Can cmath.sqrt be applied to a NumPy array?

Question

I want to calculate the square root of a numpy array of negative numbers.

I tried with np.sqrt() but it gives error beacuse the domain.

Then, I found that for complex numbers you can use cmath.sqrt(x) but it also gives me an error.

Here's my code

import numpy as np
import cmath
from cmath import sqrt

x = np.arange(-10, 10, 0.01)
E = 1

p1 = cmath.sqrt(E - x**2)

And got this error

Traceback (most recent call last):
  File "C:\Users\os.py", line 49, in <module>
    p1 = cmath.sqrt(E - x**2)
TypeError: only length-1 arrays can be converted to Python scalars

Later I tried to use a for loop and it's not possible either. Here's the code:

import numpy as np
import cmath
from cmath import sqrt

x = np.arange(-10, 10, 0.01)
E = 1

for i in range(0, len(x)):
    p1 = cmath.sqrt(E - x(i)**2)

and the message error

Traceback (most recent call last):
  File "C:\Users\os.py", line 48, in <module>
    p1 = cmath.sqrt(E - x(i)**2)
TypeError: 'numpy.ndarray' object is not callable

I don't know what am I dpoing wrong, can anyone help me?, please. I need to calculate the square root of an numpy array of negative numbers, does anyone know how to do this?

I added and answer using emath but want to also mention the problem you are having in the loop version is just a syntax error. x(i) should probably be x[i]. It seems like you are trying to index the array, but the syntax is calling it like a function. — Mark
– Mark, Commented Oct 23, 2022 at 2:08

hpaulj · Accepted Answer · 2022-10-23 02:12:41Z

Yes, you can iterate to perform cmath on individual elements of x:

In [254]: np.array([cmath.sqrt(E-i**2) for i in x])
Out[254]: 
array([0.+9.94987437j, 0.+9.93982394j, 0.+9.92977341j, ...,
       0.+9.91972278j, 0.+9.92977341j, 0.+9.93982394j])

but np.sqrt will work if you give it a complex dtype array:

In [255]: np.sqrt(E-x.astype(complex)**2)
Out[255]: 
array([0.+9.94987437j, 0.+9.93982394j, 0.+9.92977341j, ...,
       0.+9.91972278j, 0.+9.92977341j, 0.+9.93982394j])

some comparative timings

In [259]: timeit np.emath.sqrt(E-x**2)
166 µs ± 336 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)

In [260]: timeit np.sqrt(E-x.astype(complex)**2)
129 µs ± 82.1 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)

In [261]: timeit np.array([cmath.sqrt(E-i**2) for i in x])
2.54 ms ± 4.36 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Mark · Accepted Answer · 2022-10-23 01:56:48Z

3

You could use numpy.emath.sqrt() which will work with negative numbers and return the complex square roots:

import numpy as np

x = np.array([-4, -16, -25, 100, 64])

np.emath.sqrt(x)
# array([ 0.+2.j,  0.+4.j,  0.+5.j, 10.+0.j,  8.+0.j])

answered Oct 23, 2022 at 1:56

Mark

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