如何与邻国值的数据帧替换NA(遗漏值)
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12-09-2019 - |
题
862 2006-05-19 6.241603 5.774208
863 2006-05-20 NA NA
864 2006-05-21 NA NA
865 2006-05-22 6.383929 5.906426
866 2006-05-23 6.782068 6.268758
867 2006-05-24 6.534616 6.013767
868 2006-05-25 6.370312 5.856366
869 2006-05-26 6.225175 5.781617
870 2006-05-27 NA NA
我有一个数据帧x像上面一些NA,这是我要填写使用相邻非NA值像2006-05-20这将是平均的19 22
如何做到这一点是这个问题?
解决方案
正确格式化您的数据看起来像这样
862 2006-05-19 6.241603 5.774208
863 2006-05-20 NA NA
864 2006-05-21 NA NA
865 2006-05-22 6.383929 5.906426
866 2006-05-23 6.782068 6.268758
867 2006-05-24 6.534616 6.013767
868 2006-05-25 6.370312 5.856366
869 2006-05-26 6.225175 5.781617
870 2006-05-27 NA NA
和是时间序列的性质。因此,我将加载到类zoo
的对象(从的动物园强> 一>封装)的,允许您挑选了一些战略 - 见下文。哪一个你选择取决于您的数据和应用程序的性质。一般来说,“辩别丢失数据去”的字段被称为数据插补
并有一个相当大的文献。
R> x <- zoo(X[,3:4], order.by=as.Date(X[,2]))
R> x
x y
2006-05-19 6.242 5.774
2006-05-20 NA NA
2006-05-21 NA NA
2006-05-22 6.384 5.906
2006-05-23 6.782 6.269
2006-05-24 6.535 6.014
2006-05-25 6.370 5.856
2006-05-26 6.225 5.782
2006-05-27 NA NA
R> na.locf(x) # last observation carried forward
x y
2006-05-19 6.242 5.774
2006-05-20 6.242 5.774
2006-05-21 6.242 5.774
2006-05-22 6.384 5.906
2006-05-23 6.782 6.269
2006-05-24 6.535 6.014
2006-05-25 6.370 5.856
2006-05-26 6.225 5.782
2006-05-27 6.225 5.782
R> na.approx(x) # approximation based on before/after values
x y
2006-05-19 6.242 5.774
2006-05-20 6.289 5.818
2006-05-21 6.336 5.862
2006-05-22 6.384 5.906
2006-05-23 6.782 6.269
2006-05-24 6.535 6.014
2006-05-25 6.370 5.856
2006-05-26 6.225 5.782
R> na.spline(x) # spline fit ...
x y
2006-05-19 6.242 5.774
2006-05-20 5.585 5.159
2006-05-21 5.797 5.358
2006-05-22 6.384 5.906
2006-05-23 6.782 6.269
2006-05-24 6.535 6.014
2006-05-25 6.370 5.856
2006-05-26 6.225 5.782
2006-05-27 5.973 5.716
R>
其他提示
根据数据tidyr::fill()
可能是一个选项:
library(tidyverse)
df %>% fill(x) # single column x
df %>% fill(x, y) # multiple columns, x and y
df %>% fill(x, .direction = 'up') # filling from the bottom up rather than top down
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