如何对日期时间和浮点数进行插值 [英] How to do interpolation on datetime and float

问题描述

我正在使用 scipy 在时间序列上.我的x轴数据为日期时间格式，y轴为浮点型，如:

I am doing 1d interpolation using scipy on time-series. My x-axis data is in datetime format and y axis is in float like:

3/15/2012 16:00:00  32.94
3/16/2012 16:00:00  32.95
3/19/2012 16:00:00  32.61

现在在计算坡度slope = (y_hi-y_lo) / (x_hi-x_lo)时出现错误TypeError: unsupported operand type(s) for /: 'float' and 'datetime.timedelta'，这是一个明显的错误.有人可以指出我正确的方向，如何处理吗?

Now during slope calculation slope = (y_hi-y_lo) / (x_hi-x_lo) i am getting the error TypeError: unsupported operand type(s) for /: 'float' and 'datetime.timedelta' which is an obvious error. Can someone point me toward the right direction, How to handle it ?

推荐答案

您的问题是，您试图将浮点数除以datetime.timedelta对象，正如您所说的那样，该对象显然会引发TypeError.

Your issue is that you are trying to divide a float by a datetime.timedelta object which is, as you said, obviously throwing a TypeError.

您可以使用

You can convert datetime.timedelta objects to a float representing the total number of seconds within that timedelta using the datetime.timedelta.total_seconds() instance method.

在这种情况下，您可以将代码修改为:

In that case you would modify your code to something like:

slope_numerator = y_hi - y_lo
slope_denominator = (x_hi - x_lo).total_seconds()
slope = slope_numerator / slope_denominator

请注意，这将使您以秒为单位倾斜.您可以修改分母以使其符合小时，天数等条件，以适应您的目的.

Note that this will give you a slope in terms of seconds. You could modify the denominator to give it in terms of hours, days, etc to suit your purposes.

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