python – 是否有可能在给定百分位值而不是原始输入的情况下绘制matplotlib boxplot？

从我所看到的,boxplot()方法需要一系列原始值(数字)作为输入,然后从中计算百分位数以绘制箱线图.

我想有一种方法可以通过它传递百分位数并获得相应的箱线图.

例如：

假设我已经运行了几个基准测试,并且对于每个基准测试我都测量了延迟(浮点值).另外,我已经预先计算了这些值的百分位数.

因此,对于每个基准测试,我有第25,第50,第75百分位数以及最小值和最大值.

现在给出这些数据,我想绘制基准的方框图.

解决方法:

为了使用百分位数值和异常值(如果有的话)绘制方框图,我创建了一个customize_box_plot函数,该函数基本上修改了基本框图(从微小的样本数据生成)中的属性,以使其符合您的百分位值.

customized_box_plot函数

def customized_box_plot(percentiles, axes, redraw = True, *args, **kwargs):
    """
    Generates a customized boxplot based on the given percentile values
    """

    box_plot = axes.boxplot([[-9, -4, 2, 4, 9],]*n_box, *args, **kwargs) 
    # Creates len(percentiles) no of box plots

    min_y, max_y = float('inf'), -float('inf')

    for box_no, (q1_start, 
                 q2_start,
                 q3_start,
                 q4_start,
                 q4_end,
                 fliers_xy) in enumerate(percentiles):

        # Lower cap
        box_plot['caps'][2*box_no].set_ydata([q1_start, q1_start])
        # xdata is determined by the width of the box plot

        # Lower whiskers
        box_plot['whiskers'][2*box_no].set_ydata([q1_start, q2_start])

        # Higher cap
        box_plot['caps'][2*box_no + 1].set_ydata([q4_end, q4_end])

        # Higher whiskers
        box_plot['whiskers'][2*box_no + 1].set_ydata([q4_start, q4_end])

        # Box
        box_plot['boxes'][box_no].set_ydata([q2_start, 
                                             q2_start, 
                                             q4_start,
                                             q4_start,
                                             q2_start])

        # Median
        box_plot['medians'][box_no].set_ydata([q3_start, q3_start])

        # Outliers
        if fliers_xy is not None and len(fliers_xy[0]) != 0:
            # If outliers exist
            box_plot['fliers'][box_no].set(xdata = fliers_xy[0],
                                           ydata = fliers_xy[1])

            min_y = min(q1_start, min_y, fliers_xy[1].min())
            max_y = max(q4_end, max_y, fliers_xy[1].max())

        else:
            min_y = min(q1_start, min_y)
            max_y = max(q4_end, max_y)

        # The y axis is rescaled to fit the new box plot completely with 10% 
        # of the maximum value at both ends
        axes.set_ylim([min_y*1.1, max_y*1.1])

    # If redraw is set to true, the canvas is updated.
    if redraw:
        ax.figure.canvas.draw()

    return box_plot

用法

使用逆逻辑(最后的代码)我从这个example中提取了百分位数值

>>> percentiles
(-1.0597368367634488, 0.3977683984966961, 1.0298955252405229, 1.6693981537742526, 3.4951447843464449)
(-0.90494930553559483, 0.36916539612108634, 1.0303658700697103, 1.6874542731392828, 3.4951447843464449)
(0.13744105279440233, 1.3300645202649739, 2.6131540656339483, 4.8763411136047647, 9.5751914834437937)
(0.22786243898199182, 1.4120860286080519, 2.637650402506837, 4.9067126578493259, 9.4660357513550899)
(0.0064696168078617741, 0.30586770128093388, 0.70774153557312702, 1.5241965711101928, 3.3092932063051976)
(0.007009744579241136, 0.28627373934008982, 0.66039691869500572, 1.4772725266672091, 3.221716765477217)
(-2.2621660374110544, 5.1901313713883352, 7.7178532139979357, 11.277744848353247, 20.155971739152388)
(-2.2621660374110544, 5.1884411864079532, 7.3357079047721054, 10.792299385806913, 18.842012119715388)
(2.5417888074435702, 5.885996170695587, 7.7271286220368598, 8.9207423361593179, 10.846938621419374)
(2.5971767318505856, 5.753551925927133, 7.6569980004033464, 8.8161056254143233, 10.846938621419374)

请注意,为了保持这个简短,我没有显示异常值向量,它将是每个百分位数组的第6个元素.

