arraysetops.py

"""
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

"""
from __future__ import division, absolute_import, print_function

import numpy as np


__all__ = [
    'ediff1d', 'intersect1d', 'setxor1d', 'union1d', 'setdiff1d', 'unique',
    'in1d', 'isin'
    ]


def ediff1d(ary, to_end=None, to_begin=None):
    """
    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])

    """
    # force a 1d array
    ary = np.asanyarray(ary).ravel()

    # fast track default case
    if to_begin is None and to_end is None:
        return ary[1:] - ary[:-1]

    if to_begin is None:
        l_begin = 0
    else:
        to_begin = np.asanyarray(to_begin).ravel()
        l_begin = len(to_begin)

    if to_end is None:
        l_end = 0
    else:
        to_end = np.asanyarray(to_end).ravel()
        l_end = len(to_end)

    # do the calculation in place and copy to_begin and to_end
    l_diff = max(len(ary) - 1, 0)
    result = np.empty(l_diff + l_begin + l_end, dtype=ary.dtype)
    result = ary.__array_wrap__(result)
    if l_begin > 0:
        result[:l_begin] = to_begin
    if l_end > 0:
        result[l_begin + l_diff:] = to_end
    np.subtract(ary[1:], ary[:-1], result[l_begin:l_begin + l_diff])
    return result


def unique(ar, return_index=False, return_inverse=False,
           return_counts=False, axis=None):
    """
    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. If an integer,
        the subarrays indexed by the given axis will be flattened and treated
        as the elements of a 1-D array with the dimension of the given axis,
        see the notes for more details.  Object arrays or structured arrays
        that contain objects are not supported if the `axis` kwarg is used. The
        default is None.

        .. 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.

    Notes
    -----
    When an axis is specified the subarrays indexed by the axis are sorted.
    This is done by making the specified axis the first dimension of the array
    and then flattening the subarrays in C order. The flattened subarrays are
    then viewed as a structured type with each element given a label, with the
    effect that we end up with a 1-D array of structured types that can be
    treated in the same way as any other 1-D array. The result is that the
    flattened subarrays are sorted in lexicographic order starting with the
    first element.

    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])

    """
    ar = np.asanyarray(ar)
    if axis is None:
        return _unique1d(ar, return_index, return_inverse, return_counts)
    if not (-ar.ndim <= axis < ar.ndim):
        raise ValueError('Invalid axis kwarg specified for unique')

    ar = np.swapaxes(ar, axis, 0)
    orig_shape, orig_dtype = ar.shape, ar.dtype
    # Must reshape to a contiguous 2D array for this to work...
    ar = ar.reshape(orig_shape[0], -1)
    ar = np.ascontiguousarray(ar)

    dtype = [('f{i}'.format(i=i), ar.dtype) for i in range(ar.shape[1])]

    try:
        consolidated = ar.view(dtype)
    except TypeError:
        # There's no good way to do this for object arrays, etc...
        msg = 'The axis argument to unique is not supported for dtype {dt}'
        raise TypeError(msg.format(dt=ar.dtype))

    def reshape_uniq(uniq):
        uniq = uniq.view(orig_dtype)
        uniq = uniq.reshape(-1, *orig_shape[1:])
        uniq = np.swapaxes(uniq, 0, axis)
        return uniq

    output = _unique1d(consolidated, return_index,
                       return_inverse, return_counts)
    if not (return_index or return_inverse or return_counts):
        return reshape_uniq(output)
    else:
        uniq = reshape_uniq(output[0])
        return (uniq,) + output[1:]

def _unique1d(ar, return_index=False, return_inverse=False,
              return_counts=False):
    """
    Find the unique elements of an array, ignoring shape.
    """
    ar = np.asanyarray(ar).flatten()

    optional_indices = return_index or return_inverse
    optional_returns = optional_indices or return_counts

    if ar.size == 0:
        if not optional_returns:
            ret = ar
        else:
            ret = (ar,)
            if return_index:
                ret += (np.empty(0, np.intp),)
            if return_inverse:
                ret += (np.empty(0, np.intp),)
            if return_counts:
                ret += (np.empty(0, np.intp),)
        return ret

    if optional_indices:
        perm = ar.argsort(kind='mergesort' if return_index else 'quicksort')
        aux = ar[perm]
    else:
        ar.sort()
        aux = ar
    flag = np.concatenate(([True], aux[1:] != aux[:-1]))

