在pandas中对DataFrame进行插值

Question

我有一个DataFrame,比如一个波动率表面,索引为时间,列为罢工.我该如何进行二维插值？我可以,reindex但我该如何处理NaN？我知道我们可以,fillna(method='pad')但它甚至不是线性插值.有没有办法可以插入我们自己的插值方法？

Answer 1

您可以使用DataFrame.interpolate线性插值.

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

对于任何更复杂的事情,您需要推出自己的函数来处理Series对象并根据需要填充NaN值并返回另一个Series对象.

Answer 2

旧线程,但我想我会用2d外推/插值分享我的解决方案,尊重索引值,这也适用于需求.代码结束有点奇怪,所以让我知道是否有更好的解决方案:

import pandas
from   numpy import nan
import numpy

dataGrid = pandas.DataFrame({1: {1: 1, 3: 2},
                             2: {1: 3, 3: 4}})


def getExtrapolatedInterpolatedValue(x, y):
    global dataGrid
    if x not in dataGrid.index:
        dataGrid.ix[x] = nan
        dataGrid = dataGrid.sort()
        dataGrid = dataGrid.interpolate(method='index', axis=0).ffill(axis=0).bfill(axis=0)

    if y not in dataGrid.columns.values:
        dataGrid = dataGrid.reindex(columns=numpy.append(dataGrid.columns.values, y))
        dataGrid = dataGrid.sort_index(axis=1)
        dataGrid = dataGrid.interpolate(method='index', axis=1).ffill(axis=1).bfill(axis=1)

    return dataGrid[y][x]


print getExtrapolatedInterpolatedValue(2, 1.4)
>>2.3