How to calculate percent differences in a table in R - r

Question

I have a csv file where rows 1-5 represent one state, 5-10 another, etc... I also have a column with years 1970,1980,..,2010 repeated for each state. In R (although I'm not opposed to a solution in Excel if that is easier), I want for each state to calculate the percent difference between that year and 1970, i.e. for Alabama 1990 it would be (AL 1990 - AL 1970)/(AL 1970), and add it to a new column in the data table so I can export it to a csv.
State, Year, Num
AL, 1970, 1
AL, 1980, 2
AL, 1990, 3
AL, 2000, 4
AL, 2010, 6
Output would be a column
pct_change
0
1
2
3
5

Answer 1

The dplyr package includes the function first which provides an easy method for getting the first value of a group. So if we arrange by Year to make it so that 1970 will be the first value of each group, when we group_by(State), we can use first(Num) to get that first value of Num which represents the value from 1970:
# Example data with 2 states
df <- structure(list(State = c("AL", "AL", "AL", "AL", "AL", "TX",
"TX", "TX", "TX", "TX"), Year = c(1970L, 1980L, 1990L, 2000L,
2010L, 1970L, 1980L, 1990L, 2000L, 2010L), Num = c(1, 2, 3, 4,
6, 5, 2, 10, 12, 6)), class = "data.frame", row.names = c(NA,
-10L))
library(dplyr)
df %>%
arrange(State, Year) %>%
group_by(State) %>%
mutate(perc_diff = 100 * (Num - first(Num))/first(Num))
# A tibble: 10 x 4
# Groups: State [2]
State Year Num perc_diff
<chr> <int> <dbl> <dbl>
1 AL 1970 1 0
2 AL 1980 2 100
3 AL 1990 3 200
4 AL 2000 4 300
5 AL 2010 6 500
6 TX 1970 5 0
7 TX 1980 2 -60
8 TX 1990 10 100
9 TX 2000 12 140
10 TX 2010 6 20

Answer 2

We can use data.table. Convert the 'data.frame' to 'data.table' (setDT(df)), order by 'State', 'Year' in the i, grouped by 'State', get the difference of the 'Num' with the first value of 'Num' and assign (:=) to create the 'perc_diff'
library(data.table)
setDT(df)[order(State, Year), perc_diff :=
100 * (Num - first(Num))/first(Num), State][]
# State Year Num perc_diff
# 1: AL 1970 1 0
# 2: AL 1980 2 100
# 3: AL 1990 3 200
# 4: AL 2000 4 300
# 5: AL 2010 6 500
# 6: TX 1970 5 0
# 7: TX 1980 2 -60
# 8: TX 1990 10 100
# 9: TX 2000 12 140
#10: TX 2010 6 20
Or using base R
v1 <- with(df, ave(Num, State, FUN = function(x) x[1]))
df$perc_diff <- with(df, 100 * (Num - v1)/v1)
data
df <- structure(list(State = c("AL", "AL", "AL", "AL", "AL", "TX",
"TX", "TX", "TX", "TX"), Year = c(1970L, 1980L, 1990L, 2000L,
2010L, 1970L, 1980L, 1990L, 2000L, 2010L), Num = c(1, 2, 3, 4,
6, 5, 2, 10, 12, 6)), class = "data.frame", row.names = c(NA,
-10L))

Answer 3

Base R solution using tapply
df <- df[with(df, order(State, Year)), ]
df$pct_change <- unlist( tapply(df$Num, df$State, function(x) 100 * (x - x[1]) / x[1]) )
> df
State Year Num pct_change
1 AL 1970 1 0
2 AL 1980 2 100
3 AL 1990 3 200
4 AL 2000 4 300
5 AL 2010 6 500
6 TX 1970 5 0
7 TX 1980 2 -60
8 TX 1990 10 100
9 TX 2000 12 140
10 TX 2010 6 20