I want to take two lists with dict value and find the specific values that appear only in first list.
In this case, only compare 'name' key.
a = [
{'name': 'joseph', 'age': 33},
{'name': 'Emma', 'age': 11},
{'name': 'apple', 'age': 44}
]
b = [
{'name': 'apple', 'age': 44},
{'name': 'Emma', 'age': 22}
]
returnOnlyOne(a, b)
would return [{'name': 'joseph', 'age': 33}], for instance.
The set() solution is not for this case.
For efficiency, we first make a set of the names in b, then filter the list a:
from operator import itemgetter
def returnOnlyOne(a, b):
b_names = set(map(itemgetter('name'), b))
only_in_a = list(filter(lambda item: item['name'] not in b_names, a))
return only_in_a
Sample output:
a = [
{'name': 'joseph', 'age': 33},
{'name': 'Emma', 'age': 11},
{'name': 'apple', 'age': 44}
]
b = [
{'name': 'apple', 'age': 44},
{'name': 'Emma', 'age': 22}
]
print(returnOnlyOne(a, b))
# [{'name': 'joseph', 'age': 33}]
If you don't like itemgetter, filter and the like, you can write the same using comprehensions:
def returnOnlyOne(a, b):
b_names = set(item['name'] for item in b)
return [ item for item in a if item['name'] not in b_names]
Use list comprehension with map. (BTW, what's inside your list is called dict):
[d for d in a if d.get('name') not in list(map(lambda x:x.get('name'), b))]
# [{'age': 33, 'name': 'joseph'}]
Explanation:
list(map(lambda x:x.get('name'), b)): gets all name from bd.get('name') not in: checks if name from a doesn't exist in b. (i.e. appear only in first list)Nearly the same as the others.
print([ item for item in a if item['name'] not in set(item['name'] for item in b)])
