"""
Set operations for 1D numeric arrays based on sorting.
:Contains:
ediff1d,
unique,
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'
]
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):
"""
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. This will be flattened if it is not already 1-D.
return_index : bool, optional
If True, also return the indices of `ar` that result in the unique
array.
return_inverse : bool, optional
If True, also return the indices of the unique array that can be used
to reconstruct `ar`.
return_counts : bool, optional
If True, also return the number of times each unique value comes up
in `ar`.
.. versionadded:: 1.9.0
Returns
-------
unique : ndarray
The sorted unique values.
unique_indices : ndarray, optional
The indices of the first occurrences of the unique values in the
(flattened) original array. Only provided if `return_index` is True.
unique_inverse : ndarray, optional
The indices to reconstruct the (flattened) 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 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).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.bool),)
if return_inverse:
ret += (np.empty(0, np.bool),)
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 = ediff1d( aux, to_end = 1, to_begin = 1 ) == 0
flag = np.concatenate(([True], aux[1:] != aux[:-1], [True]))
# flag2 = ediff1d( flag ) == 0
flag2 = flag[1:] == flag[:-1]
return aux[flag2]
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.
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
--------
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])
"""
# Ravel both arrays, behavior for the first array could be different
ar1 = np.asarray(ar1).ravel()
ar2 = np.asarray(ar2).ravel()
# This code is significantly faster when the condition is satisfied.
if len(ar2) < 10 * len(ar1) ** 0.145:
if invert:
mask = np.ones(len(ar1), dtype=np.bool)
for a in ar2:
mask &= (ar1 != a)
else:
mask = np.zeros(len(ar1), dtype=np.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 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)))
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)]