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up vote 7 down vote favorite

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?

share|improve this question
    
You have to decide whether you want to treat x as int or as a numpy.array. Or use ininstance() to check what type is being passed –  pajton Nov 7 '11 at 13:10
    
What do you mean by x<0? x is an array, so it is not clear what this is supposed to mean. –  Björn Pollex Nov 7 '11 at 13:10
    
@Bjoern Pollex I know that is why python is getting confused, I want to apply this function to each element of the array individually (i.e. elementwise) –  Dan Nov 7 '11 at 13:12
    
@pajton Indeed but is there a way to make the second handle this gracefully like the first example? –  Dan Nov 7 '11 at 13:13
    
So what exactly do you want the output to be, for the illustrated input? –  Karl Knechtel Nov 7 '11 at 13:52

4 Answers 4

up vote 4 down vote accepted

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

from numpy import vectorize
vfunc = vectorize(func)
y = vfunc(x)
share|improve this answer
    
Neat. I do believe however, that my solution is faster as long as you use operations for which Numpy has it's own implementations (have not tested it though, just a feeling). –  Björn Pollex Nov 7 '11 at 13:28
    
Sure. All the built-in ops, such as those you compose, will be faster than pure python. I was just targeting the "is there a nice way to make the function handle arrays element wise?" question. –  Chris Kuklewicz Nov 7 '11 at 13:35
    
I wonder if there is a way to make this efficient when using functions written in C (e.g. by using Cython). I have a feeling there might be. –  Björn Pollex Nov 7 '11 at 13:50
1  
Accepted this answer since I was after readable code more than speed (this is only a little script after all) :) –  Dan Nov 7 '11 at 14:10
up vote 1 down vote

(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).

share|improve this answer
up vote 1 down vote 
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

share|improve this answer
1  
It would be even nicer (and correct) if you'd make y = zeroes(len(x)) and then y[mask] = x[mask] * 10. –  Björn Pollex Nov 7 '11 at 13:32
up vote 9 down vote

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.

share|improve this answer
    
Thanks, this seems to work. I'm curious as to why x[x<0]=0 etc doesn't seem to work? –  Dan Nov 7 '11 at 13:24
    
@Dan: It does work, but the way I show it, the potentially expensive operation x<0 has to be executed only once (on second thought, invert is probably just as expensive, you'd have to test that). –  Björn Pollex Nov 7 '11 at 13:26
    
Well, "x[x<0]=0" works for me. (Python 2.6.1 numpy 2.0.0.dev-3071eab) –  Chris Kuklewicz Nov 7 '11 at 13:28
2  
I think using np.where is even better. –  tillsten Nov 7 '11 at 20:20

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