在Python中查找两个列表/数组中最近的项目

Question

我有两个numpy数组x并y包含浮点值.对于每个值x,我想找到最接近的元素y,而不重用元素y.输出应该是1-1的元素索引到y元素索引的映射.这是一种依赖于排序的坏方法.它删除从列表中配对的每个元素.如果没有排序,这将是不好的,因为配对将取决于原始输入数组的顺序.

def min_i(values):
    min_index, min_value = min(enumerate(values),
                               key=operator.itemgetter(1))
    return min_index, min_value

# unsorted elements
unsorted_x = randn(10)*10
unsorted_y = randn(10)*10

# sort lists
x = sort(unsorted_x)
y = sort(unsorted_y)

pairs = []
indx_to_search = range(len(y))

for x_indx, x_item in enumerate(x):
    if len(indx_to_search) == 0:
        print "ran out of items to match..."
        break
    # until match is found look for closest item
    possible_values = y[indx_to_search]
    nearest_indx, nearest_item = min_i(possible_values)
    orig_indx = indx_to_search[nearest_indx]
    # remove it
    indx_to_search.remove(orig_indx)
    pairs.append((x_indx, orig_indx))
print "paired items: "
for k,v in pairs:
    print x[k], " paired with ", y[v]

我宁愿做没有第一排序的元素,但如果他们进行排序,然后我想在原有的无序列表索引unsorted_x,unsorted_y.在numpy/scipy/Python或使用pandas的最佳方法是什么？谢谢.

编辑:澄清我并不是想要找到所有元素的最佳匹配(例如,不是最小化距离的总和),而是最适合每个元素,并且如果它有时以牺牲其他元素为代价也没关系.我认为这y通常比x上面的例子大得多,所以对于每个xin 值,通常有很多很好的拟合y,我只想有效地找到它.

有人可以为此展示一个scipy kdtrees的例子吗？文档非常稀疏

kdtree = scipy.spatial.cKDTree([x,y])
kdtree.query([-3]*10) # ?? unsure about what query takes as arg

Answer 1

编辑2KDTree如果您可以选择一些邻居来保证您的阵列中的每个项目都有一个唯一的邻居,那么使用的解决方案可以很好地执行.使用以下代码:

def nearest_neighbors_kd_tree(x, y, k) :
    x, y = map(np.asarray, (x, y))
    tree =scipy.spatial.cKDTree(y[:, None])    
    ordered_neighbors = tree.query(x[:, None], k)[1]
    nearest_neighbor = np.empty((len(x),), dtype=np.intp)
    nearest_neighbor.fill(-1)
    used_y = set()
    for j, neigh_j in enumerate(ordered_neighbors) :
        for k in neigh_j :
            if k not in used_y :
                nearest_neighbor[j] = k
                used_y.add(k)
                break
    return nearest_neighbor

和n=1000点样本,我得到:

In [9]: np.any(nearest_neighbors_kd_tree(x, y, 12) == -1)
Out[9]: True

In [10]: np.any(nearest_neighbors_kd_tree(x, y, 13) == -1)
Out[10]: False

所以最优的是k=13,然后时间是:

In [11]: %timeit nearest_neighbors_kd_tree(x, y, 13)
100 loops, best of 3: 9.26 ms per loop

但在最坏的情况下,你可能需要k=1000,然后:

In [12]: %timeit nearest_neighbors_kd_tree(x, y, 1000)
1 loops, best of 3: 424 ms per loop

这比其他选项慢:

In [13]: %timeit nearest_neighbors(x, y)
10 loops, best of 3: 60 ms per loop

In [14]: %timeit nearest_neighbors_sorted(x, y)
10 loops, best of 3: 47.4 ms per loop

编辑在搜索之前对数组进行排序可以获得超过1000个项目的数组:

def nearest_neighbors_sorted(x, y) :
    x, y = map(np.asarray, (x, y))
    y_idx = np.argsort(y)
    y = y[y_idx]
    nearest_neighbor = np.empty((len(x),), dtype=np.intp)
    for j, xj in enumerate(x) :
        idx = np.searchsorted(y, xj)
        if idx == len(y) or idx != 0 and y[idx] - xj > xj - y[idx-1] :
            idx -= 1
        nearest_neighbor[j] = y_idx[idx]
        y = np.delete(y, idx)
        y_idx = np.delete(y_idx, idx)
    return nearest_neighbor

使用10000个元素的长数组:

In [2]: %timeit nearest_neighbors_sorted(x, y)
1 loops, best of 3: 557 ms per loop

In [3]: %timeit nearest_neighbors(x, y)
1 loops, best of 3: 1.53 s per loop

对于较小的阵列,它表现稍差.

如果只是为了丢弃重复项,您将不得不遍历所有项目以实现贪婪的最近邻居算法.考虑到这一点,这是我能够提出的最快速度:

def nearest_neighbors(x, y) :
    x, y = map(np.asarray, (x, y))
    y = y.copy()
    y_idx = np.arange(len(y))
    nearest_neighbor = np.empty((len(x),), dtype=np.intp)
    for j, xj in enumerate(x) :
        idx = np.argmin(np.abs(y - xj))
        nearest_neighbor[j] = y_idx[idx]
        y = np.delete(y, idx)
        y_idx = np.delete(y_idx, idx)

    return nearest_neighbor

而现在:

n = 1000
x = np.random.rand(n)
y = np.random.rand(2*n)

我明白了:

In [11]: %timeit nearest_neighbors(x, y)
10 loops, best of 3: 52.4 ms per loop