问题
Suppose I have
>>> v
array([1, 1, 1, 1, 1, 2, 2, 2, 3, 4, 3, 4, 3, 4, 3, 4, 5, 5, 5])
Is there an efficient numpy way to find each index where the value changes? For instance, I would want some result like,
>>> index_of_changed_values(v)
[0, 5, 8, 9, 10, 11, 12, 13, 14, 15, 16]
If this is not possible with some numpy routine, what is a fast way to do it in python? It would also be useful to me to be referred to some good numpy tutorials since I am a numpy beginner.
回答1:
You can get this functionality in numpy by comparing each element with it's neighbor;
v[:-1] != v[1:]
array([False, False, False, False, True, False, False, True, True,
True, True, True, True, True, True, True, False, False], dtype=bool)
to get the indices you use the "where" function
np.where(v[:-1] != v[1:])[0]
array([ 4, 7, 8, 9, 10, 11, 12, 13, 14, 15])
From here you can prepend the first element and add a one to get to the same indexing scheme you have in your question.
回答2:
Similar to @kith answer, but requires less massaging of the result:
np.where(np.roll(v,1)!=v)[0]
No need to prepend 0 or add 1. Example:
>>> v=np.array([1, 1, 1, 2, 2, 3, 3, 4, 4, 4])
>>> np.where(np.roll(v,1)!=v)[0]
array([0, 3, 5, 7])
EDIT: as @Praveen mentioned, this fails when the last and the first elements are equal.
回答3:
Maybe It's because Python 3.5 but the above codes did not work for me.
Looks like v[:-1] != v[1:] does not return an iterable but a single bool.
I came up with following list comprehension using zip and enumerate
[ i for i, (x, y) in enumerate(zip(v[:-1],v[1:])) if x!=y]
Someone looking for solution in py3.5 might find this useful!
来源：https://stackoverflow.com/questions/19125661/find-index-where-elements-change-value-numpy