有没有办法使用numpy.percentile函数来计算加权百分位数？或者是否有人知道替代python函数来计算加权百分位数？
谢谢！
不幸的是，numpy没有内置的加权函数，但是，你可以随时把东西放在一起。
def weight_array(ar, weights):
zipped = zip(ar, weights)
weighted = []
for i in zipped:
for j in range(i[1]):
weighted.append(i[0])
return weighted
np.percentile(weight_array(ar, weights), 25)
这是我正在使用的代码。它不是最佳的（我无法在numpy中编写），但仍比接受的解决方案更快，更可靠
def weighted_quantile(values, quantiles, sample_weight=None,
values_sorted=False, old_style=False):
""" Very close to numpy.percentile, but supports weights.
NOTE: quantiles should be in [0, 1]!
:param values: numpy.array with data
:param quantiles: array-like with many quantiles needed
:param sample_weight: array-like of the same length as `array`
:param values_sorted: bool, if True, then will avoid sorting of
initial array
:param old_style: if True, will correct output to be consistent
with numpy.percentile.
:return: numpy.array with computed quantiles.
"""
values = np.array(values)
quantiles = np.array(quantiles)
if sample_weight is None:
sample_weight = np.ones(len(values))
sample_weight = np.array(sample_weight)
assert np.all(quantiles >= 0) and np.all(quantiles <= 1), \
'quantiles should be in [0, 1]'
if not values_sorted:
sorter = np.argsort(values)
values = values[sorter]
sample_weight = sample_weight[sorter]
weighted_quantiles = np.cumsum(sample_weight) - 0.5 * sample_weight
if old_style:
# To be convenient with numpy.percentile
weighted_quantiles -= weighted_quantiles[0]
weighted_quantiles /= weighted_quantiles[-1]
else:
weighted_quantiles /= np.sum(sample_weight)
return np.interp(quantiles, weighted_quantiles, values)
例子：
weighted_quantile（[1,2,9,3.2,4]，[0.0,0.5,1。]）
数组（[1.，3.2,9。]）
weighted_quantile（[1,2,9,3.2,4]，[0.0,0.5,1]，sample_weight = [2,1,2,4,1]）
数组（[1.，3.2,9。]）
快速解决方案，首先排序然后插值：
def weighted_percentile(data, percents, weights=None):
''' percents in units of 1%
weights specifies the frequency (count) of data.
'''
if weights is None:
return np.percentile(data, percents)
ind=np.argsort(data)
d=data[ind]
w=weights[ind]
p=1.*w.cumsum()/w.sum()*100
y=np.interp(percents, p, d)
return y
为额外的（非原创）答案道歉（没有足够的代表对@nayyarv的评论）。他的解决方案对我有用（即它复制了np.percentage的默认行为），但我认为你可以用原始np.percentage的编写方式消除for循环。
def weighted_percentile(a, q=np.array([75, 25]), w=None):
"""
Calculates percentiles associated with a (possibly weighted) array
Parameters
----------
a : array-like
The input array from which to calculate percents
q : array-like
The percentiles to calculate (0.0 - 100.0)
w : array-like, optional
The weights to assign to values of a. Equal weighting if None
is specified
Returns
-------
values : np.array
The values associated with the specified percentiles.
