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Set operations for arrays based on sorting.
:Contains:
unique,
isin,
ediff1d,
intersect1d,
setxor1d,
in1d,
union1d,
setdiff1d
:Notes:
For floating point arrays, inaccurate results may appear due to usual round-off
and floating point comparison issues.
Speed could be gained in some operations by an implementation of
sort(), that can provide directly the permutation vectors, avoiding
thus calls to argsort().
To do: Optionally return indices analogously to unique for all functions.
:Author: Robert Cimrman
� )�division�absolute_import�print_functionN�ediff1d� intersect1d�setxor1d�union1d� setdiff1d�unique�in1d�isinc C sr t j | � j � } | d k rI | d k rI | d d � | d d � S| d k r^ d } n! t j | � j � } t | � } | d k r� d } n! t j | � j � } t | � } t t | � d d � } t j | | | d | j �} | j | � } | d k r| | d | � <| d k r9| | | | d � <t j | d d � | d d � | | | | � � | S)a?
The differences between consecutive elements of an array.
Parameters
----------
ary : array_like
If necessary, will be flattened before the differences are taken.
to_end : array_like, optional
Number(s) to append at the end of the returned differences.
to_begin : array_like, optional
Number(s) to prepend at the beginning of the returned differences.
Returns
-------
ediff1d : ndarray
The differences. Loosely, this is ``ary.flat[1:] - ary.flat[:-1]``.
See Also
--------
diff, gradient
Notes
-----
When applied to masked arrays, this function drops the mask information
if the `to_begin` and/or `to_end` parameters are used.
Examples
--------
>>> x = np.array([1, 2, 4, 7, 0])
>>> np.ediff1d(x)
array([ 1, 2, 3, -7])
>>> np.ediff1d(x, to_begin=-99, to_end=np.array([88, 99]))
array([-99, 1, 2, 3, -7, 88, 99])
The returned array is always 1D.
>>> y = [[1, 2, 4], [1, 6, 24]]
>>> np.ediff1d(y)
array([ 1, 2, -3, 5, 18])
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Find the unique elements of an array.
Returns the sorted unique elements of an array. There are three optional
outputs in addition to the unique elements: the indices of the input array
that give the unique values, the indices of the unique array that
reconstruct the input array, and the number of times each unique value
comes up in the input array.
Parameters
----------
ar : array_like
Input array. Unless `axis` is specified, this will be flattened if it
is not already 1-D.
return_index : bool, optional
If True, also return the indices of `ar` (along the specified axis,
if provided, or in the flattened array) that result in the unique array.
return_inverse : bool, optional
If True, also return the indices of the unique array (for the specified
axis, if provided) that can be used to reconstruct `ar`.
return_counts : bool, optional
If True, also return the number of times each unique item appears
in `ar`.
.. versionadded:: 1.9.0
axis : int or None, optional
The axis to operate on. If None, `ar` will be flattened beforehand.
Otherwise, duplicate items will be removed along the provided axis,
with all the other axes belonging to the each of the unique elements.
Object arrays or structured arrays that contain objects are not
supported if the `axis` kwarg is used.
.. versionadded:: 1.13.0
Returns
-------
unique : ndarray
The sorted unique values.
unique_indices : ndarray, optional
The indices of the first occurrences of the unique values in the
original array. Only provided if `return_index` is True.
unique_inverse : ndarray, optional
The indices to reconstruct the original array from the
unique array. Only provided if `return_inverse` is True.
unique_counts : ndarray, optional
The number of times each of the unique values comes up in the
original array. Only provided if `return_counts` is True.
.. versionadded:: 1.9.0
See Also
--------
numpy.lib.arraysetops : Module with a number of other functions for
performing set operations on arrays.
