Find index where elements change value numpy

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.

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.

Almost ten years later, but I came across this one today.

@kith answer is good, but may not be as neat as we want (also taking into account the steps not explicit in the answer).

that answer in the complete form would be,

v = np.array([1, 1, 1, 1, 1, 2, 2, 2, 3, 4, 3, 4, 3, 4, 3, 4, 5, 5, 5])np.concatenate((np.array([0]),np.where(v[:-1] != v[1:])[0]+1),axis=0)

An alternative I like more is,

np.where(np.diff(v,prepend=np.nan))[0]

which also returns

array([ 0,  5,  8,  9, 10, 11, 12, 13, 14, 15, 16], dtype=int64)

As I said, the idea is the same as @kith's but,

• I replace v[:-1] != v[1:] for np.diff(), then in np.where the array is casted to boolean, this does not change much but seems neater.
• I removed the extra step of adding 1 and prepending 0. This is done by prepending np.nan before doing np.diff(). The first element of the diff output will then be np.nan, and in python np.nan always evaluates True.