How can I repeat the following example sequence:
l = np.array([3,4,5,6,7])
Up to n times, doubling the values on each repetition. So for n=3:
[3, 4, 5, 6, 7, 6, 8, 10, 12, 14, 12, 16, 20, 24, 28]
Is there a simple way avoiding loops with numpy perhaps?
numpy.outer + numpy.ndarray.ravel:
>>> a = np.array([3,4,5,6,7])
>>> n = 3
>>> factors = 2**np.arange(n)
>>> np.outer(factors, a).ravel()
array([ 3, 4, 5, 6, 7, 6, 8, 10, 12, 14, 12, 16, 20, 24, 28])
Details:
>>> factors
array([1, 2, 4])
>>> np.outer(factors, a)
array([[ 3, 4, 5, 6, 7], # 1*a
[ 6, 8, 10, 12, 14], # 2*a
[12, 16, 20, 24, 28]]) # 4*a
You can use concatenate and list comprehension using powers of 2 as the multiplicative factor. Here 3 is the number of repetitions you need.
l = np.array([3,4,5,6,7])
final = np.concatenate([l*2**(i) for i in range(3)])
print (final)
array([ 3, 4, 5, 6, 7, 6, 8, 10, 12, 14, 12, 16, 20, 24, 28])
Here is one way:
>>> (l * [[1], [2], [4]]).flatten()
array([ 3, 4, 5, 6, 7, 6, 8, 10, 12, 14, 12, 16, 20, 24, 28])
(Only works if you want multiples of the input.)
You could do:
import numpy as np
l = np.array([3, 4, 5, 6, 7])
rows = np.tile(l, 3).reshape(-1, len(l)) * np.power(2, np.arange(3)).reshape(-1, 1)
print(rows.flatten())
Output
[ 3 4 5 6 7 6 8 10 12 14 12 16 20 24 28]
You can use np.power with np.ndarray.ravel:
A = np.array([3,4,5,6,7])
res = (A * np.power(2, np.arange(3))[:, None]).ravel()
print(res)
array([ 3, 4, 5, 6, 7, 6, 8, 10, 12, 14, 12, 16, 20, 24, 28])
You can to loop over n, and concatenate a new list each time:
l = [3,4,5,6,7]
n = 3
new_l = []
for i in range(n):
new_l += [j*2**i for j in l]
new_l = np.array(new_l)
Output
array([3, 4, 5, 6, 7, 6, 8, 10, 12, 14, 12, 16, 20, 24, 28])
You can do that in pure python using list comprehensions:
l = [3,4,5,6,7]
n = 3
mult = [1 * 2**i for i in range(n)]
final = list([x*m for m in mult for x in l])