43

有没有办法使用 numpy.percentile 函数来计算加权百分位数?或者有人知道计算加权百分位数的替代python函数吗?

谢谢!

4

12 回答 12

66

完全矢量化的 numpy 解决方案

这是我使用的代码。它不是一个最佳方案(我无法用它来写numpy),但仍然比公认的解决方案更快、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. ])

于 2015-04-16T14:22:35.833 回答
11

一个快速的解决方案,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
于 2015-07-21T13:00:34.687 回答
9

我不知道加权百分位数是什么意思,但从@Joan Smith 的回答看来,您似乎只需要重复 中的每个元素ar,您可以使用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])
于 2014-02-18T05:43:46.463 回答
7

为额外的(非原创的)答案道歉(没有足够的代表评论@nayyarv's)。他的解决方案对我有用(即它复制了 的默认行为),但我认为您可以通过原始编写np.percentage方式的线索来消除 for 循环。np.percentage

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))
于 2015-08-26T00:08:23.927 回答
6

将此参考用于加权百分位数方法更清洁、更简单。

import numpy as np

def weighted_percentile(data, weights, perc):
    """
    perc : percentile in [0-1]!
    """
    ix = np.argsort(data)
    data = data[ix] # sort data
    weights = weights[ix] # sort weights
    cdf = (np.cumsum(weights) - 0.5 * weights) / np.sum(weights) # 'like' a CDF function
    return np.interp(perc, cdf, data)
于 2020-04-21T12:47:15.977 回答
4

这似乎现在在 statsmodels 中实现

from statsmodels.stats.weightstats import DescrStatsW
wq = DescrStatsW(data=np.array([1, 2, 9, 3.2, 4]), weights=np.array([0.0, 0.5, 1.0, 0.3, 0.5]))
wq.quantile(probs=np.array([0.1, 0.9]), return_pandas=False)
# array([2., 9.])

DescrStatsW 对象还实现了其他方法,例如加权平均等。https://www.statsmodels.org/stable/generated/statsmodels.stats.weightstats.DescrStatsW.html

于 2020-08-16T17:47:17.857 回答
2

我使用这个功能来满足我的需要:

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
  • 它尽可能地被矢量化。
  • 它有很多健全性检查。
  • 它使用浮点数作为权重。
  • 它可以在没有重量的情况下工作(→ 相等的重量)。
  • 它可以一次计算多个分位数。

如果您想要百分位数而不是分位数,请将结果乘以 100。

于 2015-02-25T13:35:48.350 回答
2

正如评论中提到的,对于浮点权重来说,简单地重复值是不可能的,对于非常大的数据集也是不切实际的。这里有一个加权百分位数的库:http: //kochanski.org/gpk/code/speechresearch/gmisclib/gmisclib.weighted_percentile-module.html 它对我有用。

于 2015-01-22T07:06:51.300 回答
2
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,它给出相同的行为

np.percentile(np.repeat(a, weights), percentile)

内存开销更少。np.percentile 是一个 O(n) 实现,因此对于小权重它可能更快。它整理了所有边缘情况 - 这是一个精确的解决方案。上面的插值答案假设是线性的,当它在大多数情况下是一个步骤时,除了权重为 1 时。

假设我们有权重为 [3, 11, 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] 作为匹配项,并且插值给出 1.28,尽管完全在值为 2 的第二个 bin 中。

于 2015-08-16T10:32:27.263 回答
2

weightedcalcs软件包支持分位数

import weightedcalcs as wc
import pandas as pd

df = pd.DataFrame({'v': [1, 2, 3], 'w': [3, 2, 1]})
calc = wc.Calculator('w')  # w designates weight

calc.quantile(df, 'v', 0.5)
# 1.5
于 2019-12-12T03:42:04.883 回答
1

不幸的是,numpy 并没有为所有东西提供内置的加权函数,但是,你总是可以把一些东西放在一起。

def weight_array(ar, weights):
     zipped = zip(ar, weights)
     weighted = []
     for a, w in zipped:
         for j in range(w):
             weighted.append(a)
     return weighted


np.percentile(weight_array(ar, weights), 25)
于 2014-02-18T04:16:13.980 回答
0

这是我的解决方案:

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,然后用于估计加权百分位数。

于 2016-08-30T11:31:39.697 回答