"""
# Standardize and sort based on values in a
q = np.array(q) / 100.0
if w is None:
w = np.ones(a.size)
idx = np.argsort(a)
a_sort = a[idx]
w_sort = w[idx]
# Get the cumulative sum of weights
ecdf = np.cumsum(w_sort)
# Find the percentile index positions associated with the percentiles
p = q * (w.sum() - 1)
# Find the bounding indices (both low and high)
idx_low = np.searchsorted(ecdf, p, side='right')
idx_high = np.searchsorted(ecdf, p + 1, side='right')
idx_high[idx_high > ecdf.size - 1] = ecdf.size - 1
# Calculate the weights
weights_high = p - np.floor(p)
weights_low = 1.0 - weights_high
# Extract the low/high indexes and multiply by the corresponding weights
x1 = np.take(a_sort, idx_low) * weights_low
x2 = np.take(a_sort, idx_high) * weights_high
# Return the average
return np.add(x1, x2)
# Sample data
a = np.array([1.0, 2.0, 9.0, 3.2, 4.0], dtype=np.float)
w = np.array([2.0, 1.0, 3.0, 4.0, 1.0], dtype=np.float)
# Make an unweighted "copy" of a for testing
a2 = np.repeat(a, w.astype(np.int))
# Tests with different percentiles chosen
q1 = np.linspace(0.0, 100.0, 11)
q2 = np.linspace(5.0, 95.0, 10)
q3 = np.linspace(4.0, 94.0, 10)
for q in (q1, q2, q3):
assert np.all(weighted_percentile(a, q, w) == np.percentile(a2, q))
我不知道加权百分位是什么意思，但是从@Joan Smith的回答，看来你只需要重复ar中的每个元素，你可以使用numpy.repeat()：
import numpy as np
np.repeat([1,2,3], [4,5,6])
结果是：
array([1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3])
我根据自己的需要使用此功能：
def quantile_at_values(values, population, weights=None):
values = numpy.atleast_1d(values).astype(float)
population = numpy.atleast_1d(population).astype(float)
# if no weights are given, use equal weights
if weights is None:
weights = numpy.ones(population.shape).astype(float)
normal = float(len(weights))
# else, check weights
else:
weights = numpy.atleast_1d(weights).astype(float)
assert len(weights) == len(population)
assert (weights >= 0).all()
normal = numpy.sum(weights)
assert normal > 0.
quantiles = numpy.array([numpy.sum(weights[population <= value]) for value in values]) / normal
assert (quantiles >= 0).all() and (quantiles <= 1).all()
return quantiles
如果您想要百分位数而不是分位数，则将结果乘以100。
正如评论中所提到的，对于浮点权重而言，简单地重复值是不可能的，对于非常大的数据集而言则不切实际。有一个图书馆在这里做加权百分位数：http://kochanski.org/gpk/code/speechresearch/gmisclib/gmisclib.weighted_percentile-module.html它对我有用。
def weighted_percentile(a, percentile = np.array([75, 25]), weights=None):
"""
O(nlgn) implementation for weighted_percentile.
"""
percentile = np.array(percentile)/100.0
if weights is None:
weights = np.ones(len(a))
a_indsort = np.argsort(a)
a_sort = a[a_indsort]
weights_sort = weights[a_indsort]
ecdf = np.cumsum(weights_sort)
percentile_index_positions = percentile * (weights.sum()-1)+1
# need the 1 offset at the end due to ecdf not starting at 0
locations = np.searchsorted(ecdf, percentile_index_positions)
out_percentiles = np.zeros(len(percentile_index_positions))
for i, empiricalLocation in enumerate(locations):
# iterate across the requested percentiles
if ecdf[empiricalLocation-1] == np.floor(percentile_index_positions[i]):
# i.e. is the percentile in between 2 separate values
uppWeight = percentile_index_positions[i] - ecdf[empiricalLocation-1]
lowWeight = 1 - uppWeight
out_percentiles[i] = a_sort[empiricalLocation-1] * lowWeight + \
a_sort[empiricalLocation] * uppWeight
else:
# i.e. the percentile is entirely in one bin
out_percentiles[i] = a_sort[empiricalLocation]
return out_percentiles
这是我的功能，它给出了相同的行为
np.percentile(np.repeat(a, weights), percentile)
内存开销较少。 np.percentile是一个O（n）实现，所以它对于小权重可能更快。它将所有边缘案例整理出来 - 这是一个精确的解决方案。上面的插值答案假定是线性的，当它是大多数情况下的一个步骤时，除非权重为1。
假设我们有权重[1,2,1,7]的数据[1,2,3]，我想要25％的百分位数。我的ecdf将是[3,10,21]，我正在寻找第5个值。插值将看到[3,1]和[10,2]作为匹配并插值给出1.28，尽管完全在第二个bin中值为2。
我的解决方案：
def my_weighted_perc(data,perc,weights=None):
if weights==None:
return nanpercentile(data,perc)
else:
d=data[(~np.isnan(data))&(~np.isnan(weights))]
ix=np.argsort(d)
d=d[ix]
wei=weights[ix]
wei_cum=100.*cumsum(wei*1./sum(wei))
return interp(perc,wei_cum,d)
它只是计算数据的加权CDF，然后用它来估计加权百分位数。