如何在时间序列线图上绘制回归线 [英] How to plot a regression line on a timeseries line plot

问题描述

我有一个关于斜率值的问题，我计算如下:

将pandas导入为pd将 yfinance 导入为 yf导入 matplotlib.pyplot 作为 plt将日期时间导入为 dt将 numpy 导入为 npdf = yf.download('aapl', '2015-01-01', '2021-01-01')df.rename(columns = {'Adj Close' : 'Adj_close'}, inplace= True)x1 = pd.Timestamp('2019-01-02')x2 = df.index[-1]y1 = df[df.index == x1].Adj_close[0]y2 = df[df.index == x2].Adj_close[0]斜率 = (y2 - y1)/(x2 - x1). 天角度 = 圆(np.rad2deg(np.arctan2(y2 - y1，(x2 - x1))，1)图, ax1 = plt.subplots(figsize= (15, 6))ax1.grid(True, linestyle=':')ax1.set_zorder(1)ax1.set_frame_on(假)ax1.plot(df.index, df.Adj_close, c='k', lw= 0.8)ax1.plot([x1, x2], [y1, y2], c='k')ax1.set_xlim(df.index[0], df.index[-1])plt.show()

它返回坡度的角度值为 7.3 度.看图表，哪个看起来不真实:

看起来接近 45 度.这里有什么问题?

这是我需要计算角度的直线:

解决方案

• OP 中的实现不是确定或绘制线性模型的正确方法.因此，绕过了确定绘制直线的角度的问题，并展示了一种更严格的绘制回归线的方法.
• 可以通过将日期时间日期转换为序数来添加回归线. 可以使用 sklearn 计算模型，或使用 seaborn 添加到图中.regplot，如下所示.
• 使用

  计算线性模型

  • 使用

    斜坡的角度

    • 这是axes 方面的一个工件，对于x 和y 来说是不相等的.当坡向相等时，看到坡度为 7.0 度.

    x = x2 - x1y = y2[0][0] - y1[0][0]斜率 = y/x打印(轮(斜率，7)==轮(model.coef_[0][0]，7))[出去]:真的角度 = 圆(np.rad2deg(np.arctan2(y，x))，1)打印(角度)[出去]:7.0# 给定现有图ax1 = df.plot(y='Adj Close', c='k', figsize=(15, 6), grid=True,legend=False,title='根据 2019 年 1 月 2 日的回归线调整收盘价')ax1.plot([x1, x2], [y1[0][0], y2[0][0]], label='线性模型', c='magenta')# 使方面相等ax1.set_aspect('equal',adjustable='box')

    I have a question about the value of the slope in degrees which I have calculated below:

    import pandas as pd
    import yfinance as yf
    import matplotlib.pyplot as plt
    import datetime as dt
    import numpy as np
    
    df = yf.download('aapl', '2015-01-01', '2021-01-01')
    df.rename(columns = {'Adj Close' : 'Adj_close'}, inplace= True)
    
    x1 = pd.Timestamp('2019-01-02')
    x2 = df.index[-1]
    y1 = df[df.index == x1].Adj_close[0]
    y2 = df[df.index == x2].Adj_close[0]
    
    slope = (y2 - y1)/ (x2 - x1).days
    angle = round(np.rad2deg(np.arctan2(y2 - y1, (x2 - x1).days)), 1)
    
    fig, ax1 = plt.subplots(figsize= (15, 6))
    ax1.grid(True, linestyle= ':')
    ax1.set_zorder(1)
    ax1.set_frame_on(False)
    ax1.plot(df.index, df.Adj_close, c= 'k', lw= 0.8)
    ax1.plot([x1, x2], [y1, y2], c= 'k')
    
    
    ax1.set_xlim(df.index[0], df.index[-1])
    plt.show()
    

    It returns the value of the angle of the slope as 7.3 degrees. Which doesnt look true looking at the chart:

    It looks close to 45 degrees. What is wrong here?

