在 pandas 中对DataFrame进行插值 [英] Interpolation on DataFrame in pandas

问题描述

我有一个DataFrame，比如说一个波动性表面，其索引为时间，而列为行权.如何进行二维插值?我可以reindex，但是如何处理NaN?我知道我们可以fillna(method='pad')，但它甚至不是线性插值.有没有办法插入我们自己的方法进行插值?

I have a DataFrame, say a volatility surface with index as time and column as strike. How do I do two dimensional interpolation? I can reindex but how do i deal with NaN? I know we can fillna(method='pad') but it is not even linear interpolation. Is there a way we can plug in our own method to do interpolation?

推荐答案

您可以使用DataFrame.interpolate获得线性插值.

You can use DataFrame.interpolate to get a linear interpolation.

In : df = pandas.DataFrame(numpy.random.randn(5,3), index=['a','c','d','e','g'])

In : df
Out:
          0         1         2
a -1.987879 -2.028572  0.024493
c  2.092605 -1.429537  0.204811
d  0.767215  1.077814  0.565666
e -1.027733  1.330702 -0.490780
g -1.632493  0.938456  0.492695

In : df2 = df.reindex(['a','b','c','d','e','f','g'])

In : df2
Out:
          0         1         2
a -1.987879 -2.028572  0.024493
b       NaN       NaN       NaN
c  2.092605 -1.429537  0.204811
d  0.767215  1.077814  0.565666
e -1.027733  1.330702 -0.490780
f       NaN       NaN       NaN
g -1.632493  0.938456  0.492695

In : df2.interpolate()
Out:
          0         1         2
a -1.987879 -2.028572  0.024493
b  0.052363 -1.729055  0.114652
c  2.092605 -1.429537  0.204811
d  0.767215  1.077814  0.565666
e -1.027733  1.330702 -0.490780
f -1.330113  1.134579  0.000958
g -1.632493  0.938456  0.492695

对于更复杂的事情，您需要推出自己的函数来处理Series对象，并根据需要填充NaN值，然后返回另一个Series对象.

For anything more complex, you need to roll-out your own function that will deal with a Series object and fill NaN values as you like and return another Series object.

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