U Dx`b@s6dZddlZddlZddlmZejejddZdddd d d d d gZd+ddZ ee d,ddZ ddZ d-ddZ ee d.dd Z d/ddZd0ddZeed1ddZd2ddZeed3ddZd4dd Zeed5d!d Zd6d"d#Zeed7d$d Zd%d&Zeed'd Zd8d(d)Zeed9d*d ZdS):a~ 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 N) overridesnumpy)moduleediff1d intersect1dsetxor1dunion1d setdiff1duniquein1disincCs |||fSN)aryto_endto_beginrr>> 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]) NrZ same_kind)ZcastingzSdtype of `to_begin` must be compatible with input `ary` under the `same_kind` rule.zQdtype of `to_end` must be compatible with input `ary` under the `same_kind` rule.dtype) np asanyarrayravelrZcan_cast TypeErrorlenmaxemptyZ__array_wrap__subtract)rrrZ dtype_reqZl_beginZl_endZl_diffresultrrrr%s6-      *cCst|dkr|dS|SdS)z5 Unpacks one-element tuples for use as return values rrN)r)xrrr _unpack_tuple}s r"cCs|fSr r)ar return_indexreturn_inverse return_countsaxisrrr_unique_dispatchersr(Fc shtdkr(t|||}t|SztdWn&tjk r`tjdYnXjj dtj ddtj dt fddt jdD}z0jddkrֈ|}ntjt|d}Wn<tk r&}zd} t| jjd|W5d}~XYnXfd d } t||||} | | df| dd} t| S) a  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. 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 Nrrrcsg|]}dj|djfqS)zf{i})i)formatr).0r))r#rr szunique..z;The axis argument to unique is not supported for dtype {dt})dtcs<t|}|}|j|fdd}t|d}|S)Nrr)rviewreshapermoveaxis)Zuniqn)r' orig_dtype orig_shaperr reshape_uniq)s  zunique..reshape_uniq)rr _unique1dr"r0Z AxisErrorndimshaperr/prodintpZascontiguousarrayranger.rrrr*) r#r$r%r&r'retrZ consolidatedemsgr4outputr)r#r'r2r3rr s4z $  &c Cst|}|p|}|r8|j|r&dndd}||}n ||}tj|jtjd}d|dd<|dd|ddk|dd<||f}|r|||f7}|rt|d} tj|jtj d} | | |<|| f7}|rt t ||j gf} |t | f7}|S) z? Find the unique elements of an array, ignoring shape. mergesortZ quicksortkindrTNrr)rrflattenargsortsortrr7Zbool_Zcumsumr9 concatenateZnonzerosizeZdiff) r#r$r%r&Zoptional_indicespermauxmaskr;ZimaskZinv_idxidxrrrr56s,     r5cCs||fSr r)ar1ar2 assume_uniquereturn_indicesrrr_intersect1d_dispatcherWsrOc Cs t|}t|}|sP|r>t|dd\}}t|dd\}}q`t|}t|}n|}|}t||f}|rtj|dd}||}n||dd|ddk}|dd|} |r|dd|} |dd||j} |s|| } || } | | | fS| SdS)a Find the intersection of two arrays. Return the sorted, unique values that are in both of the input arrays. Parameters ---------- ar1, ar2 : array_like Input arrays. Will be flattened if not already 1D. assume_unique : bool If True, the input arrays are both assumed to be unique, which can speed up the calculation. If True but ``ar1`` or ``ar2`` are not unique, incorrect results and out-of-bounds indices could result. Default is False. return_indices : bool If True, the indices which correspond to the intersection of the two arrays are returned. The first instance of a value is used if there are multiple. Default is False. .. versionadded:: 1.15.0 Returns ------- intersect1d : ndarray Sorted 1D array of common and unique elements. comm1 : ndarray The indices of the first occurrences of the common values in `ar1`. Only provided if `return_indices` is True. comm2 : ndarray The indices of the first occurrences of the common values in `ar2`. Only provided if `return_indices` is True. See Also -------- numpy.lib.arraysetops : Module with a number of other functions for performing set operations on arrays. Examples -------- >>> np.intersect1d([1, 3, 4, 3], [3, 1, 2, 1]) array([1, 3]) To intersect more than two arrays, use functools.reduce: >>> from functools import reduce >>> reduce(np.intersect1d, ([1, 3, 4, 3], [3, 1, 2, 1], [6, 3, 4, 2])) array([3]) To return the indices of the values common to the input arrays along with the intersected values: >>> x = np.array([1, 1, 2, 3, 4]) >>> y = np.array([2, 1, 4, 6]) >>> xy, x_ind, y_ind = np.intersect1d(x, y, return_indices=True) >>> x_ind, y_ind (array([0, 2, 4]), array([1, 0, 2])) >>> xy, x[x_ind], y[y_ind] (array([1, 2, 4]), array([1, 2, 4]), array([1, 2, 4])) T)r$r?r@rNr)rrr rrErCrDrF) rKrLrMrNZind1Zind2rHZaux_sort_indicesrIZint1dZ ar1_indicesZ ar2_indicesrrrr\s2?     cCs||fSr rrKrLrMrrr_setxor1d_dispatchersrQcCs||st|}t|}t||f}|jdkr0|S|tdg|dd|ddkdgf}||dd|dd@S)a 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]) rTrNr)r rrErFrD)rKrLrMrHflagrrrrs (cCs||fSr r)rKrLrMinvertrrr_in1d_dispatchersrTcCsft|}t|}|jjp*|jj}t|dt|dksH|r|rvtjt|td}|D]}|||kM}qbn(tjt|td}|D]}|||kO}q|S|stj |dd\}}t |}t ||f}|j dd} || } |r| dd | d d k} n| dd | d d k} t | |gf} tj |j td} | | | <|rZ| d t|S| |Sd S) a 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]) g(\?rT)r%r?r@rNr)rasarrayrrZ hasobjectrZonesboolzerosr rErCrr7)rKrLrMrSZcontains_objectrIaZrev_idxr#orderZsarZbool_arrRr;rrrr s6C  cCs||fSr relementZ test_elementsrMrSrrr_isin_dispatchercsr]cCs"t|}t||||d|jS)a 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]]) rMrS)rrVr r/r7r[rrrr gs ^ cCs||fSr rrKrLrrr_union1d_dispatchersr`cCsttj||fddS)a= 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]) N)r')r rrEr_rrrrs"cCs||fSr rrPrrr_setdiff1d_dispatchersracCs8|rt|}nt|}t|}|t||dddS)a 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]) Tr^)rrVrr r rPrrrr s %)NN)NN)NNNN)FFFN)FFF)NN)FF)N)F)NN)FF)NN)FF)N)F)__doc__ functoolsrrZ numpy.corerpartialZarray_function_dispatch__all__rrr"r(r r5rOrrQrrTr r]r r`rrar rrrrsj   W  , "  b  *  p  b $