Applying a function along a numpy array

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)
>>> t
0.006894525984534994
>>> t = timeit.timeit("sigmoid(arr)", "from __main__ import sigmoid, np; arr = np.random.randn(100)", number=100)
>>> t
0.0007238480029627681

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)
>>> t
0.3823414359940216
>>> t = timeit.timeit("sigmoid(arr)", "from __main__ import sigmoid, np; arr = np.random.randn(10000)", number=100)
>>> t
0.011259705002885312

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)
>>> t
3.7680041620042175
>>> t = timeit.timeit("sigmoid(arr)", "from __main__ import sigmoid, np; arr = np.random.randn(100000)", number=100)
>>> t
0.09544878199812956

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.html This 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 np
import math

# custom function
def sigmoid(x):
  return 1 / (1 + math.exp(-x))

# define vectorized sigmoid
sigmoid_v = np.vectorize(sigmoid)

# test
scores = 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 np
import math
import timeit

def 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                 expit
N=100:   0.00179314613342 0.000460863113403 0.000132083892822
N=1000:  0.0122890472412  0.00084114074707  0.000464916229248
N=10000: 0.109477043152   0.00530695915222  0.00424313545227