Function Of Numpy Array With If-statement
Solution 1:
Use numpy.vectorize
to wrap func before applying it to array x
:
from numpy import vectorize
vfunc = vectorize(func)
y = vfunc(x)
Solution 2:
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.
Solution 3:
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
.
Solution 4:
(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).
Solution 5:
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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