    Here is the line for which I need to calculate the angle:

    解决方案

    • The implementation in the OP is not the correct way to determine, or plot a linear model. As such, the question about determining the angle to plot the line is bypassed, and a more rigorous approach to plotting the regression line is shown.
    • A regression line can be added by converting the datetime dates to ordinal. The model can be calculated with sklearn, or added to the plot with seaborn.regplot, as show below.
    • Plot the full data with pandas.DataFrame.plot
    • Tested in python 3.8.11, pandas 1.3.2, matplotlib 3.4.3, seaborn 0.11.2, sklearn 0.24.2

    Imports and Data

    import yfinance as yf
    import pandas as pd
    import seaborn as sns
    import matplotlib.pyplot as plt
    import numpy as np
    from sklearn.linear_model import LinearRegression
    
    # download the data
    df = yf.download('aapl', '2015-01-01', '2021-01-01')
    
    # convert the datetime index to ordinal values, which can be used to plot a regression line
    df.index = df.index.map(pd.Timestamp.toordinal)
    
    # display(df.iloc[:5, [4]])
            Adj Close
    Date             
    735600  24.782110
    735603  24.083958
    735604  24.086227
    735605  24.423975
    735606  25.362394
    
    # convert the regression line start date to ordinal
    x1 = pd.to_datetime('2019-01-02').toordinal()
    
    # data slice for the regression line
    data=df.loc[x1:].reset_index()
    

    Plot a Regression Line with seaborn

    • Using seaborn.regplot no calculations are required to add the regression line to the line plot of the data.
    • Convert the x-axis labels to datetime format
    • Play around with the xticks and labels if you need the endpoints adjusted.

    # plot the Adj Close data
    ax1 = df.plot(y='Adj Close', c='k', figsize=(15, 6), grid=True, legend=False,
                  title='Adjusted Close with Regression Line from 2019-01-02')
    
    # add a regression line
    sns.regplot(data=data, x='Date', y='Adj Close', ax=ax1, color='magenta', scatter_kws={'s': 7}, label='Linear Model', scatter=False)
    
    ax1.set_xlim(df.index[0], df.index[-1])
    
    # convert the axis back to datetime
    xticks = ax1.get_xticks()
    labels = [pd.Timestamp.fromordinal(int(label)).date() for label in xticks]
    ax1.set_xticks(xticks)
    ax1.set_xticklabels(labels)
    
    ax1.legend()
    
    plt.show()
    

    Calculate the Linear Model

    • Use sklearn.linear_model.LinearRegression to calculate any desired points from the linear model, and then plot the corresponding line with matplotlib.pyplot.plot
    • In regards to your other question, Extending the trendline of a stock chart to the right, you would calculate the model over a specified range, and then extend the line by predicting y1 and y2, given x1 and x2.
    • This answer shows how to convert the ordinal axis values back to a date format.

    # create the model
    model = LinearRegression()
    
    # extract x and y from dataframe data
    x = data[['Date']]
    y = data[['Adj Close']]
    
    # fit the mode
    model.fit(x, y)
    
    # print the slope and intercept if desired
    print('intercept:', model.intercept_)
    print('slope:', model.coef_)
    
    intercept: [-90078.45713565]
    slope: [[0.1222514]]
    
    # calculate y1, given x1
    y1 = model.predict(np.array([[x1]]))
    
    print(y1)
    array([[28.27904095]])
    
    # calculate y2, given the last date in data
    x2 = data.Date.iloc[-1]
    y2 = model.predict(np.array([[x2]]))
    
    print(y2)
    array([[117.40030862]])
    
    # this can be added to `ax1` with
    ax1 = df.plot(y='Adj Close', c='k', figsize=(15, 6), grid=True, legend=False,
                  title='Adjusted Close with Regression Line from 2019-01-02')
    ax1.plot([x1, x2], [y1[0][0], y2[0][0]], label='Linear Model', c='magenta')
    ax1.legend()
    

    Angle of the Slope

    • This is an artifact of the aspect of the axes, which is not equal for x and y. When the aspect is equal, see that the slope is 7.0 deg.

    x = x2 - x1
    y = y2[0][0] - y1[0][0]
    slope = y / x
    
    print(round(slope, 7) == round(model.coef_[0][0], 7))
    [out]:
    True
    
    angle = round(np.rad2deg(np.arctan2(y, x)), 1)
    print(angle)
    [out]:
    7.0
    
    # given the existing plot
    ax1 = df.plot(y='Adj Close', c='k', figsize=(15, 6), grid=True, legend=False,
                  title='Adjusted Close with Regression Line from 2019-01-02')
    ax1.plot([x1, x2], [y1[0][0], y2[0][0]], label='Linear Model', c='magenta')
    
    # make the aspect equal
    ax1.set_aspect('equal', adjustable='box')
    

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