将股票图表的趋势线向右延伸

Question

我想将我绘制的这条趋势线延伸到右侧。 到目前为止的代码如下：

import matplotlib.pyplot as plt
import yfinance as yf

df = yf.download('aapl', '2020-01-01', '2021-01-01')

df1 = df[df.index > '2020-06-01']
df2 = df[df.index < '2020-06-02']
lowest2 = df1[df1.Close == df1.Close.min()]
lowest1 = df2[df2.Close == df2.Close.min()]

fig, ax = plt.subplots(figsize= (10, 6))
ax.plot(df.index, df.Close)
ax.plot([lowest1.index[0], lowest2.index[0]], [lowest1.Low[0], lowest2.Low[0]])
ax.set_xlim(df.index[0], df.index[-1])
plt.show()

这是输出图像：

如果有人能指出我正确的方向，我将不胜感激。

Answer 1

这是结合使用时间戳的非常简单的数学：

import matplotlib.pyplot as plt
import yfinance as yf

df = yf.download('aapl', '2020-01-01', '2021-01-01')

df1 = df[df.index > '2020-06-01']
df2 = df[df.index < '2020-06-02']
lowest2 = df1[df1.Close == df1.Close.min()]
lowest1 = df2[df2.Close == df2.Close.min()]

fig, ax = plt.subplots(figsize= (10, 6))
ax.plot(df.index, df.Close)

## calculating slope
x1, x2, y1, y2 = lowest1.index[0], lowest2.index[0], lowest1.Low[0], lowest2.Low[0]
slope = (y2-y1)/(x2-x1).total_seconds()

## extending line
x3 = pd.Timestamp('2020-10-01')
y3 = y1+(x3-x1).total_seconds()*slope

## plotting lines
ax.plot([x1, x2], [y1, y2])
ax.plot([x1, x3], [y1, y3], ls=':')

ax.set_xlim(df.index[0], df.index[-1])

输出：