将每个列表值映射到其相应的百分位数 [英] Map each list value to its corresponding percentile

问题描述

我想创建一个函数，该函数将一个(排序的)列表作为其参数，并输出一个包含每个元素的相应百分位数的列表.

I'd like to create a function that takes a (sorted) list as its argument and outputs a list containing each element's corresponding percentile.

例如，fn([1,2,3,4,17])返回[0.0, 0.25, 0.50, 0.75, 1.00].

任何人都可以:

1. 请帮助我更正下面的代码?或
2. 提供比我的代码更好的替代方法，用于将列表中的值映射到它们对应的百分位数?

我当前的代码:

def median(mylist):
    length = len(mylist)
    if not length % 2:
        return (mylist[length / 2] + mylist[length / 2 - 1]) / 2.0
    return mylist[length / 2]

###############################################################################
# PERCENTILE FUNCTION
###############################################################################

def percentile(x):
    """
    Find the correspoding percentile of each value relative to a list of values.
    where x is the list of values
    Input list should already be sorted!
    """

    # sort the input list
    # list_sorted = x.sort()

    # count the number of elements in the list
    list_elementCount = len(x)

    #obtain set of values from list

    listFromSetFromList = list(set(x))

    # count the number of unique elements in the list
    list_uniqueElementCount = len(set(x))

    # define extreme quantiles
    percentileZero    = min(x)
    percentileHundred = max(x)

    # define median quantile
    mdn = median(x) 

    # create empty list to hold percentiles
    x_percentile = [0.00] * list_elementCount 

    # initialize unique count
    uCount = 0

    for i in range(list_elementCount):
        if x[i] == percentileZero:
            x_percentile[i] = 0.00
        elif x[i] == percentileHundred:
            x_percentile[i] = 1.00
        elif x[i] == mdn:
            x_percentile[i] = 0.50 
        else:
            subList_elementCount = 0
            for j in range(i):
                if x[j] < x[i]:
                    subList_elementCount = subList_elementCount + 1 
            x_percentile[i] = float(subList_elementCount / list_elementCount)
            #x_percentile[i] = float(len(x[x > listFromSetFromList[uCount]]) / list_elementCount)
            if i == 0:
                continue
            else:
                if x[i] == x[i-1]:
                    continue
                else:
                    uCount = uCount + 1
    return x_percentile

当前，如果我提交percentile([1,2,3,4,17])，则返回列表[0.0, 0.0, 0.5, 0.0, 1.0].

Currently, if I submit percentile([1,2,3,4,17]), the list [0.0, 0.0, 0.5, 0.0, 1.0] is returned.

推荐答案

我认为您的示例输入/输出与计算百分位数的典型方式不符.如果将百分位数计算为数据点的比例严格小于此值"，则最高值应为0.8(因为5个值中的4个小于最大值).如果将其计算为数据点的百分比小于或等于此值"，则底值应为0.2(因为5个值中的1个等于最小值).因此，百分位数将为[0, 0.2, 0.4, 0.6, 0.8]或[0.2, 0.4, 0.6, 0.8, 1].您的定义似乎是数据点的数量严格小于此值，被视为不等于该值的数据点的数量的比例"，但是根据我的经验，这不是一个常见的定义(例如，参见维基百科).

I think your example input/output does not correspond to typical ways of calculating percentile. If you calculate the percentile as "proportion of data points strictly less than this value", then the top value should be 0.8 (since 4 of 5 values are less than the largest one). If you calculate it as "percent of data points less than or equal to this value", then the bottom value should be 0.2 (since 1 of 5 values equals the smallest one). Thus the percentiles would be [0, 0.2, 0.4, 0.6, 0.8] or [0.2, 0.4, 0.6, 0.8, 1]. Your definition seems to be "the number of data points strictly less than this value, considered as a proportion of the number of data points not equal to this value", but in my experience this is not a common definition (see for instance wikipedia).

