CODEWITHSUNDEEP
pythondictionarylist-comprehension
Sarang Manjrekar
Sarang Manjrekar

Reputation: 1991

list comprehension on a dictionary giving empty list

I have a dictionary as below :

 'give': (('VBP', 6), ('VB', 15)),
 'discounts': ('NNS', 1),
 'maintaining': ('VBG', 4),
 'increasing': ('VBG', 18),
 'spending': (('NN', 24), ('VBG', 2)),
 'become': ((('VBN', 7), ('VB', 15)), ('VBP', 1)),
 'permanent': ('JJ', 2),
 'fixtures': ('NNS', 1),
 'news': ('NN', 24),
 'weeklies': ('NNS', 2),
 'underscore': ('VBP', 1),
 'fierce': ('JJ', 2),
 'competition': ('NN', 10)

I am writing a list comprehension as below :

result = [x for x in mydict.items() if type(x[1][0]) == 'str']

But this results in an empty list, whereas if I see there are many elements in dictionary, for which this condition satisfies.

Upvotes: 1

Views: 94

Answers (1)

niraj
niraj

Reputation: 18208

Either you can change 'str' to str i.e. 

result = [x for x in mydict.items() if type(x[1][0]) == str]

or may be you can try with isinstance method to check if it is the instance of string (details):

result = [x for x,value in mydict.items() if isinstance(x[1][0],str)]
print(result)

Result:

['increasing', 'maintaining', 'fierce', 'permanent', 'fixtures', 'underscore', 'news', 'weeklies', 'discounts', 'become', 'give', 'competition', 'spending']

Upvotes: 3

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