如何将其转换回日期时间格式？

Question

我想在日期执行算术运算，所以我转换了这些日期

idx_1 = 2017-06-07 00:00:00
idx_2 = 2017-07-27 00:00:00 

使用浮动，

x1 = time.mktime(idx_1.timetuple())  # returns float of dates
>>> 1496773800.0
x2 = time.mktime(idx_2.timetuple())
>>> 1501093800.0
y1 = 155.98
y2 = 147.07

使用以下代码绘制：

    import datetime as dt
    import time
    import numpy as np
    import matplotlib.pyplot as plt
    x = [x1, x2]
    y = [y1, y2]
    Difference = x2 - x1 #this helps to end the plotted line at specific point
    coefficients = np.polyfit(x, y, 1)
    polynomial = np.poly1d(coefficients)
    # the np.linspace lets you set number of data points, line length.
    x_axis = np.linspace(x1, x2 + Difference, 3)  # linspace(start, end, num)
    y_axis = polynomial(x_axis)
    plt.plot(x_axis, y_axis)
    plt.plot(x[0], y[0], 'go')
    plt.plot(x[1], y[1], 'go')
    plt.show()

哪个情节：

如何让matplotlib在x轴而不是浮点上绘制实际日期？

任何形式的帮助都会受到极大的赞赏。

Answer 1

从datetime对象开始，您可以使用matplotlib的date2num和num2date函数来转换数值和从数值转换。优点是matplotlib.dates定位器和格式化程序可以理解数值数据。

import datetime
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates

idx_1 = datetime.datetime(2017,06,07,0,0,0)
idx_2 = datetime.datetime(2017,07,27,0,0,0)

idx = [idx_1, idx_2]

y1 = 155.98
y2 = 147.07

x = matplotlib.dates.date2num(idx)
y = [y1, y2]
Difference = x[1] - x[0] #this helps to end the plotted line at specific point
coefficients = np.polyfit(x, y, 1)
polynomial = np.poly1d(coefficients)
# the np.linspace lets you set number of data points, line length.
x_axis = np.linspace(x[0], x[1] + Difference, 3)  # linspace(start, end, num)
y_axis = polynomial(x_axis)
plt.plot(x_axis, y_axis)
plt.plot(x[0], y[0], 'go')
plt.plot(x[1], y[1], 'go')

loc= matplotlib.dates.AutoDateLocator()
plt.gca().xaxis.set_major_locator(loc)
plt.gca().xaxis.set_major_formatter(matplotlib.dates.AutoDateFormatter(loc))
plt.gcf().autofmt_xdate()
plt.show()