The most efficient way to create an array of cos and sin in Numpy

Question

I need to save an array the size nwith values cos(x)and sin(x)let's say

array[[cos(0.9), sin(0.9)],
      [cos(0.35),sin(0.35)],
      ...]

The arguments for each cos and sin pair are randomly selected. My code, as far as I improve it, is as follows:

def randvector():
""" Generates random direction for n junctions in the unitary circle """
    x = np.empty([n,2])
    theta = 2 * np.pi * np.random.random_sample((n))
    x[:,0] = np.cos(theta)
    x[:,1] = np.sin(theta)
    return x

Is there a shorter way or a more efficient way to achieve this?

Answer 1

Your code is quite efficient. And the fair answer is not bad, I think.

For efficient and short, how about this code?

def randvector(n):
    theta = 2 * np.pi * np.random.random_sample((n))
    return np.vstack((np.cos(theta), np.sin(theta))).T

UPDATE

Add cProfile result.

justhalf's

      5 function calls in 4.707 seconds

Ordered by: standard name

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
     1    0.001    0.001    4.707    4.707 <string>:1(<module>)
     1    2.452    2.452    4.706    4.706 test.py:6(randvector1)
     1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
     1    0.010    0.010    0.010    0.010 {method 'random_sample' of 'mtrand.RandomState' objects}
     1    2.244    2.244    2.244    2.244 {numpy.core.multiarray.array}

Op's

      5 function calls in 0.088 seconds

Ordered by: standard name

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
     1    0.000    0.000    0.088    0.088 <string>:1(<module>)
     1    0.079    0.079    0.088    0.088 test.py:9(randvector2)
     1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
     1    0.009    0.009    0.009    0.009 {method 'random_sample' of 'mtrand.RandomState' objects}
     1    0.000    0.000    0.000    0.000 {numpy.core.multiarray.empty}

mine

      21 function calls in 0.087 seconds

Ordered by: standard name

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
     1    0.000    0.000    0.087    0.087 <string>:1(<module>)
     2    0.000    0.000    0.000    0.000 numeric.py:322(asanyarray)
     1    0.000    0.000    0.002    0.002 shape_base.py:177(vstack)
     2    0.000    0.000    0.000    0.000 shape_base.py:58(atleast_2d)
     1    0.076    0.076    0.087    0.087 test.py:17(randvector3)
     6    0.000    0.000    0.000    0.000 {len}
     1    0.000    0.000    0.000    0.000 {map}
     2    0.000    0.000    0.000    0.000 {method 'append' of 'list' objects}
     1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
     1    0.009    0.009    0.009    0.009 {method 'random_sample' of 'mtrand.RandomState' objects}
     2    0.000    0.000    0.000    0.000 {numpy.core.multiarray.array}
     1    0.002    0.002    0.002    0.002 {numpy.core.multiarray.concatenate}

Answer 2

Your code already looks good to me, but here are a few more thoughts.

. , .

def randvector2(n):
    return np.exp((2.0j * np.pi) * np.random.rand(n, 1)).view(dtype=np.float64)

n = 10000

Yours:

1000 loops, best of 3: 716 µs per loop

:

1000 loops, best of 3: 834 µs per loop

, , . , hstack. . , .

def randvector3(n):
    x = np.empty([n,2])
    theta = (2 * np.pi) * np.random.rand(n)
    np.cos(theta, out=x[:,0])
    np.sin(theta, out=x[:,1])
    return x

:

1000 loops, best of 3: 698 µs per loop

numexpr, ( , ).

import numexpr as ne
def randvector3(n):
    sample = np.random.rand(n, 1)
    c = 2.0j * np.pi
    return ne.evaluate('exp(c * sample)').view(dtype=np.float64)

:

1000 loops, best of 3: 366 µs per loop

, -, , , . . hstack .

: n = 10, . n = 10000000, pure-numpy .

Answer 3

, :

def randvector(n):
    return np.array([(np.cos(theta), np.sin(theta)) for theta in 2*np.pi*np.random.random_sample(n)])

, IanH, . , 5 , NumPy.

, :

?

, , , ( !)

( , "" ) ?

, , , , numpy ( cos sin )

Timing

:

OP randvector: 0,002131

My randvector: 0.013218 s

mskimm randvector: 0,003175

, , mskimm randvector = D