""" Set operations for arrays based on sorting. 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 `numpy.sort`, that can provide directly the permutation vectors, thus avoiding calls to `numpy.argsort`. Original author: Robert Cimrman """ import functools import numpy as np from numpy.core import overrides array_function_dispatch = functools.partial( overrides.array_function_dispatch, module='numpy') __all__ = [ 'ediff1d', 'intersect1d', 'setxor1d', 'union1d', 'setdiff1d', 'unique', 'in1d', 'isin' ] def _ediff1d_dispatcher(ary, to_end=None, to_begin=None): return (ary, to_end, to_begin) @array_function_dispatch(_ediff1d_dispatcher) 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, ..., -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 that the dtype of `ary` is used for the output 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) if not np.can_cast(to_begin, dtype_req, casting="same_kind"): raise TypeError("dtype of `to_begin` must be compatible " "with input `ary` under the `same_kind` rule.") 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) if not np.can_cast(to_end, dtype_req, casting="same_kind"): raise TypeError("dtype of `to_end` must be compatible " "with input `ary` under the `same_kind` rule.") 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 def _unpack_tuple(x): """ Unpacks one-element tuples for use as return values """ if len(x) == 1: return x[0] else: return x def _unique_dispatcher(ar, return_index=None, return_inverse=None, return_counts=None, axis=None): return (ar,) @array_function_dispatch(_unique_dispatcher) 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 * 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. repeat : Repeat elements of an array. 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 (move the axis to the first dimension to keep the order of the other axes) 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. .. versionchanged: NumPy 1.21 If nan values are in the input array, a single nan is put to the end of the sorted unique values. Also for complex arrays all NaN values are considered equivalent (no matter whether the NaN is in the real or imaginary part). As the representant for the returned array the smallest one in the lexicographical order is chosen - see np.sort for how the lexicographical order is defined for complex 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='>> indices array([0, 1, 3]) >>> a[indices] array(['a', 'b', 'c'], dtype='>> 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]) Reconstruct the input values from the unique values and counts: >>> a = np.array([1, 2, 6, 4, 2, 3, 2]) >>> values, counts = np.unique(a, return_counts=True) >>> values array([1, 2, 3, 4, 6]) >>> counts array([1, 3, 1, 1, 1]) >>> np.repeat(values, counts) array([1, 2, 2, 2, 3, 4, 6]) # original order not preserved """ ar = np.asanyarray(ar) if axis is None: ret = _unique1d(ar, return_index, return_inverse, return_counts) return _unpack_tuple(ret) # axis was specified and not None try: ar = np.moveaxis(ar, axis, 0) except np.AxisError: # this removes the "axis1" or "axis2" prefix from the error message raise np.AxisError(axis, ar.ndim) from None # 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], np.prod(orig_shape[1:], dtype=np.intp)) ar = np.ascontiguousarray(ar) dtype = [('f{i}'.format(i=i), ar.dtype) for i in range(ar.shape[1])] # At this point, `ar` has shape `(n, m)`, and `dtype` is a structured # data type with `m` fields where each field has the data type of `ar`. # In the following, we create the array `consolidated`, which has # shape `(n,)` with data type `dtype`. try: if ar.shape[1] > 0: consolidated = ar.view(dtype) else: # If ar.shape[1] == 0, then dtype will be `np.dtype([])`, which is # a data type with itemsize 0, and the call `ar.view(dtype)` will # fail. Instead, we'll use `np.empty` to explicitly create the # array with shape `(len(ar),)`. Because `dtype` in this case has # itemsize 0, the total size of the result is still 0 bytes. consolidated = np.empty(len(ar), dtype=dtype) except TypeError as e: # 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)) from e def reshape_uniq(uniq): n = len(uniq) uniq = uniq.view(orig_dtype) uniq = uniq.reshape(n, *orig_shape[1:]) uniq = np.moveaxis(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) 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 if optional_indices: perm = ar.argsort(kind='mergesort' if return_index else 'quicksort') aux = ar[perm] else: ar.sort() aux = ar mask = np.empty(aux.shape, dtype=np.bool_) mask[:1] = True if aux.shape[0] > 0 and aux.dtype.kind in "cfmM" and np.isnan(aux[-1]): if aux.dtype.kind == "c": # for complex all NaNs are considered equivalent aux_firstnan = np.searchsorted(np.isnan(aux), True, side='left') else: aux_firstnan = np.searchsorted(aux, aux[-1], side='left') if aux_firstnan > 0: mask[1:aux_firstnan] = ( aux[1:aux_firstnan] != aux[:aux_firstnan - 1]) mask[aux_firstnan] = True mask[aux_firstnan + 1:] = False else: mask[1:] = aux[1:] != aux[:-1] ret = (aux[mask],) if return_index: ret += (perm[mask],) if return_inverse: imask = np.cumsum(mask) - 1 inv_idx = np.empty(mask.shape, dtype=np.intp) inv_idx[perm] = imask ret += (inv_idx,) if return_counts: idx = np.concatenate(np.nonzero(mask) + ([mask.size],)) ret += (np.diff(idx),) return ret def _intersect1d_dispatcher( ar1, ar2, assume_unique=None, return_indices=None): return (ar1, ar2) @array_function_dispatch(_intersect1d_dispatcher) def intersect1d(ar1, ar2, assume_unique=False, return_indices=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. 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. If True but ``ar1`` or ``ar2`` are not unique, incorrect results and out-of-bounds indices could result. 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 def _setxor1d_dispatcher(ar1, ar2, assume_unique=None): return (ar1, ar2) @array_function_dispatch(_setxor1d_dispatcher) 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_dispatcher(ar1, ar2, assume_unique=None, invert=None): return (ar1, ar2) @array_function_dispatch(_in1d_dispatcher) 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() # Ensure that iteration through object arrays yields size-1 arrays if ar2.dtype == object: ar2 = ar2.reshape(-1, 1) # 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_dispatcher(element, test_elements, assume_unique=None, invert=None): return (element, test_elements) @array_function_dispatch(_isin_dispatcher) 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]) 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) def _union1d_dispatcher(ar1, ar2): return (ar1, ar2) @array_function_dispatch(_union1d_dispatcher) 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_dispatcher(ar1, ar2, assume_unique=None): return (ar1, ar2) @array_function_dispatch(_setdiff1d_dispatcher) def setdiff1d(ar1, ar2, assume_unique=False): """ 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)]