I have a quite basic R question. I have the following data frame with a year column that has no 1-year steps.
year <- c(1991,1993,1996)
value <-c(3, NA, 4)
However, for plotting a line chart, I want to fill the missing years so that I have a series from 1990 to 2000 in 1-year steps. The additional years shall be filled with NA values.
Is there a smart solution to this problem?
We can use complete from tidyr.
dat <- data.frame(
year = c(1991,1993,1996),
value = c(3, NA, 4)
)
library(dplyr)
library(tidyr)
dat2 <- dat %>%
complete(year = 1990:2000)
print(dat2)
# # A tibble: 11 x 2
# year value
# <dbl> <dbl>
# 1 1990 NA
# 2 1991 3
# 3 1992 NA
# 4 1993 NA
# 5 1994 NA
# 6 1995 NA
# 7 1996 4
# 8 1997 NA
# 9 1998 NA
# 10 1999 NA
# 11 2000 NA
Using base R to generate a sequence from 1990 to 2000 and merge with original data.frame.
df1 <- data.frame(year = c(1991, 1993, 1996),
value = c(3, NA, 4))
merge(df1,
data.frame(full = seq(1990, 2000))
by.x = "year",
by.y = "full",
all = TRUE)
year value
1 1990 NA
2 1991 3
3 1992 NA
4 1993 NA
5 1994 NA
6 1995 NA
7 1996 4
8 1997 NA
9 1998 NA
10 1999 NA
11 2000 NA
We assume that what you have is:
dd <- data.frame(year, value)
This is a time series so it makes sense to represent it using a time series representation such as ts, zoo or xts. We convert it to zoo and then to ts. The latter conversion will fill in the missing years.
library(zoo)
z <- read.zoo(dd)
tt <- as.ts(z)
tt
## Time Series:
## Start = 1991
## End = 1996
## Frequency = 1
## [1] 3 NA NA NA NA 4
If you really want to convert it to a data frame then use fortify.zoo(tt) .
If the only reason to do this is for plotting a line chart then alternately just remove the missing values. Any of these will work.
plot(na.omit(dd), type = "l", xlab = "year", ylab = "value")
plot(na.omit(z), xlab = "year", ylab = "value")
library(ggplot2)
autoplot(na.omit(z)) + xlab("year") + ylab("value")
The last plot is shown here:
