Constant Lambda Expressions, Shape Of Return Data

A list of lambda expressions given to me (by Sympy's lambdify), some explicitly depending on a variable x, some constant. I would like to evaluate those consistently with Numpy arr

Solution 1:

You could use ones_like:

>>>X = numpy.array([1, 2, 3])>>>defg(x): return numpy.ones_like(x)>>>g(X)
array([1, 1, 1])

Note that this returns integers, not floats, because that was the input dtype; you could specify dtype=float or multiply by 1.0 if you prefer to always get floats out.

PS: It's a little odd to use a lambda and then immediately give it a name. It's like wearing a mask but handing out business cards.

PPS: back before ones_like I tended to use x*0+1 when I wanted something appropriately shaped.

Solution 2:

I don't see the problem, just do:

import numpy as np

X = np.array([1, 2, 3])

f = lambda x: 1.0 + x**2print(f(X))

g = lambda x: np.ones(shape=(len(X),))
print(g(X))

Which prints:

[  2.   5.  10.][ 1.  1.  1.]

Notice that using np.ones(shape=(len(X),)) is the same that using np.ones_like(X)

Solution 3:

Use ones_like:

g = lambda x: np.ones_like(x) * 1.0

There's also this slightly hackier solution:

g = lambda x: 1.0 + (x*0)

Solution 4:

You seem to want an array of ones:

>>>import numpy>>>numpy.ones(3)
array([ 1.,  1.,  1.])

If you want to set scalars, it's easy to do so

g = lambda x: numpy.ones(shape=x.shape) * 2

g(X)

returns

array([ 2.,  2.,  2.])

So for an arbitrary array:

g = lambda x: numpy.ones(shape=x.shape) * 1n = numpy.array([1,2,3,4,5])

g(n) is

array([ 1.,  1.,  1.,  1.,  1.])