Examples
--------
>>> np.unique([1, 1, 2, 2, 3, 3])
array([1, 2, 3])
>>> a = np.array([[1, 1], [2, 3]])
>>> np.unique(a)
array([1, 2, 3])
Return the unique rows of a 2D array
>>> a = np.array([[1, 0, 0], [1, 0, 0], [2, 3, 4]])
>>> np.unique(a, axis=0)
array([[1, 0, 0], [2, 3, 4]])
Return the indices of the original array that give the unique values:
>>> a = np.array(['a', 'b', 'b', 'c', 'a'])
>>> u, indices = np.unique(a, return_index=True)
>>> u
array(['a', 'b', 'c'],
dtype='|S1')
>>> indices
array([0, 1, 3])
>>> a[indices]
array(['a', 'b', 'c'],
dtype='|S1')
Reconstruct the input array from the unique values:
>>> a = np.array([1, 2, 6, 4, 2, 3, 2])
>>> u, indices = np.unique(a, return_inverse=True)
>>> u
array([1, 2, 3, 4, 6])
>>> indices
array([0, 1, 4, 3, 1, 2, 1])
>>> u[indices]
array([1, 2, 6, 4, 2, 3, 2])
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Find the unique elements of an array, ignoring shape.
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Find the intersection of two arrays.
Return the sorted, unique values that are in both of the input arrays.
Parameters
----------
ar1, ar2 : array_like
Input arrays.
assume_unique : bool
If True, the input arrays are both assumed to be unique, which
can speed up the calculation. Default is False.
Returns
-------
intersect1d : ndarray
Sorted 1D array of common and unique elements.
See Also
--------
numpy.lib.arraysetops : Module with a number of other functions for
performing set operations on arrays.
Examples
--------
>>> np.intersect1d([1, 3, 4, 3], [3, 1, 2, 1])
array([1, 3])
To intersect more than two arrays, use functools.reduce:
>>> from functools import reduce
>>> reduce(np.intersect1d, ([1, 3, 4, 3], [3, 1, 2, 1], [6, 3, 4, 2]))
array([3])
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Find the set exclusive-or of two arrays.
Return the sorted, unique values that are in only one (not both) of the
input arrays.
Parameters
----------
ar1, ar2 : array_like
Input arrays.
assume_unique : bool
If True, the input arrays are both assumed to be unique, which
can speed up the calculation. Default is False.
Returns
-------
setxor1d : ndarray
Sorted 1D array of unique values that are in only one of the input
arrays.
Examples
--------
>>> a = np.array([1, 2, 3, 2, 4])
>>> b = np.array([2, 3, 5, 7, 5])
>>> np.setxor1d(a,b)
array([1, 4, 5, 7])
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Test whether each element of a 1-D array is also present in a second array.
Returns a boolean array the same length as `ar1` that is True
where an element of `ar1` is in `ar2` and False otherwise.
We recommend using :func:`isin` instead of `in1d` for new code.
Parameters
----------
ar1 : (M,) array_like
Input array.
ar2 : array_like
The values against which to test each value of `ar1`.
assume_unique : bool, optional
If True, the input arrays are both assumed to be unique, which
can speed up the calculation. Default is False.
invert : bool, optional
If True, the values in the returned array are inverted (that is,
False where an element of `ar1` is in `ar2` and True otherwise).
Default is False. ``np.in1d(a, b, invert=True)`` is equivalent
to (but is faster than) ``np.invert(in1d(a, b))``.
.. versionadded:: 1.8.0
Returns
-------
in1d : (M,) ndarray, bool
The values `ar1[in1d]` are in `ar2`.
See Also
--------
isin : Version of this function that preserves the
shape of ar1.
numpy.lib.arraysetops : Module with a number of other functions for
performing set operations on arrays.
Notes
-----
`in1d` can be considered as an element-wise function version of the
python keyword `in`, for 1-D sequences. ``in1d(a, b)`` is roughly
equivalent to ``np.array([item in b for item in a])``.
However, this idea fails if `ar2` is a set, or similar (non-sequence)
container: As ``ar2`` is converted to an array, in those cases
``asarray(ar2)`` is an object array rather than the expected array of
contained values.
.. versionadded:: 1.4.0
Examples
--------
>>> test = np.array([0, 1, 2, 5, 0])
>>> states = [0, 2]
>>> mask = np.in1d(test, states)
>>> mask
array([ True, False, True, False, True], dtype=bool)
>>> test[mask]
array([0, 2, 0])
>>> mask = np.in1d(test, states, invert=True)
>>> mask
array([False, True, False, True, False], dtype=bool)
>>> test[mask]
array([1, 5])
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Calculates `element in test_elements`, broadcasting over `element` only.
