Aside: Fast Permutation Tests

import numpy as np
import pandas as pd
import os

import plotly.express as px
pd.options.plotting.backend = 'plotly'

Speeding things up 🏃

Speeding up permutation tests

• A permutation test, like all simulation-based hypothesis tests, generates an approximation of the distribution of the test statistic.
  • If we found all permutations, the distribution would be exact!
  • If there are $a$ elements in one group and $b$ in the other, the total number of permutations is ${a + b \choose a}$.
  • If $a = 100$ and $b = 150$, there are more than ${250 \choose 100} \approx 6 \cdot 10^{71}$ permutations!

• The more repetitions we use, the better our approximation will be.
• Unfortunately, our code is pretty slow, so we can't use many repetitions.

Example: Birth weight and smoking 🚬

baby_fp = os.path.join('data', 'baby.csv')
baby = pd.read_csv(baby_fp)
baby = baby[['Maternal Smoker', 'Birth Weight']]

baby.head()

	Maternal Smoker	Birth Weight
0	False	120
1	False	113
2	True	128
3	True	108
4	False	136

Recall our permutation test from last class:

• Null Hypothesis: In the population, birth weights of smokers' babies and non-smokers' babies have the same distribution, and the observed differences in our samples are due to random chance.
• Alternative Hypothesis: In the population, smokers' babies have lower birth weights than non-smokers' babies, on average. The observed difference in our samples cannot be explained by random chance alone.

Timing the birth weights example ⏰

We'll use 3000 repetitions instead of 500.

%%time

n_repetitions = 3000

differences = []
for _ in range(n_repetitions):
    
    # Step 1: Shuffle the weights and store them in a DataFrame.
    with_shuffled = baby.assign(Shuffled_Weights=np.random.permutation(baby['Birth Weight']))

    # Step 2: Compute the test statistic.
    # Remember, alphabetically, False comes before True,
    # so this computes True - False.
    group_means = (
        with_shuffled
        .groupby('Maternal Smoker')
        .mean()
        .loc[:, 'Shuffled_Weights']
    )
    difference = group_means.diff().iloc[-1]
    
    # Step 3: Store the result.
    differences.append(difference)

CPU times: user 2.04 s, sys: 3.15 ms, total: 2.05 s
Wall time: 2.05 s

fig = px.histogram(pd.DataFrame(differences), x=0, nbins=50, histnorm='probability', 
                   title='Empirical Distribution of the Test Statistic, Original Approach')
fig.update_layout(xaxis_range=[-5, 5])

A faster approach

• Our previous approach involved calling groupby inside of a loop.
• We can avoid groupby entirely!
• Let's start by generating a Boolean array of size (3000, 1174).
  • Each row will correspond to a single permutation of the 'Maternal Smoker' (bool) column.

is_smoker = baby['Maternal Smoker'].values
weights = baby['Birth Weight'].values

is_smoker

array([False, False,  True, ...,  True, False, False])

%%time

np.random.seed(24) # So that we get the same results each time (for lecture).

# We are still using a for-loop!
is_smoker_permutations = np.column_stack([
    np.random.permutation(is_smoker)
    for _ in range(3000)
]).T

CPU times: user 89.4 ms, sys: 2.69 ms, total: 92 ms
Wall time: 91.7 ms

In is_smoker_permutatons, each row is a new simulation.

• False means that baby comes from a non-smoking mother.
• True means that baby comes from a smoking mother.

is_smoker_permutations

array([[ True,  True, False, ...,  True,  True, False],
       [ True, False,  True, ...,  True, False, False],
       [False,  True, False, ...,  True, False,  True],
       ...,
       [False, False,  True, ..., False, False,  True],
       [ True,  True, False, ...,  True, False,  True],
       [ True,  True,  True, ..., False,  True,  True]])

is_smoker_permutations.shape

(3000, 1174)

Note that each row has 459 Trues and 715 Falses – it's just the order of them that's different.

is_smoker_permutations.sum(axis=1)

array([459, 459, 459, ..., 459, 459, 459])

The first row of is_smoker_permutations tells us that in this permutation, we'll assign baby 1 to "smoker", baby 2 to "smoker", baby 3 to "non-smoker", and so on.

is_smoker_permutations[0]

array([ True,  True, False, ...,  True,  True, False])

Broadcasting

• If we multiply is_smoker_permutations by weights, we will create a new (3000, 1174) array, in which:
  • the weights of babies assigned to "smoker" are present, and
  • the weights of babies assigned to "non-smoker" are 0.
• is_smoker_permutations is a "mask".

