CODEWITHSUNDEEP
pythonpython-3.xlistloopsdictionary
user7131669
user7131669

Reputation: 61

Filtering Dictionary with List of Lists as Value

I have a dictionary with values being a list of lists, e.g.:

Review_dict["0002"]

[['0002', '0292', '4', datetime.datetime(1998, 2, 27, 11, 1, 34)],
 ['0002', '0251', '5', datetime.datetime(1998, 2, 27, 12, 2, 24)],
 ['0002', '0050', '5', datetime.datetime(1998, 2, 27, 12, 3, 24)],
 ['0002', '0314', '1', datetime.datetime(1998, 3, 4, 10, 5, 45)]]

I want to remove certain list (within the "big" list), where 2nd element = '0251' or '0314'.

e.g. Review_dict["0002"] should become:

[['0002', '0292', '4', datetime.datetime(1998, 2, 27, 11, 1, 34)],
 ['0002', '0050', '5', datetime.datetime(1998, 2, 27, 12, 3, 24)]]

I have come up with this code but I want to know if there's a simpler/ more elegant way to do so.

remove = ['0251', '0314']

for key in list(Review_dict):
    for ID in remove:
        for num in range(len(Review_dict[key])):
            if Review_dict[key][num][1] == ID:
                del Review_dict[key][num]

Upvotes: 1

Views: 443

Answers (3)

user2390182
user2390182

Reputation: 73460

You could just mutate the values of your outer dict:

remove = ['0251', '0314']

for v in Review_dict.values():
    v[:] = (lst for lst in v if lst[1] not in remove)

Generally, one should not repeatedly remove from a list as each remove has linear complexity. Building from scratch is the better option. Whether this performs better depends on your data and how much there actually is to remove.

As a mutation, the slice assignment v[:] = ... allows to ignore the keys of the outer dict entirely as you don't need to rebind them.

Upvotes: 1

Kays Kadhi
Kays Kadhi

Reputation: 560

You can use the filter method and pass a lambda function

Review_dict["0002"] = list(filter(lambda x:x[1] not in remove,Review_dict["0002"]))

Upvotes: 0

Krishna Chaurasia
Krishna Chaurasia

Reputation: 9572

You can use list comprehension as:

import datetime
Review_dict = dict()

Review_dict["0002"] = [['0002', '0292', '4', datetime.datetime(1998, 2, 27, 11, 1, 34)],
 ['0002', '0251', '5', datetime.datetime(1998, 2, 27, 12, 2, 24)],
 ['0002', '0050', '5', datetime.datetime(1998, 2, 27, 12, 3, 24)],
 ['0002', '0314', '1', datetime.datetime(1998, 3, 4, 10, 5, 45)]]

print(Review_dict)
Review_dict["0002"] = [l for l in Review_dict["0002"] if l[1] not in ['0251', '0314']]
print(Review_dict)

Output:

{'0002': [['0002', '0292', '4', datetime.datetime(1998, 2, 27, 11, 1, 34)], ['0002', '0050', '5', datetime.datetime(1998, 2, 27, 12, 3, 24)]]}

Here you only take elements which satisfy the criteria i.e. first element of inner list is not present in some other list of elements.

Upvotes: 0

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