Suppose I have a 1D array A,
A = [0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.51, 1.52, 1.6, 2, 3, 4, 5, 6, 7, 8, 9, 10]
and I have a value a = 1.5 and I need to find the smallest index of the entry where the value would fit in the array. In this case it should be 5.
import numpy as np
A = np.array([0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.51, 1.52, 1.6, 2, 3, 4, 5, 6, 7, 8, 9, 10])
a = 1.5
print A[np.where(A >= a >= A)]
I know this would not work but can np.where find such indexes?
Assuming A is sorted, you can do this in O(log n) time with np.searchsorted (A can be an array or a list):
>>> A = [0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.51, 1.52, 1.6, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> np.searchsorted(A, 1.5)
5
The bisect module does exactly that (in O(log n) time):
>>> from bisect import bisect_left
>>> bisect_left(A, 1.5)
5
You're looking basically for the minimum index where the value is greater than or equal to what you're trying to insert.
One way to do that is via a call to min but with the proviso that you need to handle an empty array if the value is beyond the last element:
>>> import numpy as np
>>> A = np.array([0, 0.1, 0.2, 0.3, 0.4, 0.5, 1, 2, 3, 4, 5])
>>> min(np.append(np.where(A >= -1)[0],len(A))
0
>>> min(np.append(np.where(A >= 0)[0],len(A))
0
>>> min(np.append(np.where(A >= 0.01)[0],len(A))
1
>>> min(np.append(np.where(A >= 3.5)[0],len(A))
9
>>> min(np.append(np.where(A >= 999)[0],len(A))
11
In all cases, that gives you the index of the element you need to insert before (or one beyond the highest index if you need to append to the list).
Aalways sorted?