Question

我有两个N个浮点数（作为(x,y)坐标，可能有重复）和z个N个浮点数组（作为权重）为了坐标）。

对于每个(x,y)个浮点数，我需要选择具有最小关联z值的对。我已经定义了一个selectMinz()函数来执行此操作（请参阅下面的代码），但这需要太长时间。

我怎样才能提高这个功能的性能？

import numpy as np
import time


def getData():
    N = 100000
    x = np.arange(0.0005, 0.03, 0.001)
    y = np.arange(6., 10., .05)
    # Select N values for x,y, where values can be repeated
    x = np.random.choice(x, N)
    y = np.random.choice(y, N)
    z = np.random.uniform(10., 15., N)
    return x, y, z


def selectMinz(x, y, z):
    """
    Select the minimum z for each (x,y) pair.
    """
    xy_unq, z_unq = [], []
    # For each (x,y) pair
    for i, xy in enumerate(zip(*[x, y])):
        # If this xy pair was already stored in the xy_unq list
        if xy in xy_unq:
            # If the stored z value associated with this xy pair is
            # larger than this new z[i] value
            if z_unq[xy_unq.index(xy)] > z[i]:
                # Store this smaller value instead
                z_unq[xy_unq.index(xy)] = z[i]
        else:
            # Store the xy pair, and its associated z value
            xy_unq.append(xy)
            z_unq.append(z[i])

    return xy_unq, z_unq


# Define data with the proper format.
x, y, z = getData()

s = time.clock()
xy_unq, z_unq = selectMinz(x, y, z)  # <-- TAKES TOO LONG (~15s in my system)
print(time.clock() - s)

Answer 1

将xy_unq和z_unq的数据结构更改为包含两条信息的字典会使我的系统上的时间从~7s降低到~0.1s。

def selectMinz(x, y, z):
    """
    Select the minimum z for each (x,y) pair.
    """
    xy_unq = {}
    # For each (x,y) pair
    for i, xy in enumerate(zip(*[x, y])):
        # If this xy pair was already stored in the xy_unq list
        if xy in xy_unq:
            # If the stored z value associated with this xy pair is
            # larger than this new z[i] value
            if xy_unq[xy] > z[i]:
                # Store this smaller value instead
                xy_unq[xy] = z[i]
        else:
            # Store the xy pair, and its associated z value
            xy_unq[xy] = z[i]

    return xy_unq.keys(), xy_unq.values()

上述方法的时间范围为~0.106s至~0.11s。这是一种替代方法，代码行数较少，但需要的时间稍长（~0.14）：

def selectMinz(x, y, z):
    """
    Select the minimum z for each (x,y) pair.
    """
    xy_unq = {}
    # For each (x,y) pair
    for i, xy in enumerate(zip(*[x, y])):
        # If this xy pair was already stored in the xy_unq list
        if xy in xy_unq:
            # Store the value that is smaller between the current stored value and the new z[i]
            xy_unq[xy] = min(xy_unq[xy], z[i])
        else:
            # Store the xy pair, and its associated z value
            xy_unq[xy] = z[i]

    return xy_unq.keys(), xy_unq.values()

Answer 2

步骤：

使用lex-sort按顺序排列x-y对。或者，我们可以使用缩放方法根据另一个数组的值范围缩放其中一个数组，然后将其与另一个数组求和，最后使用argsort获取lex排序的等效索引。
使用np.minimum.reduceat获取间隔中的最小值，由对分组定义。

因此，我们将有一个矢量化解决方案，如此 -

def selectMinz_vectorized(x, y, z):
    # Get grouped lex-sort indices
    sidx = (y + x*(y.max() - y.min() + 1)).argsort()
    # or sidx = np.lexsort([x, y])

