imshow can take an array of [r, g, b] entries. So you can convert the absolute values to intensities and phases – to hues. I will use as an example complex numbers, because for it it makes the most sense. If needed, you can always add numpy arrays Z = X + 1j * Y. … Read more
well lets start with Real domain and integer b only: a^^b = a^a^a^a^a…^a // a is there b times this can be evaluated like this in C++: double tetration(double a,int b) // a^^b = a^a^a^a… b times { double c; if (b<=0) return 0; for (c=a;b>1;b–) c=pow(a,c); return c; } as You already got the … Read more
You could do it as is shown below using the str.format() method: >>> n = 3.4+2.3j >>> n (3.4+2.3j) >>> ‘({0.real:.2f} + {0.imag:.2f}i)’.format(n) ‘(3.40 + 2.30i)’ >>> ‘({c.real:.2f} + {c.imag:.2f}i)’.format(c=n) ‘(3.40 + 2.30i)’ To make it handle both positive and negative imaginary portions properly, you would need a (even more) complicated formatting operation: >>> n … Read more
This seems to do what you want: numpy.apply_along_axis(lambda args: [complex(*args)], 3, Data) Here is another solution: # The ellipsis is equivalent here to “:,:,:”… numpy.vectorize(complex)(Data[…,0], Data[…,1]) And yet another simpler solution: Data[…,0] + 1j * Data[…,1] PS: If you want to save memory (no intermediate array): result = 1j*Data[…,1]; result += Data[…,0] devS’ solution below … Read more
What’s wrong with just separating it out into real and imaginary parts? scipy.integrate.quad requires the integrated function return floats (aka real numbers) for the algorithm it uses. import scipy from scipy.integrate import quad def complex_quadrature(func, a, b, **kwargs): def real_func(x): return scipy.real(func(x)) def imag_func(x): return scipy.imag(func(x)) real_integral = quad(real_func, a, b, **kwargs) imag_integral = quad(imag_func, … Read more
In python, you can put ‘j’ or ‘J’ after a number to make it imaginary, so you can write complex literals easily: >>> 1j 1j >>> 1J 1j >>> 1j * 1j (-1+0j) The ‘j’ suffix comes from electrical engineering, where the variable ‘i’ is usually used for current. (Reasoning found here.) The type of … Read more
I assume 1D DFT/IDFT … All DFT’s use this formula: X(k) is transformed sample value (complex domain) x(n) is input data sample value (real or complex domain) N is number of samples/values in your dataset This whole thing is usually multiplied by normalization constant c. As you can see for single value you need N … Read more
This code will help you, and it’s fairly self-explanatory: #include <stdio.h> /* Standard Library of Input and Output */ #include <complex.h> /* Standard Library of Complex Numbers */ int main() { double complex z1 = 1.0 + 3.0 * I; double complex z2 = 1.0 – 4.0 * I; printf(“Working with complex numbers:\n\v”); printf(“Starting values: … Read more