CODEWITHSUNDEEP
pythonpython-3.xdictionary
Fusdev
Fusdev

Reputation: 127

How to selectively create a dictionary from an existing one?

I have two dictionaries like below which have common keys:

dictionary1 = {1: 'a', 2: 'b' , 3: 'c'}
dictionary2 = {1: 'no', 2: 'yes' ,3:'yes'}

I want to create a new dictionary with the key and values of dictionary1 only if the corresponding values of the dictionary2 key has "yes".

Expected output:

{2: 'b', 3: 'c'}

What I have tried:

dictionary1 = {1: 'a', 2: 'b' , 3: 'c'}
dictionary2 = {1: 'no', 2: 'yes' ,3:'yes'}
common_pairs = dict()

for key,value in dictionary2.items():
  for key,v in dictionary1.items():
    if(value == "yes"):
      common_pairs[key] =  v

Upvotes: 2

Views: 61

Answers (3)

Alain T.
Alain T.

Reputation: 42133

You could do this in a dictionary comprehension:

common_pairs = { key:value for key,value in dictionary1.items()
                           if dictionary2.get(key,"") == "Yes" }

The get function provides a default value if the key is not present. this will prevent errors if a key in dictionary1 is absent from dictionary2.

Upvotes: 1

fignick
fignick

Reputation: 11

You can just iterate through dict2, check if the element is in dict1 and that the key is yes, and then add it to the dictionary.

dictionary1 = {1: 'a', 2: 'b' , 3: 'c'}
dictionary2 = {1: 'no', 2: 'yes' ,3:'yes'}
common_pairs = dict()

for key in dictionary2:
    if key in dictionary1 and dictionary2[key] == 'yes':
        common_pairs[key] = dictionary1[key]

Upvotes: 0

Brad Solomon
Brad Solomon

Reputation: 40888

You don't need a nested for loop. Just a single iteration over dictionary1 items with corresponding O(1) lookup in dictionary2:

With a dict comprehension this would look like:

>>> dictionary1 = {1: 'a', 2: 'b' , 3: 'c'}
>>> dictionary2 = {1: 'no', 2: 'yes' ,3:'yes'}
>>> new = {k: v for k, v in dictionary1.items() if dictionary2[k] == 'yes'}
>>> new
{2: 'b', 3: 'c'}

With a traditional for loop:

>>> new = {}
>>> for k, v in dictionary1.items():
...     if dictionary2[k] == 'yes':
...         new[k] = v
... 
>>> new
{2: 'b', 3: 'c'}

Upvotes: 4

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