Calculating cosine values for an array in Python

Question

I have this array named a of 1242 numbers. I need to get the cosine value for all the numbers in Python.

When I use : cos_ra = math.cos(a) I get an error stating:

TypeError: only length-1 arrays can be converted to Python scalars

How can I solve this problem??

Thanks in advance

...because math.cos isn't vectorized, but np.cos is. (Also, better for performance since only one function call overhead) — smci
– smci, Commented Apr 29, 2020 at 2:56

Ashwini Chaudhary · Accepted Answer · 2014-01-10 12:02:33Z

6

Problem is you're using numpy.math.cos here, which expects you to pass a scalar. Use numpy.cos if you want to apply cos to an iterable.

In [30]: import numpy as np

In [31]: np.cos(np.array([1, 2, 3]))                                                             
Out[31]: array([ 0.54030231, -0.41614684, -0.9899925 ])

Error:

In [32]: np.math.cos(np.array([1, 2, 3]))                                                        
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-32-8ce0f3c0df04> in <module>()
----> 1 np.math.cos(np.array([1, 2, 3]))

TypeError: only length-1 arrays can be converted to Python scalars

answered Jan 10, 2014 at 12:02

Ashwini Chaudhary

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2 Comments

Srivatsan Over a year ago

my reuptation is only at 11 as I joined stackexachange just now

Ashwini Chaudhary Over a year ago

@user1469380 Accepting an answer doesn't require any reputation ;-), but upvoting an answer does.

zhangxaochen · Accepted Answer · 2014-01-10 11:54:37Z

3

use numpy:

In [178]: from numpy import *

In [179]: a=range(1242)

In [180]: b=np.cos(a)

In [181]: b
Out[181]: 
array([ 1.        ,  0.54030231, -0.41614684, ...,  0.35068442,
       -0.59855667, -0.99748752])

besides, numpy array operations are very fast:

In [182]: %timeit b=np.cos(a)  #numpy is the fastest
10000 loops, best of 3: 165 us per loop

In [183]: %timeit cos_ra = [math.cos(i) for i in a]
1000 loops, best of 3: 225 us per loop

In [184]: %timeit map(math.cos, a)
10000 loops, best of 3: 173 us per loop

answered Jan 10, 2014 at 11:54

zhangxaochen

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Comments

Danstahr · Accepted Answer · 2014-01-10 11:59:30Z

3

The problem is that math.cos expect to get a number as argument while you are trying to pass a list. You need to call math.cos on each of the list elements.

Try using map :

map(math.cos, a)

edited Jan 10, 2014 at 11:59

answered Jan 10, 2014 at 11:53

Danstahr

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Comments

Roshan · Accepted Answer · 2019-02-28 06:15:06Z

2

Easy Way Motivated by the answer of zhangxaochen.

np.cos(np.arange(start, end, step))

Hope this helps!

edited Feb 28, 2019 at 6:15

Roshan

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answered Feb 28, 2019 at 4:57

Choung young jae

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Comments

jayelm · Accepted Answer · 2014-01-10 11:55:27Z

1

math.cos() can only be called on individual values, not lists.

Another alternative, using list comprehension:

cos_ra = [math.cos(i) for i in a]

answered Jan 10, 2014 at 11:55

jayelm

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