将值数字化到 "floor" bin python

Question

我需要将一些值数字化，以便返回的索引是 "floor" 或 "ceiling" bin。

例如，对于 bins = numpy.array([0.0, 0.5, 1.0, 1.5, 2.0]) 和值 0.2，我希望索引为 0，对于值 0.26，返回的索引应为 1, 等等。

我有以下丑陋的函数来做我想做的事：

import numpy

def get_bin_index(value, bins):
    bin_diff = bins[1]-bins[0]
    index = numpy.digitize(value, bins)
    if bins[index] - value > bin_diff/2.0:
        index -= 1
    return index

有什么巧妙的（阅读 better/efficient）方法可以做到这一点吗？

编辑：包括时间值（满足我的好奇心！）

In [1]: def get_bin_index(value, bins):
    ...:     bin_diff = bins[1]-bins[0]
    ...:     index = numpy.digitize(value, bins)
    ...:     if bins[index] - value > bin_diff/2.0:
    ...:         index -= 1
    ...:     return index
    ...:

In [2]: def get_bin_index_c(value, bins):
    ...:     return numpy.rint((value-bins[0])/(bins[1]-bins[0]))
    ...:

In [3]: def get_bin_index_mid_digitized(value, bins):
    ...:     return numpy.digitize(0.6, (bins[1:] + bins[:-1])/2.0)
    ...:

In [4]: bin_halfs = numpy.array([0.0, 0.5, 1.0, 1.5, 2.0])

In [5]: %timeit get_bin_index(0.9, bin_halfs)
The slowest run took 5.71 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 4.93 µs per loop

In [6]: %timeit get_bin_index_c(0.9, bin_halfs)
The slowest run took 14.60 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 2.34 µs per loop

In [7]: %timeit get_bin_index_mid_digitized(0.9, bin_halfs)
The slowest run took 4.09 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 8.37 µs per loop

Answer 1

您可以简单地获取垃圾箱的中间并与 np.digitize -

一起使用

np.digitize(value, (bins[1:] + bins[:-1])/2.0)

Answer 2

如果 bin_diffs 都相同，您可以通过以下方式在常数时间内完成此操作：

def get_bin_index2(value, bins):
    return numpy.rint((value - bins[0])/(bins[1]-bins[0]))

将值数字化到 "floor" bin python

Digitizing value to "floor" bin python

python

numpy

digitization