Reputation: 115
I have a random numpy array
import numpy as np
a = np.random.randn(10000*5).reshape((10000,5))
and I want to transform as efficiently as possible each column by the function
def lintransform(interval,x):
return (interval[1]-interval[0])*x + interval[0]
Where interval
is one of five sorted array of length 2 to transform the columns of a
.
(for example listofintervals = [[0,3],[1,9],[0.5,3],[4,10],[1,2.7]]
)
What is the most efficient way to apply each this function for each column respectively and generate a new array, changing the interval used according to its position in the listofintervals
?
Upvotes: 0
Views: 496
Reputation: 6485
Using numpy vectorization you can do:
import numpy as np
a = np.random.randn(10000, 5)
intervals = np.array([[0,3],
[1,9],
[0.5,3],
[4,10],
[1,2.7]])
r = (intervals[:,1] - intervals[:,0]) * a + intervals[:,0]
Which takes:
%timeit (intervals[:,1] - intervals[:,0]) * a + intervals[:,0]
131 µs ± 1.35 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Upvotes: 3
Reputation: 19555
Looping through the columns, and applying your function to each row should work:
for col in range(a.shape[1]):
a[col] = lintransform(listofintervals[col], a[col])
Output:
a
array([[-5.80231737, -3.1056331 , -1.3878622 , 3.2891958 , -1.35495844],
[-7.93085499, 18.46079707, 13.81923528, -3.18486045, -0.31541526],
[ 1.53477244, 2.61705202, -2.14505552, 0.14751953, 4.70029497],
...,
[ 1.13798389, -0.6765344 , -0.1364982 , -1.0443724 , 0.06717867],
[-1.78251012, 0.11171333, 1.28247762, 0.52285423, 0.16057854],
[-0.59513499, -0.76866946, -0.37233491, -1.08463643, -0.45660967]])
Upvotes: 1