic, lst):
  # initialize count
  count = 0  
  for item in lst:
      if item in dic:
          # increase count
          count += 1
  # return final count
  return count

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How do I create the right members function?

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