First, let's try this on just the first permutation (i.e. the first row of is_smoker_permutations).

weights * is_smoker_permutations[0]

array([120, 113,   0, ..., 130, 125,   0])

Now, on all of is_smoker_permutations:

weights * is_smoker_permutations

array([[120, 113,   0, ..., 130, 125,   0],
       [120,   0, 128, ..., 130,   0,   0],
       [  0, 113,   0, ..., 130,   0, 117],
       ...,
       [  0,   0, 128, ...,   0,   0, 117],
       [120, 113,   0, ..., 130,   0, 117],
       [120, 113, 128, ...,   0, 125, 117]])

The mean of the non-zero entries in a row is the mean of the weights of "smoker" babies in that permutation.

Why can't we use .mean(axis=1)?

n_smokers = is_smoker.sum()
mean_smokers = (weights * is_smoker_permutations).sum(axis=1) / n_smokers
mean_smokers

array([118.94117647, 118.07625272, 120.38562092, ..., 119.76688453,
       119.2745098 , 120.22222222])

mean_smokers.shape

(3000,)

We also need to get the weights of the non-smokers in our permutations. We can do this by "inverting" the is_smoker_permutations mask and performing the same calculations.

n_non_smokers = 1174 - n_smokers
mean_non_smokers = (weights * ~is_smoker_permutations).sum(axis=1) / n_non_smokers
mean_non_smokers

array([119.7972028 , 120.35244755, 118.86993007, ..., 119.26713287,
       119.58321678, 118.97482517])

test_statistics = mean_smokers - mean_non_smokers
test_statistics

array([-0.85602633, -2.27619483,  1.51569085, ...,  0.49975166,
       -0.30870698,  1.24739705])

Putting it all together

%%time

is_smoker = baby['Maternal Smoker'].values
weights = baby['Birth Weight'].values
n_smokers = is_smoker.sum()
n_non_smokers = 1174 - n_smokers

is_smoker_permutations = np.column_stack([
    np.random.permutation(is_smoker)
    for _ in range(3000)
]).T

mean_smokers = (weights * is_smoker_permutations).sum(axis=1) / n_smokers
mean_non_smokers = (weights * ~is_smoker_permutations).sum(axis=1) / n_non_smokers
fast_differences = mean_smokers - mean_non_smokers

CPU times: user 97.5 ms, sys: 3.59 ms, total: 101 ms
Wall time: 100 ms

fig = px.histogram(pd.DataFrame(fast_differences), x=0, nbins=50, histnorm='probability', 
                   title='Empirical Distribution of the Test Statistic, Faster Approach')
fig.update_layout(xaxis_range=[-5, 5])

The distribution of test statistics with the fast simulation is similar to the original distribution of test statistics.

Other performance considerations

np.random.permutation (fast) vs df.sample (slow)

In lecture, we mentioned the fact that np.random.permutation is faster than using the df.sample method. It's because df.sample has to shuffle the index as well.

How fast does a single shuffle take for each approach?

to_shuffle = baby.copy()
weights = to_shuffle['Birth Weight']

%%timeit
np.random.permutation(weights.values)

20.1 µs ± 75 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

%%timeit
weights.sample(frac=1)

53.9 µs ± 320 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Adding columns in place (fast) vs. assign (slow)

Don't use assign; instead, add the new column in-place.

Why? This way, we don't create a new copy of our DataFrame on each iteration.

%%timeit
to_shuffle['Shuffled_Weights'] = np.random.permutation(weights.values)

38.7 µs ± 451 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

%%timeit
to_shuffle.assign(Shuffled_Weights=np.random.permutation(weights.values))

121 µs ± 1.91 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)