Numpy Pure Functions for performance, caching

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Question :

Numpy Pure Functions for performance, caching

I’m writing some moderately performance critical code in numpy.
This code will be in the inner most loop, of a computation that’s run time is measured in hours.
A quick calculation suggest that this code will be executed up something like 10^12 times, in some variations of the calculation.

So the function is to calculate sigmoid(X) and another to calculate its derivative (gradient).
Sigmoid has the property that for
y=sigmoid(x), dy/dx= y(1-y)
In python for numpy this looks like:

sigmoid = vectorize(lambda(x): 1.0/(1.0+exp(-x)))
grad_sigmoid = vectorize(lambda (x): sigmoid(x)*(1-sigmoid(x)))

As can be seen, both functions are pure (without side effects),
so they are ideal candidates for memoization,
at least for the short term, I have some worries about caching every single call to sigmoid ever made: Storing 10^12 floats which would take several terabytes of RAM.

Is there a good way to optimise this?
Will python pick up that these are pure functions and cache them for me, as appropriate?
Am I worrying over nothing?

Answer #1:

These functions already exist in scipy. The sigmoid function is available as scipy.special.expit.

In [36]: from scipy.special import expit

Compare expit to the vectorized sigmoid function:

In [38]: x = np.linspace(-6, 6, 1001)

In [39]: %timeit y = sigmoid(x)
100 loops, best of 3: 2.4 ms per loop

In [40]: %timeit y = expit(x)
10000 loops, best of 3: 20.6 µs per loop

expit is also faster than implementing the formula yourself:

In [41]: %timeit y = 1.0 / (1.0 + np.exp(-x))
10000 loops, best of 3: 27 µs per loop

The CDF of the logistic distribution is the sigmoid function. It is available as the cdf method of scipy.stats.logistic, but cdf eventually calls expit, so there is no point in using that method. You can use the pdf method to compute the derivative of the sigmoid function, or the _pdf method which has less overhead, but “rolling your own” is faster:

In [44]: def sigmoid_grad(x):
   ....:     ex = np.exp(-x)
   ....:     y = ex / (1 + ex)**2
   ....:     return y

Timing (x has length 1001):

In [45]: from scipy.stats import logistic

In [46]: %timeit y = logistic._pdf(x)
10000 loops, best of 3: 73.8 µs per loop

In [47]: %timeit y = sigmoid_grad(x)
10000 loops, best of 3: 29.7 µs per loop

Be careful with your implementation if you are going to use values that are far into the tails. The exponential function can overflow pretty easily. logistic._cdf is a bit more robust than my quick implementation of sigmoid_grad:

In [60]: sigmoid_grad(-500)
/home/warren/anaconda/bin/ipython:3: RuntimeWarning: overflow encountered in double_scalars
  import sys
Out[60]: 0.0

In [61]: logistic._pdf(-500)
Out[61]: 7.1245764067412855e-218

An implementation using sech**2 (1/cosh**2) is a bit slower than the above sigmoid_grad:

In [101]: def sigmoid_grad_sech2(x):
   .....:     y = (0.5 / np.cosh(0.5*x))**2
   .....:     return y
   .....: 

In [102]: %timeit y = sigmoid_grad_sech2(x)
10000 loops, best of 3: 34 µs per loop

But it handles the tails better:

In [103]: sigmoid_grad_sech2(-500)
Out[103]: 7.1245764067412855e-218

In [104]: sigmoid_grad_sech2(500)
Out[104]: 7.1245764067412855e-218
Answered By: Warren Weckesser

Answer #2:

Just expanding on my comment, here is a comparison between your sigmoid through vectorize and using numpy directly:

In [1]: x = np.random.normal(size=10000)

In [2]: sigmoid = np.vectorize(lambda x: 1.0 / (1.0 + np.exp(-x)))

In [3]: %timeit sigmoid(x)
10 loops, best of 3: 63.3 ms per loop

In [4]: %timeit 1.0 / (1.0 + np.exp(-x))
1000 loops, best of 3: 250 us per loop

As you can see, not only does vectorize make it much slower, the fact is that you can calculate 10000 sigmoids in 250 microseconds (that is, 25 nanoseconds for each). A single dictionary look-up in Python is slower than that, let alone all the other code to get the memoization in place.

The only way to optimize this that I can think of is writing a sigmoid ufunc for numpy, which basically will implement the operation in C. That way, you won’t have to do each operation in the sigmoid to the entire array, even though numpy does this really fast.

Answered By: Gustav Larsson

Answer #3:

If you are looking to memoize this process, I’d wrap that code in a function and decorate with functools.lru_cache(maxsize=n). Experiment with the maxsize value to find the appropriate size for your application. For best results, use a maxsize argument that is a power of two.

from functools import lru_cache

lru_cache(maxsize=8096)
def sigmoids(x):
    sigmoid = vectorize(lambda(x): 1.0/(1.0+exp(-x)))
    grad_sigmoid = vectorize(lambda (x): sigmoid(x)*(1-sigmoid(x)))
    return sigmoid, grad_sigmoid

If you’re on 2.7 (which I expect you are since you’re using numpy), you can take a look at https://pypi.python.org/pypi/repoze.lru/ for a memoization library with identical syntax.

You can install it via pip: pip install repoze.lru

from repoze.lru import lru_cache

lru_cache(maxsize=8096)
def sigmoids(x):
    sigmoid = vectorize(lambda(x): 1.0/(1.0+exp(-x)))
    grad_sigmoid = vectorize(lambda (x): sigmoid(x)*(1-sigmoid(x)))
    return sigmoid, grad_sigmoid
Answered By: Madison May

Answer #4:

Mostly I agree with Warren Weckesser and his answer above.
But for derivative of sigmoid the following can be used:

In [002]: def sg(x):
     ...: s = scipy.special.expit(x)
     ...: return s * (1.0 - s) 

Timings:

In [003]: %timeit y = logistic._pdf(x)
10000 loops, best of 3: 45 µs per loop

In [004]: %timeit y = sg(x)
10000 loops, best of 3: 20.4 µs per loop

The only problem is accuracy:

In [005]: sg(37)
Out[005]: 0.0

In [006]: logistic._pdf(37)
Out[006]: 8.5330476257440658e-17    
Answered By: Sklavit

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