Very basic manipulation of numpy in python functions

Question

I was always thinking to avoid loop operation in my python code. Numpy really helps, but in some slightly complicated cases, I felt stuck on how to utilize numpy array wisely.

Below is an simple example illustrating my inability, a will be a parameter, and b is an numpy array.

def f(a,b):
    return np.sum( a * b)

so there is no problem if I wish to evaluate this function by a given single parameter and an array.

x = 2
y = np.arange(3)
print (f(x,y))

But sometimes I want to evaluate different parameter value of the function altogether with a fixed array value.

I would try:

x2 = np.array([1,4,5,2,8])
print (f(x2,y))

What I wish to get is surely an array with value:

[f(1,y),f(4,y),f(5,y),f(2,y),f(8,y)]

However, python will try to evaluate the dot product of x and y, since now they are both np arrays and It will report

ValueError: operands could not be broadcast together with shapes (5,) (3,)

How should I overcome this, in numpy array-wise fashion, producing the sequence

[f(1,y),f(4,y),f(5,y),f(2,y),f(8,y)]

without using loops?

(In this example, I could resolved problem by modify f by:

def f(a,b):
    return a * np.sum(b)

But in most general cases, we cannot factor the parameter out.)

in [f(1,y),f(4,y),f(5,y),f(2,y),f(8,y)] y would be an array of length 3. So what numbers would you expect to appear? Should this become a 3x5 array? — Fabian Rost
– Fabian Rost, Commented Nov 12, 2015 at 21:32
For the summation calculation, you can just do - x2*y.sum(). — Divakar
– Divakar, Commented Nov 12, 2015 at 21:34
To @FabianRost, I am expecting to get 1x5 array, the i'th entry gives the value of f(x2[i],y) — Draditate
– Draditate, Commented Nov 12, 2015 at 21:38
@Divakar, I understand this way work, because we can factor the x out of the summation. In general, x and y is not separable before taking summation. For instance if f(a,b) = np.sum (sin( a * b)). — Draditate
– Draditate, Commented Nov 12, 2015 at 21:42
@LienongXu In that case do - (np.sin(y[:,None]*x2)).sum(0). As long there is a ufunc for it, its doable that way. — Divakar
– Divakar, Commented Nov 12, 2015 at 21:44

philngo · Accepted Answer · 2015-11-12 21:53:10Z

2

np.newaxis is a very handy tool for cases like this, and gives you a bit more control over broadcasting. In this case, you'll want to add a new axes to give numpy some hints about where to broadcast:

>>> x = np.array([1,4,5,2,8])
>>> y = np.arange(3)
>>> x[:,np.newaxis] * y
array([[ 0,  1,  2],
       [ 0,  4,  8],
       [ 0,  5, 10],
       [ 0,  2,  4],
       [ 0,  8, 16]])

If you'd like the sums along the second axis, you can sum like this:

>>> (x[:, np.newaxis] * y).sum(axis=1)
array([ 3, 12, 15,  6, 24])

edited Nov 12, 2015 at 21:53

answered Nov 12, 2015 at 21:43

philngo

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5 Comments

Divakar Over a year ago

np.newaxis is redundant in y[np.newaxis,:].

Divakar Over a year ago

Also, I think OP is summing elements along the second axis for the desired output.

philngo Over a year ago

Thanks - didn't see all of the comments on the question.

Draditate Over a year ago

@pnhgiol, I will take it. Valuable answer.

hpaulj Over a year ago

numpy might not require y[np.newaxis, :], but dumb humans might find it clearer. I could even omit the ':', and use y[None] (but that would not add clarity).

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