[Solved] Function of Numpy Array with if-statement

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
          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:Scriptstest.py", line 17, in <module>
    y = func(x)
  File "D:Scriptstest.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?

Enquirer: Dan


Solution #1:

Use numpy.vectorize to wrap func before applying it to array x:

from numpy import vectorize
vfunc = vectorize(func)
y = vfunc(x)
Respondent: Dan

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   


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])


1) If x is a masked array, you need to use np.ma.where instead, since this works for masked arrays.

Respondent: Chris Kuklewicz

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.

Respondent: hurrdrought

Solution #4:

(I realize this is an old question, but …)

There is one more option which wasn’t mentioned here — using np.choose.

    # the boolean condition
    x < 0,
        # index 0: value if condition is False
        10 * x,
        # index 1: value if condition is True

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).

Respondent: Björn Pollex

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

Respondent: shx2

Solution #6:

not sure why you need a function

x = np.arange(-1, 1, 0.01)
y = x * np.where(x < 0, 0, 10)
Respondent: Jdog

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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