numpy bincount - 选择最大权重而不是总和所有权重

Question

是否可以使用np.bincount 但获得最大值而不是权重总和？这里，索引3 处的bbb 有两个值，11.1 和55.5。我想要55.5，而不是66.6。我怀疑我选择使用其他功能，但不确定哪一个适合此目的。

bbb = np.array([ 3, 7, 11, 13, 3])
weight = np.array([ 11.1, 22.2, 33.3, 44.4, 55.5])
print np.bincount(bbb, weight, minlength=15)

OUT >> [  0.    0.    0.   66.6   0.    0.    0.   22.2   0.    0.    0.   33.3   0.   44.4   0. ]

请注意，实际上bbb 和weight 非常大（大约5e6 元素）。

Answer 1

可能没有 Nils 的回答那么快，但 numpy_indexed 包（免责声明：我是它的作者）具有更灵活的语法来执行这些类型的操作：

import numpy_indexed as npi
unique_keys, maxima_per_key = npi.group_by(bbb).max(weight)

Answer 2

方法#1：这是使用np.maximum.reduceat 获得分箱最大值的一种方法 -

def binned_max(bbb, weight, minlength):
    sidx = bbb.argsort()
    weight_s = weight[sidx]
    bbb_s = bbb[sidx]
    cut_idx = np.flatnonzero(np.concatenate(([True], bbb_s[1:] != bbb_s[:-1])))
    bbb_unq = bbb_s[cut_idx]
    #Or bbb_unq, cut_idx = np.unique(bbb_s, return_index=1)
    max_val = np.maximum.reduceat(weight_s, cut_idx)
    out = np.zeros(minlength, dtype=weight.dtype)
    out[bbb_unq] = max_val
    return out

示例运行 -

In [36]: bbb = np.array([ 3, 7, 11, 13, 3])
    ...: weight = np.array([ 11.1, 22.2, 33.3, 44.4, 55.5])

In [37]: binned_max(bbb, weight, minlength=15)
Out[37]: 
array([  0. ,   0. ,   0. ,  55.5,   0. ,   0. ,   0. ,  22.2,   0. ,
         0. ,   0. ,  33.3,   0. ,  44.4,   0. ])

---

方法 #2： 我试图通过numba 查看/玩得开心来解决这个问题，它似乎非常有效。这是一种麻木的方式-

from numba import njit 

@njit
def numba_func(out, bins, weight, minlength):
    l = len(bins)
    for i in range(l):
        if out[bins[i]] < weight[i]:
            out[bins[i]] = weight[i]
    return out

def maxat_numba(bins, weight, minlength):
    out = np.zeros(minlength, dtype=weight.dtype)
    out[bins] = weight.min()
    numba_func(out, bins, weight, minlength)
    return out

运行时测试-

带有np.maximum.at 的内置插件看起来很整洁，在大多数情况下都是首选，因此针对它测试建议的那个 -

# @Nils Werner's soln with np.maximum.at
def maxat_numpy(bins, weight, minlength):
    out = np.zeros(minlength)
    np.maximum.at(out, bins, weight)
    return out

时间安排 -

案例#1：

In [155]: bbb = np.random.randint(1,1000, (10000))

In [156]: weight = np.random.rand(*bbb.shape)

In [157]: %timeit maxat_numpy(bbb, weight, minlength=bbb.max()+1)
1000 loops, best of 3: 686 µs per loop

In [158]: %timeit maxat_numba(bbb, weight, minlength=bbb.max()+1)
10000 loops, best of 3: 60.6 µs per loop

案例#2：

In [159]: bbb = np.random.randint(1,10000, (1000000))

In [160]: weight = np.random.rand(*bbb.shape)

In [161]: %timeit maxat_numpy(bbb, weight, minlength=bbb.max()+1)
10 loops, best of 3: 66 ms per loop

In [162]: %timeit maxat_numba(bbb, weight, minlength=bbb.max()+1)
100 loops, best of 3: 5.42 ms per loop

Answer 3

solution to your 2D question 也适用于一维情况，所以你可以使用np.maxmimum.at

out = np.zeros(15)
np.maximum.at(out, bbb, weight)
# array([  0. ,   0. ,   0. ,  55.5,   0. ,   0. ,   0. ,  22.2,   0. ,
#          0. ,   0. ,  33.3,   0. ,  44.4,   0. ])