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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?
343
const repeat = (arr, n) => Array .from ( { length: arr.length * n }, (_, i) => arr [i % arr.length]);
This method doesn’t use the other useful data provided in questions like range of numbers is between 1 to n and there are only two repeating elements. Traverse the array once.
Each time you tap the T key (while still holding down the other keys) you will get another copy with the transformation applied. 7. Repeat this all the way around. Because it’s repeating that same transformation from the pivot point, that pivot point is really important.
The repeat () is an inbuilt function in julia which is used to construct an array by repeating the specified array elements with the specified number of times. repeat (A::AbstractArray, counts::Integer…) A::AbstractArray: Specified array. counts::Integer: Specified number of times each element get repeated.
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])
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