使用典型的百分位数定义，数据点的百分位数等于其等级除以数据点的数量. (例如，在Stats SE上查看此问题，询问如何做同样的事情在R中.)如何计算百分位数与在计算排名(例如，对绑定值进行排名)上的差异有所不同. scipy.stats.percentileofscore函数提供了四种计算百分位数的方法:

With the typical percentile definitions, the percentile of a data point is equal to its rank divided by the number of data points. (See for instance this question on Stats SE asking how to do the same thing in R.) Differences in how to compute the percentile amount to differences in how to compute the rank (for instance, how to rank tied values). The scipy.stats.percentileofscore function provides four ways of computing percentiles:

>>> x = [1, 1, 2, 2, 17]
>>> [stats.percentileofscore(x, a, 'rank') for a in x]
[30.0, 30.0, 70.0, 70.0, 100.0]
>>> [stats.percentileofscore(x, a, 'weak') for a in x]
[40.0, 40.0, 80.0, 80.0, 100.0]
>>> [stats.percentileofscore(x, a, 'strict') for a in x]
[0.0, 0.0, 40.0, 40.0, 80.0]
>>> [stats.percentileofscore(x, a, 'mean') for a in x]
[20.0, 20.0, 60.0, 60.0, 90.0]

(我使用了包含联系的数据集来说明在这种情况下会发生什么.)

(I used a dataset containing ties to illustrate what happens in such cases.)

等级"方法为联系组分配的等级等于他们将覆盖的等级的平均值(即，第二名的三路并列获得3的等级，因为它占据"了等级2、3和4). 弱"方法根据小于或等于给定点的数据点的比例分配百分位数； 严格"是相同的，但计数点的比例严格小于给定点. 均值"方法是后两者的平均值.

The "rank" method assigns tied groups a rank equal to the average of the ranks they would cover (i.e., a three-way tie for 2nd place gets a rank of 3 because it "takes up" ranks 2, 3 and 4). The "weak" method assigns a percentile based on the proportion of data points less than or equal to a given point; "strict" is the same but counts proportion of points strictly less than the given point. The "mean" method is the average of the latter two.

正如Kevin H. Lin所指出的，在循环中调用percentileofscore效率低下，因为它必须在每次通过时重新计算排名.但是，可以使用 scipy.stats.rankdata ，让您一次计算所有百分比:

As Kevin H. Lin noted, calling percentileofscore in a loop is inefficient since it has to recompute the ranks on every pass. However, these percentile calculations can be easily replicated using different ranking methods provided by scipy.stats.rankdata, letting you calculate all the percentiles at once:

>>> from scipy import stats
>>> stats.rankdata(x, "average")/len(x)
array([ 0.3,  0.3,  0.7,  0.7,  1. ])
>>> stats.rankdata(x, 'max')/len(x)
array([ 0.4,  0.4,  0.8,  0.8,  1. ])
>>> (stats.rankdata(x, 'min')-1)/len(x)
array([ 0. ,  0. ,  0.4,  0.4,  0.8])

在最后一种情况下，将等级从1降低到1，而不是从1开始降低.(我省略了均值"，但是可以通过对后两种方法的结果求平均值来轻松获得.)

In the last case the ranks are adjusted down by one to make them start from 0 instead of 1. (I've omitted "mean", but it could easily be obtained by averaging the results of the latter two methods.)

我做了一些时间安排.对于像您的示例这样的小数据，使用rankdata的速度要比Kevin H. Lin的解决方案要慢一些(大概是由于scipy在将事物转换为底层的numpy数组时会产生开销)，但比在c中调用percentileofscore更快.如reptilicus的回答中那样循环:

I did some timings. With small data such as that in your example, using rankdata is somewhat slower than Kevin H. Lin's solution (presumably due to the overhead scipy incurs in converting things to numpy arrays under the hood) but faster than calling percentileofscore in a loop as in reptilicus's answer:

In [11]: %timeit [stats.percentileofscore(x, i) for i in x]
1000 loops, best of 3: 414 µs per loop

In [12]: %timeit list_to_percentiles(x)
100000 loops, best of 3: 11.1 µs per loop

In [13]: %timeit stats.rankdata(x, "average")/len(x)
10000 loops, best of 3: 39.3 µs per loop

但是，对于大型数据集，numpy的性能优势生效，使用rankdata的速度比Kevin的list_to_percentiles快10倍:

With a large dataset, however, the performance advantage of numpy takes effect and using rankdata is 10 times faster than Kevin's list_to_percentiles:

In [18]: x = np.random.randint(0, 10000, 1000)

In [19]: %timeit [stats.percentileofscore(x, i) for i in x]
1 loops, best of 3: 437 ms per loop

In [20]: %timeit list_to_percentiles(x)
100 loops, best of 3: 1.08 ms per loop

In [21]: %timeit stats.rankdata(x, "average")/len(x)
10000 loops, best of 3: 102 µs per loop

这种优势只会在越来越大的数据集上变得更加明显.

This advantage will only become more pronounced on larger and larger datasets.

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