Question

我有两个numpy整数数组，长度都是几亿。在每个数组中，值是唯一的，每个值最初都是未排序的。

我希望各自的索引产生它们的排序交集。例如：

x = np.array([4, 1, 10, 5, 8, 13, 11])
y = np.array([20, 5, 4, 9, 11, 7, 25])

然后这些排序的交集是[4, 5, 11]，所以我们想要将x和y中的每一个转换成该数组的索引，所以我们希望它返回：

mx = np.array([0, 3, 6])
my = np.array([2, 1, 4])

此后x[mx] == y[my] == np.intersect1d(x, y)

到目前为止，我们唯一的解决方案涉及三种不同的方法，因此似乎不太可能是最优的。

每个值代表一个星系，以防这个问题变得更有趣。

Answer 1

对于一个纯粹的numpy解决方案，你可以这样做：

使用np.unique分别获取x和y中的唯一值和相应索引：

# sorted unique values in x and y and the indices corresponding to their first
# occurrences, such that u_x == x[u_idx_x]
u_x, u_idx_x = np.unique(x, return_index=True)
u_y, u_idx_y = np.unique(y, return_index=True)

使用np.intersect1d：

查找唯一值的交集

# we can assume_unique, which can be faster for large arrays
i_xy = np.intersect1d(u_x, u_y, assume_unique=True)

最后，使用np.in1d仅选择与x或y中的唯一值对应的索引，这些索引恰好位于x的交叉点和y：

# it is also safe to assume_unique here
i_idx_x = u_idx_x[np.in1d(u_x, i_xy, assume_unique=True)]
i_idx_y = u_idx_y[np.in1d(u_y, i_xy, assume_unique=True)]

将所有内容整合到一个功能中：

def intersect_indices(x, y):
    u_x, u_idx_x = np.unique(x, return_index=True)
    u_y, u_idx_y = np.unique(y, return_index=True)
    i_xy = np.intersect1d(u_x, u_y, assume_unique=True)
    i_idx_x = u_idx_x[np.in1d(u_x, i_xy, assume_unique=True)]
    i_idx_y = u_idx_y[np.in1d(u_y, i_xy, assume_unique=True)]
    return i_idx_x, i_idx_y

例如：

x = np.array([4, 1, 10, 5, 8, 13, 11])
y = np.array([20, 5, 4, 9, 11, 7, 25])

i_idx_x, i_idx_y = intersect_indices(x, y)

print(i_idx_x, i_idx_y)
# (array([0, 3, 6]), array([2, 1, 4]))

速度测试：

In [1]: k = 1000000

In [2]: %%timeit x, y = np.random.randint(k, size=(2, k))
intersect_indices(x, y)
....: 
1 loops, best of 3: 597 ms per loop

更新

我最初错过了这样一个事实：在您的情况下，x和y都只包含唯一值。考虑到这一点，使用间接排序可以做得更好：

def intersect_indices_unique(x, y):
    u_idx_x = np.argsort(x)
    u_idx_y = np.argsort(y)
    i_xy = np.intersect1d(x, y, assume_unique=True)
    i_idx_x = u_idx_x[x[u_idx_x].searchsorted(i_xy)]
    i_idx_y = u_idx_y[y[u_idx_y].searchsorted(i_xy)]
    return i_idx_x, i_idx_y

这是一个更现实的测试用例，其中x和y都包含唯一（但部分重叠）的值：

In [1]: n, k = 10000000, 1000000

In [2]: %%timeit x, y = (np.random.choice(n, size=k, replace=False) for _ in range(2))
intersect_indices(x, y)
....: 
1 loops, best of 3: 593 ms per loop

In [3]: %%timeit x, y = (np.random.choice(n, size=k, replace=False) for _ in range(2))
intersect_indices_unique(x, y)
....: 
1 loops, best of 3: 453 ms per loop

