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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
"""
overrides.array_function_dispatch, module='numpy')
'ediff1d', 'intersect1d', 'setxor1d', 'union1d', 'setdiff1d', 'unique', 'in1d', 'isin' ]
return (ary, to_end, to_begin)
""" 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()
# enforce propagation of the dtype of input # ary to returned result dtype_req = ary.dtype
# 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, dtype=dtype_req) if not np.all(_to_begin == to_begin): raise ValueError("cannot convert 'to_begin' to array with dtype " "'%r' as required for input ary" % dtype_req) to_begin = _to_begin.ravel() l_begin = len(to_begin)
if to_end is None: l_end = 0 else: _to_end = np.asanyarray(to_end, dtype=dtype_req) # check that casting has not overflowed if not np.all(_to_end == to_end): raise ValueError("cannot convert 'to_end' to array with dtype " "'%r' as required for input ary" % dtype_req) to_end = _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
""" Unpacks one-element tuples for use as return values """ else: return x
return_counts=None, axis=None): return (ar,)
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 * 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])
"""
# axis was specified and not None try: ar = np.swapaxes(ar, axis, 0) except np.AxisError: # this removes the "axis1" or "axis2" prefix from the error message raise np.AxisError(axis, ar.ndim)
# Must reshape to a contiguous 2D array for this to work... orig_shape, orig_dtype = ar.shape, ar.dtype 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) output = (reshape_uniq(output[0]),) + output[1:] return _unpack_tuple(output)
return_counts=False): """ Find the unique elements of an array, ignoring shape. """
perm = ar.argsort(kind='mergesort' if return_index else 'quicksort') aux = ar[perm] else:
ret += (perm[mask],) imask = np.cumsum(mask) - 1 inv_idx = np.empty(mask.shape, dtype=np.intp) inv_idx[perm] = imask ret += (inv_idx,) idx = np.concatenate(np.nonzero(mask) + ([mask.size],)) ret += (np.diff(idx),)
ar1, ar2, assume_unique=None, return_indices=None): return (ar1, ar2)
""" 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. Will be flattened if not already 1D. assume_unique : bool If True, the input arrays are both assumed to be unique, which can speed up the calculation. Default is False. return_indices : bool If True, the indices which correspond to the intersection of the two arrays are returned. The first instance of a value is used if there are multiple. Default is False.
.. versionadded:: 1.15.0
Returns ------- intersect1d : ndarray Sorted 1D array of common and unique elements. comm1 : ndarray The indices of the first occurrences of the common values in `ar1`. Only provided if `return_indices` is True. comm2 : ndarray The indices of the first occurrences of the common values in `ar2`. Only provided if `return_indices` is True.
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])
To return the indices of the values common to the input arrays along with the intersected values: >>> x = np.array([1, 1, 2, 3, 4]) >>> y = np.array([2, 1, 4, 6]) >>> xy, x_ind, y_ind = np.intersect1d(x, y, return_indices=True) >>> x_ind, y_ind (array([0, 2, 4]), array([1, 0, 2])) >>> xy, x[x_ind], y[y_ind] (array([1, 2, 4]), array([1, 2, 4]), array([1, 2, 4]))
""" ar1 = np.asanyarray(ar1) ar2 = np.asanyarray(ar2)
if not assume_unique: if return_indices: ar1, ind1 = unique(ar1, return_index=True) ar2, ind2 = unique(ar2, return_index=True) else: ar1 = unique(ar1) ar2 = unique(ar2) else: ar1 = ar1.ravel() ar2 = ar2.ravel()
aux = np.concatenate((ar1, ar2)) if return_indices: aux_sort_indices = np.argsort(aux, kind='mergesort') aux = aux[aux_sort_indices] else: aux.sort()
mask = aux[1:] == aux[:-1] int1d = aux[:-1][mask]
if return_indices: ar1_indices = aux_sort_indices[:-1][mask] ar2_indices = aux_sort_indices[1:][mask] - ar1.size if not assume_unique: ar1_indices = ind1[ar1_indices] ar2_indices = ind2[ar2_indices]
return int1d, ar1_indices, ar2_indices else: return int1d
return (ar1, ar2)
""" 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]]
return (ar1, ar2)
""" 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]
return (element, test_elements)
""" 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])
The indices of the matched values can be obtained with `nonzero`:
>>> np.nonzero(mask) (array([0, 1]), array([1, 0]))
The test can also be inverted:
>>> 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)
return (ar1, ar2)
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))
return (ar1, ar2)
""" Find the set difference of two arrays.
Return the 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 1D array of values in `ar1` that are not in `ar2`. The result is sorted when `assume_unique=False`, but otherwise only sorted if the input is sorted.
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)]
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