Equivalent of j in NumPy

Question

What is the equivalent of Octave's j in NumPy? How can I use j in Python?

In Octave:

octave:1> j ans =  0 + 1i octave:1> j*pi/4 ans =  0.00000 + 0.78540i 

But in Python:

>>> import numpy as np >>> np.imag <function imag at 0x2368140> >>> np.imag(3) array(0) >>> np.imag(3,2) Traceback (most recent call last):   File "<stdin>", line 1, in <module> TypeError: imag() takes exactly 1 argument (2 given) >>> np.imag(32) array(0) >>>  >>> 0+np.imag(1) 1 

Answer 1

In Python, 1j or 0+1j is a literal of complex type. You can broadcast that into an array using expressions, for example

In [17]: 1j * np.arange(5) Out[17]: array([ 0.+0.j,  0.+1.j,  0.+2.j,  0.+3.j,  0.+4.j]) 

Create an array from literals:

In [18]: np.array([1j]) Out[18]: array([ 0.+1.j]) 

Note that what Michael9 posted creates a complex, not a complex array:

In [21]: np.complex(0,1) Out[21]: 1j In [22]: type(_) Out[22]: complex 

Answer 2

You can create one if needed or use 1j which instance of complex class

 >>> 1j #complex object  1j  >>> type(1j)  <class 'complex'>  >>> j = np.complex(0,1) #create complex number  >>> j  1j