    if not optional_returns:
        ret = aux[flag]
    else:
        ret = (aux[flag],)
        if return_index:
            ret += (perm[flag],)
        if return_inverse:
            iflag = np.cumsum(flag) - 1
            inv_idx = np.empty(ar.shape, dtype=np.intp)
            inv_idx[perm] = iflag
            ret += (inv_idx,)
        if return_counts:
            idx = np.concatenate(np.nonzero(flag) + ([ar.size],))
            ret += (np.diff(idx),)
    return ret

def intersect1d(ar1, ar2, assume_unique=False):
    """
    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])
    """
    if not assume_unique:
        # Might be faster than unique( intersect1d( ar1, ar2 ) )?
        ar1 = unique(ar1)
        ar2 = unique(ar2)
    aux = np.concatenate((ar1, ar2))
    aux.sort()
    return aux[:-1][aux[1:] == aux[:-1]]

def setxor1d(ar1, ar2, assume_unique=False):
    """
    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])

    """
    if not assume_unique:
        ar1 = unique(ar1)
        ar2 = unique(ar2)

    aux = np.concatenate((ar1, ar2))
    if aux.size == 0:
        return aux

    aux.sort()
    flag = np.concatenate(([True], aux[1:] != aux[:-1], [True]))
    return aux[flag[1:] & flag[:-1]]


def in1d(ar1, ar2, assume_unique=False, invert=False):
    """
    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])
    >>> test[mask]
    array([0, 2, 0])
    >>> mask = np.in1d(test, states, invert=True)
    >>> mask
    array([False,  True, False,  True, False])
    >>> test[mask]
    array([1, 5])
    """
    # Ravel both arrays, behavior for the first array could be different
    ar1 = np.asarray(ar1).ravel()
    ar2 = np.asarray(ar2).ravel()

    # Check if one of the arrays may contain arbitrary objects
    contains_object = ar1.dtype.hasobject or ar2.dtype.hasobject

    # This code is run when
    # a) the first condition is true, making the code significantly faster
    # b) the second condition is true (i.e. `ar1` or `ar2` may contain
    #    arbitrary objects), since then sorting is not guaranteed to work
    if len(ar2) < 10 * len(ar1) ** 0.145 or contains_object:
        if invert:
            mask = np.ones(len(ar1), dtype=bool)
            for a in ar2:
                mask &= (ar1 != a)
        else:
            mask = np.zeros(len(ar1), dtype=bool)
            for a in ar2:
                mask |= (ar1 == a)
        return mask

    # Otherwise use sorting
    if not assume_unique:
        ar1, rev_idx = np.unique(ar1, return_inverse=True)
        ar2 = np.unique(ar2)

    ar = np.concatenate((ar1, ar2))
    # We need this to be a stable sort, so always use 'mergesort'
    # here. The values from the first array should always come before
    # the values from the second array.
    order = ar.argsort(kind='mergesort')
    sar = ar[order]
    if invert:
        bool_ar = (sar[1:] != sar[:-1])
    else:
        bool_ar = (sar[1:] == sar[:-1])
    flag = np.concatenate((bool_ar, [invert]))
    ret = np.empty(ar.shape, dtype=bool)
    ret[order] = flag

    if assume_unique:
        return ret[:len(ar1)]
    else:
        return ret[rev_idx]


def isin(element, test_elements, assume_unique=False, invert=False):
    """
    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]])
    >>> element[mask]
    array([2, 4])
    >>> mask = np.isin(element, test_elements, invert=True)
    >>> mask
    array([[ True, False],
           [ False, True]])
    >>> 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]])

    Casting the set to a list gives the expected result:

    >>> np.isin(element, list(test_set))
    array([[ False,  True],
           [ True,  False]])
    """
    element = np.asarray(element)
    return in1d(element, test_elements, assume_unique=assume_unique,
                invert=invert).reshape(element.shape)


def union1d(ar1, ar2):
    """
    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])
    """
    return unique(np.concatenate((ar1, ar2), axis=None))

def setdiff1d(ar1, ar2, assume_unique=False):
    """
    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])

    """
    if assume_unique:
        ar1 = np.asarray(ar1).ravel()
    else:
        ar1 = unique(ar1)
        ar2 = unique(ar2)
    return ar1[in1d(ar1, ar2, assume_unique=True, invert=True)]