@Divakar 的解决方案在性能方面非常相似：

In [4]: %%timeit x, y = (np.random.choice(n, size=k, replace=False) for _ in range(2))
searchsorted_based(x, y)
....: 
1 loops, best of 3: 472 ms per loop

Answer 2

这是一个基于intersect1d实施的选项，相当简单。它需要拨打argsort。

无可否认的简单测试通过。

import numpy as np


def my_intersect(x, y):
    """my_intersect(x, y) -> xm, ym
    x, y: 1-d arrays of unique values
    xm, ym: indices into x and y giving sorted intersection
    """
    # basic idea taken from numpy.lib.arraysetops.intersect1d
    aux = np.concatenate((x, y))
    sidx = aux.argsort()
    # Note: intersect1d uses aux[:-1][aux[1:]==aux[:-1]] here - I don't know why the first [:-1] is necessary
    inidx = aux[sidx[1:]] == aux[sidx[:-1]]

    # quicksort is not stable, so must do some work to extract indices
    # (if stable, sidx[inidx.nonzero()]  would be for x)
    # interlace the two sets of indices, and check against lengths
    xym = np.vstack((sidx[inidx.nonzero()],
                     sidx[1:][inidx.nonzero()])).T.flatten()

    xm = xym[xym < len(x)]
    ym = xym[xym >= len(x)] - len(x)

    return xm, ym


def check_my_intersect(x, y):
    mx, my = my_intersect(x, y)
    assert (x[mx] == np.intersect1d(x, y)).all()

    # not really necessary: np.intersect1d returns a sorted list
    assert (x[mx] == sorted(x[mx])).all()
    assert (x[mx] == y[my]).all()


def random_unique_unsorted(n):
    while True:
        x = np.unique(np.random.randint(2*n, size=n))
        if len(x):
            break
    np.random.shuffle(x)
    return x


x = np.array([4, 1, 10, 5, 8, 13, 11])
y = np.array([20, 5, 4, 9, 11, 7, 25])

check_my_intersect(x, y)


for i in range(20):
    x = random_unique_unsorted(100+i)
    y = random_unique_unsorted(200+i)
    check_my_intersect(x, y)

编辑：＆＃34;注意＆＃34;注释令人困惑（使用...作为语音省略号，忘了它也是一个Python运算符。）

Answer 3

你也可以使用np.searchsorted，就像这样 -

def searchsorted_based(x,y):

    # Get argsort for both x and y
    xsort_idx = x.argsort()
    ysort_idx = y.argsort()

    # Sort x and y and store them
    X = x[xsort_idx]
    Y = y[ysort_idx]

    # Find positions of Y in X and the matches by the positions that 
    # shift between 'left' and 'right' based searches. 
    # Use the matches posotions to get corresponding argsort for X.
    x1 = np.searchsorted(X,Y,'left') 
    x2 = np.searchsorted(X,Y,'right') 
    out1 = xsort_idx[x1[x2 != x1]]

    # Repeat for X in Y findings
    y1 = np.searchsorted(Y,X,'left') 
    y2 = np.searchsorted(Y,X,'right')
    out2 = ysort_idx[y1[y2 != y1]]

    return out1, out2

示例运行 -

In [100]: x = np.array([4, 1, 10, 5, 8, 13, 11])
     ...: y = np.array([20, 5, 4, 9, 11, 7, 25])
     ...: 

In [101]: searchsorted_based(x,y)
Out[101]: (array([0, 3, 6]), array([2, 1, 4]))

Answer 4

使用dict的纯Python解决方案可能适合您：

def indices_from_values(a, intersect):
    idx = {value: index for index, value in enumerate(a)}
    return np.array([idx[x] for x in intersect])

intersect = np.intersect1d(x, y)
mx = indices_from_values(x, intersect)
my = indices_from_values(y, intersect)
np.allclose(x[mx], y[my]) and np.allclose(x[mx], np.intersect1d(x, y))

相交并排序两个numpy数组的索引

4 个答案:

更新