I am using Matplotlib and Numpy to produce some plots. I wish to define a function which given an array returns another array with values calculated elementwise, for example:
def func(x):
return x*10
x = numpy.arrange(-1,1,0.01)
y = func(x)
This is fine. Now however I wish to have an if-statement inside func
, for example:
def func(x):
if x<0:
return 0
else:
return x*10
x = numpy.arrange(-1,1,0.01)
y = func(x)
This unfortunately throws the following error
Traceback (most recent call last):
File "D:\Scripts\test.py", line 17, in <module>
y = func(x)
File "D:\Scripts\test.py", line 11, in func
if x<0:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
I looked at the documentation for all()
and any()
and they do not fit the bill for what I need. So is there a nice way to make the function handle arrays element wise as in the first example?
It returns a new numpy array, after filtering based on a condition, which is a numpy-like array of boolean values. For example, if condition is array([[True, True, False]]) , and our array is a = ndarray([[1, 2, 3]]) , on applying a condition to array ( a[:, condition] ), we will get the array ndarray([[1 2]]) .
A numpy array is a grid of values, all of the same type, and is indexed by a tuple of nonnegative integers. The number of dimensions is the rank of the array; the shape of an array is a tuple of integers giving the size of the array along each dimension.
where() in Python. The numpy. where() function returns the indices of elements in an input array where the given condition is satisfied.
Use numpy.vectorize
to wrap func before applying it to array x
:
from numpy import vectorize
vfunc = vectorize(func)
y = vfunc(x)
I know it is too late for this answer, but I am excited learning NumPy. You can vectorize the function on your own with numpy.where.
def func(x):
import numpy as np
x = np.where(x<0, 0., x*10)
return x
Examples
Using a scalar as data input:
x = 10
y = func(10)
y = array(100.0)
using an array as data input:
x = np.arange(-1,1,0.1)
y = func(x)
y = array([ -1.00000000e+00, -9.00000000e-01, -8.00000000e-01,
-7.00000000e-01, -6.00000000e-01, -5.00000000e-01,
-4.00000000e-01, -3.00000000e-01, -2.00000000e-01,
-1.00000000e-01, -2.22044605e-16, 1.00000000e-01,
2.00000000e-01, 3.00000000e-01, 4.00000000e-01,
5.00000000e-01, 6.00000000e-01, 7.00000000e-01,
8.00000000e-01, 9.00000000e-01])
Caveats:
1) If x
is a masked array, you need to use np.ma.where
instead, since this works for masked arrays.
This should do what you want:
def func(x):
small_indices = x < 10
x[small_indices] = 0
x[invert(small_indices)] *= 10
return x
invert
is a Numpy-function. Note that this modifies the argument. To prevent this, you'd have to modify and return a copy
of x
.
(I realize this is an old question, but ...)
There is one more option which wasn't mentioned here -- using np.choose
.
np.choose(
# the boolean condition
x < 0,
[
# index 0: value if condition is False
10 * x,
# index 1: value if condition is True
0
]
)
Though not terribly readable, this is just a single expression (not a series of statements), and does not compromize numpy's inherent speed (as np.vectorize
does).
x = numpy.arrange(-1,1,0.01)
mask = x>=0
y = numpy.zeros(len(x))
y[mask] = x[mask]*10
mask
is a boolean array that equates to True
are array indices matching the condition and False
elsewhere. The last line replaces all values in the original array with that value mulitplied by 10.
Edited to reflect Bjorn's pertinent comment
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