Efficient Filtering of Lists in a Dictionary of Lists

Question

I’m working with some reasonably large datasets (500,000 datapoints with 30 variables each) and would like to find the most efficient methods for filtering them.

For compatibility with existing code the data is structured as a dictionary of lists but can’t be converted (e.g. to pandas DataFrame) and has to be filtered in situ.

Working example:

data = {'Param0':['x1','x2','x3','x4','x5','x6'],
        'Param1':['A','A','A','B','B','C'],
        'Param2': [100,200,150,80,90,50],
        'Param3': [20,60,40,30,30,5]}

# Param0 keys to keep
keep = ['x2', 'x4']

filtered = {k: [x for i, x in enumerate(v) if data['Param0'][i] in keep] for k, v in data.items()}

The result filtered gives the desired output, but this is very slow at scale.

Are there any quicker ways of doing this?

Asked By: awenborn

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Source

Answer 1

Creating a look_up_idx first might be a better idea:

look_up_idx = [idx for idx, v in enumerate(data['Param0']) if v in keep]
filtered = {k: [v[idx] for idx in look_up_idx] for k, v in data.items()}
print(filtered)

Or plus using numpy

import numpy as np

look_up_idx = [idx for idx, v in enumerate(data['Param0']) if v in keep]
filtered = {k: list(np.asarray(v)[look_up_idx]) for k, v in data.items()}

{'Param0': ['x2', 'x4'],
 'Param1': ['A', 'B'],
 'Param2': [200, 80],
 'Param3': [60, 30]}

Answered By: I'mahdi

Answer 2

I would do:

keep_idx = [i for i, v in enumerate(data['Param0']) if v in keep]
filtered = {k: [v[i] for i in keep_idx] for k, v in data.items()}

Timing

import numpy as np
from timeit import timeit

# Solution in question
def test_1(data, keep):
    return {
        k: [x for i, x in enumerate(v) if data['Param0'][i] in keep]
        for k, v in data.items()
    }

# First solution from @I'mahdi
def test_2(data, keep):
    keep_idx = [i for i, v in enumerate(data['Param0']) if v in keep]
    return {
        k: [val for i, val in enumerate(v) if i in keep_idx]
        for k, v in data.items()
    }

# Second solution from @I'mahdi
def test_3(data, keep):
    keep_idx = [i for i, v in enumerate(data['Param0']) if v in keep]
    return {k: list(np.asarray(v)[keep_idx]) for k, v in data.items()}

# Solution in this answer
def test_4(data, keep):
    keep_idx = [i for i, v in enumerate(data['Param0']) if v in keep]
    return {k: [v[i] for i in keep_idx] for k, v in data.items()}


data = {f"Param{i}": list(range(10_000)) for i in range(20)}
keep = list(range(0, 10_000, 100))

print(test_1(data, keep) == test_2(data, keep))
print(test_2(data, keep) == test_3(data, keep))
print(test_3(data, keep) == test_4(data, keep))

for i in range(1, 5):
    t = timeit(f"test_{i}(data, keep)", globals=globals(), number=10)
    print(f"Solution {i}: {t:.3f}")

results in something like:

Solution 1: 4.571
Solution 2: 4.220
Solution 3: 0.298
Solution 4: 0.219

Answered By: Timus

Efficient Filtering of Lists in a Dictionary of Lists

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