numpy.fft.fft can not have 4 arguments?

Question

I tried to do FFT with numpy using this:

    sp=np.fft.fft(np.exp(-t), 1000, -1, "ortho")

and it returns:

    TypeError: fft() takes at most 3 arguments (4 given)

But in the manual, numpy.fft.fft() is defined as:

    numpy.fft.fft(a, n=None, axis=-1, norm=None)

n, axis, and norm are optional, and norm can be set to "ortho". Why can't I use what I wrote? Thanks.

what version of numpy do you have? `norm` is new to v1.10 so if you have an older version, there is no `norm` keyword. — R Nar, Aug 10 '16 at 19:19
Related: http://stackoverflow.com/questions/38786227/numpys-fft-with-intel-mkl — Ahmed Fasih, Aug 10 '16 at 19:29

score 0 · Answer 1 · answered Aug 10 '16 at 19:20

What version of numpy are you using?

I ask because

refers to norm as "New in version 1.10.0.", and if you were using an older version, the error message would be consistent with the behavior.

1 Answers1