Dimension of Jacobian

[TTitscher]

import numdifftools 
import numpy as np
print(numdifftools.__version__) # out: 0.9.39

def f(x):
    v = sum(x)
    return np.r_[v, 2*v, 3*v]

J = numdifftools.Jacobian(f)

input_2d = np.r_[1,2]
print(np.shape(J(input_2d))) # out: (3,2) -- OK

input_1d = np.r_[1]
print(np.shape(J(input_1d))) # out: (1,3)  -- I would expect (3,1)

The Jacobian evaluation transposes the result for one-dimensional, but vector-valued, inputs. I see that the derivative with respect to a scalar (which is almost the case for input_1d) may be defined differently. As the code above specifically requests the Jacobian, however, I was wondering if this behavior is intended or may show a inconsistency/bug.

[pbrod]

Thanks for reporting. I agree with you, I also expect shape of input_1d to give (3,1). I will look into it soon.