另请注意,可以使用所有常用的额外kwargs / args,因为它们只是传递给它内部的boxplot方法：

>>> fig, ax = plt.subplots()
>>> b = customized_box_plot(percentiles, ax, redraw=True, notch=0, sym='+', vert=1, whis=1.5)
>>> plt.show()

说明

boxplot方法返回一个字典,将boxplot的组件映射到创建的各个matplotlib.lines.Line2D实例.

引用matplotlib.pyplot.boxplot文档：

That dictionary has the following keys (assuming vertical boxplots):

boxes: the main body of the boxplot showing the quartiles and the median’s confidence intervals if enabled.

medians: horizonal lines at the median of each box.

whiskers: the vertical lines extending to the most extreme, n-outlier data points. caps: the horizontal lines at the ends of the whiskers.

fliers: points representing data that extend beyond the whiskers (outliers).

means: points or lines representing the means.

例如,观察[-9,-4,2,4,9]的微小样本数据的箱线图

>>> b = ax.boxplot([[-9, -4, 2, 4, 9],])
>>> b
{'boxes': [<matplotlib.lines.Line2D at 0x7fe1f5b21350>],
'caps': [<matplotlib.lines.Line2D at 0x7fe1f54d4e50>,
<matplotlib.lines.Line2D at 0x7fe1f54d0e50>],
'fliers': [<matplotlib.lines.Line2D at 0x7fe1f5b317d0>],
'means': [],
'medians': [<matplotlib.lines.Line2D at 0x7fe1f63549d0>],
'whiskers': [<matplotlib.lines.Line2D at 0x7fe1f5b22e10>,
             <matplotlib.lines.Line2D at 0x7fe20c54a510>]} 

>>> plt.show()

matplotlib.lines.Line2D对象有两种方法,我将在我的函数中广泛使用. set_xdata(或set_ydata)和get_xdata(或get_ydata).

使用这些方法,我们可以改变基本框图的构成线的位置,以符合您的百分位数值(这是customize_box_plot函数所做的).更改构成线的位置后,您可以使用figure.canvas.draw()重绘画布

总结从百分位数到各种Line2D对象坐标的映射.

Y坐标：

> max(q4_end – 第四四分位数的末尾)对应于最顶端的Line2D对象.
> min(q1_start – 第一个四分位数的开头)对应最下面的最大Cap2D对象.
>中位数对应于(q3_start)中位数Line2D对象.
>两个胡须位于盒子末端和极端盖子之间(q1_start和q2_start – 较低的胡须; q4_start和q4_end – 上胡须)
>盒子实际上是一个有趣的n形线,下半部分有一个帽子. n形线的极值对应于q2_start和q4_start.

X坐标：

>中心x坐标(对于多个箱形图通常为1,2,3 ……)
>库会根据指定的宽度自动计算边界x坐标.

从箱形图DICT中检索PERCENTILES的反函数：

def get_percentiles_from_box_plots(bp):
    percentiles = []
    for i in range(len(bp['boxes'])):
        percentiles.append((bp['caps'][2*i].get_ydata()[0],
                           bp['boxes'][i].get_ydata()[0],
                           bp['medians'][i].get_ydata()[0],
                           bp['boxes'][i].get_ydata()[2],
                           bp['caps'][2*i + 1].get_ydata()[0],
                           (bp['fliers'][i].get_xdata(),
                            bp['fliers'][i].get_ydata())))
    return percentiles

注意：
我没有制作完全自定义的boxplot方法的原因是因为,内置的盒子图中提供的许多功能无法完全复制.

如果我可能不必要地解释一些可能太明显的东西,也请原谅.

标签：python,python-2-7,matplotlib,percentile,boxplot
来源： https://codeday.me/bug/20191006/1857961.html