Say you have a numpy vector [0,3,1,1,1] and you run argsort
you will get [0,2,3,4,1] but all the ones are the same!
What I want is an efficient way to shuffle indices of identical values.
Any idea how to do that without a while loop with two indices on the sorted vector?
numpy.array([0,3,1,1,1]).argsort()
Use lexsort:
np.lexsort((b,a)) means Sort by a, then by b
>>> a
array([0, 3, 1, 1, 1])
>>> b=np.random.random(a.size)
>>> b
array([ 0.00673736, 0.90089115, 0.31407214, 0.24299867, 0.7223546 ])
>>> np.lexsort((b,a))
array([0, 3, 2, 4, 1])
>>> a.argsort()
array([0, 2, 3, 4, 1])
>>> a[[0, 3, 2, 4, 1]]
array([0, 1, 1, 1, 3])
>>> a[[0, 2, 3, 4, 1]]
array([0, 1, 1, 1, 3])
This is a bit of a hack, but if your array contains integers only you could add random values and argsort the result. np.random.rand gives you results in [0, 1) so in this case you're guaranteed to maintain the order for non-identical elements.
>>> import numpy as np
>>> arr = np.array([0,3,1,1,1])
>>> np.argsort(arr + np.random.rand(*arr.shape))
array([0, 4, 3, 2, 1])
>>> np.argsort(arr + np.random.rand(*arr.shape))
array([0, 3, 4, 2, 1])
>>> np.argsort(arr + np.random.rand(*arr.shape))
array([0, 3, 4, 2, 1])
>>> np.argsort(arr + np.random.rand(*arr.shape))
array([0, 2, 3, 4, 1])
>>> np.argsort(arr + np.random.rand(*arr.shape))
array([0, 2, 3, 4, 1])
>>> np.argsort(arr + np.random.rand(*arr.shape))
array([0, 4, 2, 3, 1])
Here we see index 0 is always first in the argsort result and index 1 is last, but the rest of the results are in a random order.
In general you could generate random values bounded by np.diff(np.sort(arr)).max(), but you might run into precision issues at some point.