    # Lex-sort x, y, z
    x_sorted = x[sidx]
    y_sorted = y[sidx]
    z_sorted = z[sidx]

    # Get equality mask between each sorted X and Y elem against previous ones.
    # The non-zero indices of its inverted mask gives us the indices where the 
    # new groupings start. We are calling those as cut_idx.
    seq_eq_mask = (x_sorted[1:] == x_sorted[:-1]) & (y_sorted[1:] == y_sorted[:-1])
    cut_idx = np.flatnonzero(np.concatenate(( [True], ~seq_eq_mask)))

    # Use those cut_idx to get intervalled minimum values
    minZ = np.minimum.reduceat(z_sorted, cut_idx)

    # Make tuples of the groupings of x,y and the corresponding min Z values
    return (zip(x_sorted[cut_idx], y_sorted[cut_idx]), minZ.tolist())

示例运行 -

In [120]: np.c_[x,y,z]
Out[120]: 
array([[  0.,   1.,  69.],
       [  2.,   0.,  47.],
       [  1.,   0.,  62.],
       [  0.,   2.,  33.],
       [  1.,   7.,  32.],
       [  1.,   0.,  50.],
       [  2.,   0.,  55.]])

In [121]: selectMinz(x,y,z) # original method
Out[121]: 
([(0.0, 1.0), (2.0, 0.0), (1.0, 0.0), (0.0, 2.0), (1.0, 7.0)],
 [69.0, 47.0, 50.0, 33.0, 32.0])

In [122]: selectMinz_vectorized(x,y,z)
Out[122]: 
([(1.0, 0.0), (2.0, 0.0), (0.0, 1.0), (0.0, 2.0), (1.0, 7.0)],
 [50.0, 47.0, 69.0, 33.0, 32.0])

这是我的初始方法，涉及创建堆叠数组然后执行这些操作。实现看起来像这样 -

def selectMinz_vectorized_v2(x, y, z):
    d = np.column_stack((x,y,z))
    sidx = np.lexsort(d[:,:2].T)
    b = d[sidx]  
    cut_idx = np.r_[0,np.flatnonzero(~(b[1:,:2] == b[:-1,:2]).all(1))+1]
    minZ = np.minimum.reduceat(b[:,-1], cut_idx)
    return ([tuple(i) for i in b[cut_idx,:2]], minZ.tolist())

矢量化方法的基准测试

方法 -

# Pruned version of the approach posted earlier
def selectMinz_vectorized_pruned(x, y, z):
    sidx = (y + x*(y.max() - y.min() + 1)).argsort()
    x_sorted = x[sidx]
    y_sorted = y[sidx]
    z_sorted = z[sidx]
    seq_eq_mask = (x_sorted[1:] == x_sorted[:-1]) & (y_sorted[1:] == y_sorted[:-1])
    cut_idx = np.flatnonzero(np.concatenate(( [True], ~seq_eq_mask)))
    minZ = np.minimum.reduceat(z_sorted, cut_idx)
    return x_sorted[cut_idx], y_sorted[cut_idx], minZ

def numpy_indexed_app(x,y,z): # @Eelco Hoogendoorn's soln
    return npi.group_by((x, y)).min(z)

计时 -

In [141]: x,y,z=getData(10000)

In [142]: %timeit selectMinz_vectorized_pruned(x, y, z)
     ...: %timeit numpy_indexed_app(x,y,z)
     ...: 
1000 loops, best of 3: 763 µs per loop
1000 loops, best of 3: 1.09 ms per loop

In [143]: x,y,z=getData(100000)

In [144]: %timeit selectMinz_vectorized_pruned(x, y, z)
     ...: %timeit numpy_indexed_app(x,y,z)
     ...: 
100 loops, best of 3: 8.53 ms per loop
100 loops, best of 3: 12.9 ms per loop

Answer 3

numpy_indexed包（免责声明：我是它的作者）包含以优雅高效的方式解决这类分组问题的功能：

import numpy_indexed as npi
xy_unq, z_unq = npi.group_by((x, y)).min(z)

从第三个数组中的两个数组中有效地获得每对元素的最小值

3 个答案: