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Python Python Collections (Retired) Dictionaries Membership

Parvinder Grewal
Parvinder Grewal
3,678 Points
Posted by Parvinder Grewal
Parvinder Grewal
3,678 Points

Not sure how to compare list items with dictionary keys.

def members(dictArg, keysListArg):
    count = 0
    for x in my_list:
        if my_list[x] == my_dict.keys():
            count += 1
    return count

#I am not sure how to iterate through a list and than compare it with the dictionary keys.

[MOD: added ```python markdown formatting -cf]

2 Answers

Brendan Whiting
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.a{fill-rule:evenodd;}techdegree seal-36
Brendan Whiting
Front End Web Development Techdegree Graduate 84,738 Points
Brendan Whiting
.a{fill-rule:evenodd;}techdegree seal-36
Brendan Whiting
Front End Web Development Techdegree Graduate 84,738 Points

The .keys() method on a dictionary is going to return a list. You can't just compare that whole list to each item in the other list, you have to compare item to item. Also, you have to use the parameter names that you defined, dictArg and keysListArg, not my_list and my_dict. Those might be the names of actual things you pass in, but within the function you need to stay consistent. (You could also just call your parameters my_list and my_dict).

Here's how I solved it. There's a loop within a loop. For each list_value, it's going to loop through each dict_value and compare them, and then do it all over again for the next list_value.

def members(dictArg, keysListArg): 
  count = 0 
  for list_item in keysListArg: 
    for dict_key in dictArg.keys():
      if list_item == dict_key:
        count+= 1
  return count
Chris Freeman
Chris Freeman
Treehouse Moderator 68,454 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,454 Points

This works but can be improved: You don't need to check each list_item individually against each dictArg key. Instead you can use the in keyword. Also, when referring to a dictionary as an iterable, operating over the keys is assumed, so the .keys() is unnecessary:

def members(dictArg, keysListArg): 
  count = 0 
  for list_item in keysListArg: 
    if list_item in dictArg:
        count+= 1
  return count
Chris Freeman
Chris Freeman
Treehouse Moderator 68,454 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,454 Points

No worries. Since my derivative work was based on yours, I wanted to extend your answer so that you would get points when this gets "Best Answer". Everybody wins!

Parvinder Grewal
Parvinder Grewal
3,678 Points
Parvinder Grewal
Parvinder Grewal
3,678 Points

Brendan Whiting: Thank you for your response, excuse my typo with the parameter names.

Chris Freeman Thank for your help once again. I'm glad my answer was somewhere close.