Returns a boolean array of the same shape as `element` that is True
where an element of `element` is in `test_elements` and False otherwise.
Parameters
----------
element : array_like
Input array.
test_elements : array_like
The values against which to test each value of `element`.
This argument is flattened if it is an array or array_like.
See notes for behavior with non-array-like parameters.
assume_unique : bool, optional
If True, the input arrays are both assumed to be unique, which
can speed up the calculation. Default is False.
invert : bool, optional
If True, the values in the returned array are inverted, as if
calculating `element not in test_elements`. Default is False.
``np.isin(a, b, invert=True)`` is equivalent to (but faster
than) ``np.invert(np.isin(a, b))``.
Returns
-------
isin : ndarray, bool
Has the same shape as `element`. The values `element[isin]`
are in `test_elements`.
See Also
--------
in1d : Flattened version of this function.
numpy.lib.arraysetops : Module with a number of other functions for
performing set operations on arrays.
Notes
-----
`isin` is an element-wise function version of the python keyword `in`.
``isin(a, b)`` is roughly equivalent to
``np.array([item in b for item in a])`` if `a` and `b` are 1-D sequences.
`element` and `test_elements` are converted to arrays if they are not
already. If `test_elements` is a set (or other non-sequence collection)
it will be converted to an object array with one element, rather than an
array of the values contained in `test_elements`. This is a consequence
of the `array` constructor's way of handling non-sequence collections.
Converting the set to a list usually gives the desired behavior.
.. versionadded:: 1.13.0
Examples
--------
>>> element = 2*np.arange(4).reshape((2, 2))
>>> element
array([[0, 2],
[4, 6]])
>>> test_elements = [1, 2, 4, 8]
>>> mask = np.isin(element, test_elements)
>>> mask
array([[ False, True],
[ True, False]], dtype=bool)
>>> element[mask]
array([2, 4])
>>> mask = np.isin(element, test_elements, invert=True)
>>> mask
array([[ True, False],
[ False, True]], dtype=bool)
>>> element[mask]
array([0, 6])
Because of how `array` handles sets, the following does not
work as expected:
>>> test_set = {1, 2, 4, 8}
>>> np.isin(element, test_set)
array([[ False, False],
[ False, False]], dtype=bool)
Casting the set to a list gives the expected result:
>>> np.isin(element, list(test_set))
array([[ False, True],
[ True, False]], dtype=bool)
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Find the union of two arrays.
Return the unique, sorted array of values that are in either of the two
input arrays.
Parameters
----------
ar1, ar2 : array_like
Input arrays. They are flattened if they are not already 1D.
Returns
-------
union1d : ndarray
Unique, sorted union of the input arrays.
See Also
--------
numpy.lib.arraysetops : Module with a number of other functions for
performing set operations on arrays.
Examples
--------
>>> np.union1d([-1, 0, 1], [-2, 0, 2])
array([-2, -1, 0, 1, 2])
To find the union of more than two arrays, use functools.reduce:
>>> from functools import reduce
>>> reduce(np.union1d, ([1, 3, 4, 3], [3, 1, 2, 1], [6, 3, 4, 2]))
array([1, 2, 3, 4, 6])
)r
r r? )rE rF r r r r E s !c C sS | r t j | � j � } n t | � } t | � } | t | | d d d d �S)a9
Find the set difference of two arrays.
Return the sorted, unique values in `ar1` that are not in `ar2`.
Parameters
----------
ar1 : array_like
Input array.
ar2 : array_like
Input comparison array.
assume_unique : bool
If True, the input arrays are both assumed to be unique, which
can speed up the calculation. Default is False.
Returns
-------
setdiff1d : ndarray
Sorted 1D array of values in `ar1` that are not in `ar2`.
See Also
--------
numpy.lib.arraysetops : Module with a number of other functions for
performing set operations on arrays.
Examples
--------
>>> a = np.array([1, 2, 3, 2, 4, 1])
>>> b = np.array([3, 4, 5, 6])
>>> np.setdiff1d(a, b)
array([1, 2])
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