Cosinus of an array in Python -


i want obtain results of cosine each value. 

import numpy np  import math  t = np.arange(0, 20.5, 0.5)  print(math.cos(t))

i 'typeerror: length-1 arrays can converted python scalars' error.

you want apply operation every element of array rather entire array.

solution math

>>> import numpy np >>> import math >>> t = np.arange(0, 20.5, 0.5) >>> print([math.cos(element) element in t])  [1.0, 0.8775825618903728, 0.5403023058681398, 0.0707372016677029, -0.4161468365471424, -0.8011436155469337, -0.9899924966004454, -0.9364566872907963, -0.6536436208636119, -0.2107957994307797, 0.28366218546322625, 0.70866977429126, 0.960170286650366, 0.9765876257280235, 0.7539022543433046, 0.3466353178350258, -0.14550003380861354, -0.6020119026848236, -0.9111302618846769, -0.9971721561963784, -0.8390715290764524, -0.4755369279959925, 0.004425697988050785, 0.4833047587530059, 0.8438539587324921, 0.9977982791785807, 0.9074467814501962, 0.594920663309892, 0.1367372182078336, -0.354924266788705, -0.7596879128588213, -0.9784534628188842, -0.9576594803233847, -0.7023970575027135, -0.27516333805159693, 0.2194399632114593, 0.6603167082440802, 0.939524893748256, 0.9887046181866692, 0.7958149698139441, 0.40808206181339196]

solution numpy

however better solution use np.cos accepts array:

>>> np.cos(t) array([ 1.        ,  0.87758256,  0.54030231,  0.0707372 , -0.41614684,        -0.80114362, -0.9899925 , -0.93645669, -0.65364362, -0.2107958 ,         0.28366219,  0.70866977,  0.96017029,  0.97658763,  0.75390225,         0.34663532, -0.14550003, -0.6020119 , -0.91113026, -0.99717216,        -0.83907153, -0.47553693,  0.0044257 ,  0.48330476,  0.84385396,         0.99779828,  0.90744678,  0.59492066,  0.13673722, -0.35492427,        -0.75968791, -0.97845346, -0.95765948, -0.70239706, -0.27516334,         0.21943996,  0.66031671,  0.93952489,  0.98870462,  0.79581497,         0.40808206])

speed comparison

the second solution has huge benefits in terms of speed:

in [9]: %timeit [math.cos(element) element in t] slowest run took 5.19 times longer fastest. mean intermediate result being cached. 100000 loops, best of 3: 7.54 µs per loop  in [10]: %timeit np.cos(t) slowest run took 21.73 times longer fastest. mean intermediate result being cached. 1000000 loops, best of 3: 1.1 µs per loop

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