Applying a function along a numpy array

Function numpy.apply_along_axis is not good for this purpose.Try to use numpy.vectorize to vectorize your function: https://docs.scipy.org/doc/numpy/reference/generated/numpy.vectorize.htmlThis function defines a vectorized function which takes a nested sequence of objects or numpy arrays as inputs and returns an single or tuple of numpy array as output.

import numpy as npimport math# custom functiondef sigmoid(x):  return 1 / (1 + math.exp(-x))# define vectorized sigmoidsigmoid_v = np.vectorize(sigmoid)# testscores = np.array([ -0.54761371,  17.04850603,   4.86054302])print sigmoid_v(scores)

Output: [ 0.36641822 0.99999996 0.99231327]

Performance test which shows that the scipy.special.expit is the best solution to calculate logistic function and vectorized variant comes to the worst:

import numpy as npimport mathimport timeitdef sigmoid_(x):  return 1 / (1 + math.exp(-x))sigmoidv = np.vectorize(sigmoid_)def sigmoid(x):   return 1 / (1 + np.exp(x))print timeit.timeit("sigmoidv(scores)", "from __main__ import sigmoidv, np; scores = np.random.randn(100)", number=25),\timeit.timeit("sigmoid(scores)", "from __main__ import sigmoid, np; scores = np.random.randn(100)",  number=25),\timeit.timeit("expit(scores)", "from scipy.special import expit; import numpy as np;   scores = np.random.randn(100)",  number=25)print timeit.timeit("sigmoidv(scores)", "from __main__ import sigmoidv, np; scores = np.random.randn(1000)", number=25),\timeit.timeit("sigmoid(scores)", "from __main__ import sigmoid, np; scores = np.random.randn(1000)",  number=25),\timeit.timeit("expit(scores)", "from scipy.special import expit; import numpy as np;   scores = np.random.randn(1000)",  number=25)print timeit.timeit("sigmoidv(scores)", "from __main__ import sigmoidv, np; scores = np.random.randn(10000)", number=25),\timeit.timeit("sigmoid(scores)", "from __main__ import sigmoid, np; scores = np.random.randn(10000)",  number=25),\timeit.timeit("expit(scores)", "from scipy.special import expit; import numpy as np;   scores = np.random.randn(10000)",  number=25)

Results:

size        vectorized      numpy                 expitN=100:   0.00179314613342 0.000460863113403 0.000132083892822N=1000:  0.0122890472412  0.00084114074707  0.000464916229248N=10000: 0.109477043152   0.00530695915222  0.00424313545227

Use np.exp and that will work on numpy arrays in a vectorized fashion:

>>> def sigmoid(x):...     return 1 / (1 + np.exp(-x))...>>> sigmoid(scores)array([  6.33581776e-01,   3.94391811e-08,   7.68673281e-03])>>>

You will likely not get any faster than this. Consider:

>>> def sigmoid(x):...     return 1 / (1 + np.exp(-x))...

And:

>>> def sigmoidv(x):...   return 1 / (1 + math.exp(-x))...>>> vsigmoid = np.vectorize(sigmoidv)

Now, to compare the timings. With a small (size 100) array:

>>> t = timeit.timeit("vsigmoid(arr)", "from __main__ import vsigmoid, np; arr = np.random.randn(100)", number=100)>>> t0.006894525984534994>>> t = timeit.timeit("sigmoid(arr)", "from __main__ import sigmoid, np; arr = np.random.randn(100)", number=100)>>> t0.0007238480029627681

So, still an order-of-magnitude difference with small arrays. This performance differences stays relatively constant, with a 10,000 size array:

>>> t = timeit.timeit("vsigmoid(arr)", "from __main__ import vsigmoid, np; arr = np.random.randn(10000)", number=100)>>> t0.3823414359940216>>> t = timeit.timeit("sigmoid(arr)", "from __main__ import sigmoid, np; arr = np.random.randn(10000)", number=100)>>> t0.011259705002885312

And finally with a size 100,000 array:

>>> t = timeit.timeit("vsigmoid(arr)", "from __main__ import vsigmoid, np; arr = np.random.randn(100000)", number=100)>>> t3.7680041620042175>>> t = timeit.timeit("sigmoid(arr)", "from __main__ import sigmoid, np; arr = np.random.randn(100000)", number=100)>>> t0.09544878199812956

scipy already implements the functionLuckily, Python allows us to rename things upon import:

 from scipy.special import expit as sigmoid