问题
Consider a fragmented dataset like this:
ID Date Value
1 1 2012-01-01 5065
4 1 2012-01-04 1508
5 1 2012-01-05 9489
6 1 2012-01-06 7613
7 2 2012-01-07 6896
8 2 2012-01-08 2643
11 3 2012-01-02 7294
12 3 2012-01-03 8726
13 3 2012-01-04 6262
14 3 2012-01-05 2999
15 3 2012-01-06 10000
16 3 2012-01-07 1405
18 3 2012-01-09 8372
Notice that observations are missing for (2,3,9,10,17). What I would like, is to fill out some of these gaps in the dataset with "Value" = 0, like so:
ID Date Value
1 1 2012-01-01 5920
2 1 2012-01-02 0
3 1 2012-01-03 0
4 1 2012-01-04 8377
5 1 2012-01-05 7810
6 1 2012-01-06 6452
7 2 2012-01-07 3483
8 2 2012-01-08 5426
9 2 2012-01-09 0
11 3 2012-01-02 7854
12 3 2012-01-03 1948
13 3 2012-01-04 7141
14 3 2012-01-05 5402
15 3 2012-01-06 6412
16 3 2012-01-07 7043
17 3 2012-01-08 0
18 3 2012-01-09 3270
The point is that the zeros only should be inserted if there is a past observation for the same (grouped) ID. I would like to avoid any loops, as the full dataset is quite large.
Any suggestions? To reproduce the dataframe:
df <- data.frame(matrix(0, nrow = 18, ncol = 3,
dimnames = list(NULL, c("ID","Date","Value"))) )
df[,1] = c(1,1,1,1,1,1,2,2,2,3,3,3,3,3,3,3,3,3)
df[,2] = seq(as.Date("2012-01-01"),
as.Date("2012-01-9"),
by=1)
df[,3] = sample(1000:10000,18,replace=T)
df = df[-c(2,3,9,10,17),]
回答1:
There are already some solid answers here, but I would recommend checking out the package padr
.
library(dplyr)
library(padr)
df %>%
pad(start_val = as.Date("2012-01-01"),
end_val = as.Date("2012-01-09"),
group = "ID") %>%
fill_by_value(Value)
The package gives some pretty intuitive functions for summarizing Date columns as well.
回答2:
Tidyverse has complete
which is a nice easy way to expand something like this. We can also use the fill
argument to replace the NAs
with zero in the same step.
library(tidyverse)
df %>% group_by(ID) %>%
complete(Date = seq(min(Date), max(Date), "day"), fill = list(Value = 0))
# A tibble: 16 x 3
# Groups: ID [3]
ID Date Value
<dbl> <date> <dbl>
1 1 2012-01-01 1047
2 1 2012-01-02 0
3 1 2012-01-03 0
4 1 2012-01-04 8147
5 1 2012-01-05 1359
6 1 2012-01-06 1892
7 2 2012-01-07 3362
8 2 2012-01-08 8988
9 3 2012-01-02 2731
10 3 2012-01-03 9794
...
回答3:
The following is a base R solution. It uses split
to divide the input into sub-dataframes and then lapply
to process each of them.
result <- lapply(split(df, df$ID), function(DF){
Date <- seq(min(DF$Date), max(DF$Date), by = "days")
DF2 <- data.frame(ID = rep(DF$ID[1], length.out = length(Date)))
DF2$Date <- Date
DF2$Value <- 0
DF2$Value[Date %in% DF$Date] <- DF$Value
DF2
})
result <- do.call(rbind, result)
row.names(result) <- NULL
result
来源:https://stackoverflow.com/questions/53729693/insert-rows-with-zeros-in-data-frames-in-r