I can't figure out why my relu function doesn't work but squarer works, how is it different?
import numpy as np
x = np.array([1, 2, 3, 4, 5])
squarer = lambda x: x ** 2
squarer(x)
# array([ 1, 4, 9, 16, 25])
relu = lambda x : 0 if x <= 0 else x
relu(x)
# ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
In [6]: x = np.array([1, 2, 3, 4, 5])
...:
...: squarer = lambda x: x ** 2
...: squarer(x)
Out[6]: array([ 1, 4, 9, 16, 25])
The lambda is just a function definition, and is equivalent of doing:
In [7]: x**2
Out[7]: array([ 1, 4, 9, 16, 25])
The function layer doesn't add any iteration. It's the power method of the x array that's doing the elementwise iteration.
In [8]: relu = lambda x : 0 if x <= 0 else x
Similarly the relu does not add any iteration; it's scalar python if/else clause.
In [13]: x = np.arange(-3,4)
In [14]: x
Out[14]: array([-3, -2, -1, 0, 1, 2, 3])
It can be applied to elements of x with a list comprehension:
In [15]: [relu(i) for i in x]
Out[15]: [0, 0, 0, 0, 1, 2, 3]
Arrays have a lt method, so:
In [16]: x<=0
Out[16]: array([ True, True, True, True, False, False, False])
It can be use in masked way:
In [17]: x1=x.copy()
In [18]: x1[x<=0] = 0
In [19]: x
Out[19]: array([-3, -2, -1, 0, 1, 2, 3])
In [20]: x1
Out[20]: array([0, 0, 0, 0, 1, 2, 3])
Or via a where:
In [22]: np.where(x<=0, 0,x)
Out[22]: array([0, 0, 0, 0, 1, 2, 3])
where isn't an iterator either. It is effectively the same thing as the [17][18] lines.
Using an array in a if expression amounts to trying to convert it to a scalar boolean:
In [24]: if x<=0:x
Traceback (most recent call last):
File "<ipython-input-24-6cecebf070dc>", line 1, in <module>
if x<=0:x
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
In [25]: bool(x<=0)
Traceback (most recent call last):
File "<ipython-input-25-f1a519ed746f>", line 1, in <module>
bool(x<=0)
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
It works it the array has only one element, but otherwise it raises this ambiguity error:
In [26]: bool(np.array(1)<=0)
Out[26]: False
but for "empty" array:
In [28]: bool(np.array([])<=0)
<ipython-input-28-03e1626841fc>:1: DeprecationWarning: The truth value of an empty array is ambiguous. Returning False, but in future this will result in an error. Use `array.size > 0` to check that an array is not empty.
bool(np.array([])<=0)
Out[28]: False
But testing for a 'empty' list is ok:
In [29]: bool([])
Out[29]: False
x <= 0?iftest in vectorized form, you could usenp.where:relu = lambda x : np.where(x <= 0, 0, x)reluorsquaretonp.array, python will iterate over each element and apply the lambda functionnp.whereis vectorized, while[relu(a) for a in x]goes one by one and has the overhead of creating a Python list. Especially for larger arrays,np.whereis much faster.