CODEWITHSUNDEEP
pythondictionary
Ambriador
Ambriador

Reputation: 23

filter() a dict in python - return just the key

I want to filter a dictionary, where I get the keys when they have a specific value, and just want to return the keys from the dict. Here is my code:

def smaller_than(d, value):
    result = filter(lambda x: x[1]<=3, d.items())
    return result

number_key = {35135135: 5, 60103513: 3, 10981179: 2, 18637724 : 4}
number = smaller_than(number_key, 3)
print(number)

There I get a result like [(60103513, 3), (10981179, 2)], but I just want to get it like [60103513, 10981179]. I think it is a easy answer but I don't know how to do this.

Upvotes: 2

Views: 2213

Answers (3)

Moinuddin Quadri
Moinuddin Quadri

Reputation: 48120

Alternatively, you may also use a list comprehension expression to filter your dictionary as:

>>> number_key = {35135135: 5, 60103513: 3, 10981179: 2, 18637724 : 4}

>>> [k for k, v in number_key.items() if v<= 3]
[60103513, 10981179]

Upvotes: 1

Liron Lavi
Liron Lavi

Reputation: 431

first solution, just add:

new_list = []
for i in number:
    new_list.append(i[0])

print new_list

second solution:

def smaller_than(d, value):
    result = filter(lambda x: x[1]<=3, d.items())
    result = map(lambda x: x[0], result)
    return result

number_key = {35135135: 5, 60103513: 3, 10981179: 2, 18637724 : 4}
number = smaller_than(number_key, 3)
print(number)

Upvotes: 0

Moinuddin Quadri
Moinuddin Quadri

Reputation: 48120

You do not have to pass d.items() as an iterable to filter. Just pass the dict object (same as dict.keys()) and filter the content based on dict[key] with lambda expression as:

>>> number_key = {35135135: 5, 60103513: 3, 10981179: 2, 18637724 : 4}

>>> filter(lambda x: number_key[x]<=3, number_key)
[60103513, 10981179]

Upvotes: 2

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