Octave/Matlab
Consider the following set of operations in Octave or Matlab.
Make an array.
octave:1> x=1:10
x =
1 2 3 4 5 6 7 8 9 10
Which elements of "x" are greater than 5? The result is a logical array.
octave:2> c1 = x > 5
c1 =
0 0 0 0 0 1 1 1 1 1
Which elements are less than 8?
octave:3> c2 = x < 8
c2 =
1 1 1 1 1 1 1 0 0 0
The logical combination of two logical arrays is interpreted pair-wise.
octave:4> c = c1 & c2
c =
0 0 0 0 0 1 1 0 0 0
octave:5> x(c)
ans =
6 7
Python/Numpy
Now consider doing something similar in numpy.
>>> import numpy as np
>>> x=np.arange(1,11)
>>> x
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
>>> c1 = x > 5
>>> c1
array([False, False, False, False, False, True, True, True, True, True], dtype=bool)
>>> c2 = x < 8
>>> c = c1 and c2
Traceback (most recent call last):
File "", line 1, in
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
Consider the following set of operations in Octave or Matlab.
Make an array.
octave:1> x=1:10
x =
1 2 3 4 5 6 7 8 9 10
Which elements of "x" are greater than 5? The result is a logical array.
octave:2> c1 = x > 5
c1 =
0 0 0 0 0 1 1 1 1 1
Which elements are less than 8?
c2 =
1 1 1 1 1 1 1 0 0 0
The logical combination of two logical arrays is interpreted pair-wise.
c =
0 0 0 0 0 1 1 0 0 0
octave:5> x(c)
ans =
6 7
Python/Numpy
Now consider doing something similar in numpy.
>>> import numpy as np
>>> x=np.arange(1,11)
>>> x
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
>>> c1 = x > 5
>>> c1
array([False, False, False, False, False, True, True, True, True, True], dtype=bool)
>>> c2 = x < 8
So far so good. But when I say,
Traceback (most recent call last):
File "
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
The "and" command is good only to compare two bools, not arrays of bools. Therefore numpy suggests using x.any() or x.all() to collapse the vector of logical arguments to a scalar. For example.
>>> c1.all() # are all elements of c1 True?
False
>>> c1.any() # are any elements of c1 True?
True
To do an element-by-element comparison simply use np.logical_and or np.logical_or.
>>> c = np.logical_and(c1, c2)
>>> c
array([False, False, False, False, False, True, True, False, False, False], dtype=bool)
>>> x[c]
array([6, 7])
1 comment:
I love the way you explain the material. I really enjoy and that is really helpful. Please keep posting.
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