ix_of_item_change_1d
- mdtools.numpy_helper_functions.ix_of_item_change_1d(a)[source]
Get the indices of a 1-dimensional array where its elements change.
Deprecated since version 0.0.0.dev0:
mdtools.numpy_helper_functions.ix_of_item_change_1d()
might be removed in a future release. It is replaced bymdtools.numpy_helper_functions.ix_of_item_change()
, because this function works for arrays with arbitrary dimensions and provides additional functionality.mdtools.numpy_helper_functions.ix_of_item_change()
with prepend_zero set toTrue
is equivalent tomdtools.numpy_helper_functions.ix_of_item_change_1d()
.Todo
Check for scripts using this function before removing it.
- Parameters:
a (
array_like
) – 1-dimensional array for which to get each index where its elements change.- Returns:
ix (
numpy.ndarray
) – The indices where the elements of a change. The first element of ix is always zero.
See also
mdtools.numpy_helper_functions.ix_of_item_change()
Same function for arrays of arbitrary dimension
Examples
>>> a = np.arange(3) >>> a array([0, 1, 2]) >>> mdt.nph.ix_of_item_change_1d(a) array([0, 1, 2])
>>> b = np.array([1, 2, 2, 3, 3, 3, 1]) >>> mdt.nph.ix_of_item_change_1d(b) array([0, 1, 3, 6])
>>> c = np.ones(3) >>> mdt.nph.ix_of_item_change_1d(c) array([0])
Edge cases:
>>> x = np.array([1]) >>> mdt.nph.ix_of_item_change_1d(x) array([0])
>>> x = np.array([]) >>> ix = mdt.nph.ix_of_item_change_1d(x) >>> ix array([0]) >>> x[ix] Traceback (most recent call last): ... IndexError: index 0 is out of bounds for axis 0 with size 0