我有一个包含NA值的大数据集。样本数据如下。
Data <- data.frame(col_1 = c('A','A','A','A', 'A', 'A', 'A', 'B', 'B', 'B'), col_2 = c('C','C', 'C', 'D', 'D','D', 'D', 'E', 'E', 'E'), col_3 = c(10,15,20, 10,20,25,30,5,10,15), value = c(0.9, NA, 0.6, 0.9, NA, NA,0.4, 0.8,NA,0.4))我想用线性插值来填充这些NAs。例如,在NA中填充col_1 =‘A’和col_2 =‘C’
value = 0.9 + (0.6-0.9)*(15-10)/(20-10) = 0.75对于第二个NA,col_1 =‘A’和col_2 =‘D’
value = 0.9 + (0.4-0.9)*(25-10)/(30-10) = 0.53既然我的数据很大,有没有一种有效的方法去做呢?谢谢。预期的结果是。
Data_Updated <- data.frame(col_1 = c('A','A','A','A', 'A', 'A', 'A', 'B', 'B', 'B'), col_2 = c('C','C', 'C', 'D', 'D','D', 'D', 'E', 'E', 'E'), col_3 = c(10,15,20, 10,20,25,30,5,10,15), value = c(0.9, 0.75, 0.6, 0.9, 0.65, 0.53,0.4, 0.8,0.6,0.4))发布于 2020-11-23 15:30:22
如果这足够快的话,试试:
library(data.table)
library(zoo)
setDT(Data)
Data[, value1 := na.approx(value, x = col_3), by = .(col_1, col_2)]
# col_1 col_2 col_3 value value1
# 1: A C 10 0.9 0.900
# 2: A C 15 NA 0.750
# 3: A C 20 0.6 0.600
# 4: A D 10 0.9 0.900
# 5: A D 20 NA 0.650
# 6: A D 25 NA 0.525
# 7: A D 30 0.4 0.400
# 8: B E 5 0.8 0.800
# 9: B E 10 NA 0.600
#10: B E 15 0.4 0.400发布于 2020-11-23 22:05:45
带有dplyr的选项
library(dplyr)
library(zoo)
Data %>%
group_by(col_1, col_2) %>%
mutate(value1 = na.approx(value, x = col_3))https://stackoverflow.com/questions/64971101
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