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I am brand new to R so if I'm thinking about this completely wrong feel free to tell me. I have a series of imported dataframes on power plants, one of each year (Plant1987, Plant1988 etc...) that I am trying to combine ultimately into one data frame. Prior to doing so, I'd like to add a "year" variable to each dataframe. I could do this for each individual dataframe, but would like to formalize it and do it in one step. I know how to do it in stata, but I'm struggling here.

I was thinking something along the lines of:

for (y in 1987:2008) {
     paste("Plant",y,sep="")$year <- y
}

which doesn't work because paste is obviously not the right function. Is there a smart, quick way to do this? Thanks

winitheju
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    [Store your data.frames in a list](https://stackoverflow.com/questions/17499013/how-do-i-make-a-list-of-data-frames); don't create a bunch of variables that have data in their names. Then you can apply functions over that list. `paste()` creates character vectors. Character vectors are not the same things as names/symbols. – MrFlick May 26 '17 at 19:34
  • @MrFlick can you help me think a little more about what such a function would look like. I understand the concept of applying a function to the dataframes within a list, but I don't understand how I can reference part of the dataframe name to make a new variable – winitheju May 26 '17 at 19:40

3 Answers3

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Try this ..

year=seq(1987,2008,by=1)
list_object_names = sprintf("Plant%s", 1987:2008)

list_DataFrame = lapply(list_object_names, get)

for (i in 1:length(list_DataFrame ) ){
    list_DataFrame[[i]][,'Year']=year[i]
}
BENY
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0

Here are some codes to give you some ideas. I used the mtcars data frame as an example to create a list with three data frames. After that I used two solutions to add the year (2000 to 2002) to each data frame. You will need to modify the codes for your data.

# Load the mtcars data frame
data(mtcars)

# Create a list with three data frames
ex_list <- list(mtcars, mtcars, mtcars)

# Create a list with three years: 2000 to 2002
year_list <- 2000:2002

Solution 1: Use lapply from base R

ex_list2 <- lapply(1:3, function(i) {

  dt <- ex_list[[i]]

  dt[["Year"]] <- year_list[[i]]

  return(dt)
})

Solution 2: Use map2 from purrr

library(purrr)    

ex_list3 <- map2(ex_list, year_list, .f = function(dt, year){

  dt$Year <- year

  return(dt)
})

ex_list2 and ex_list3 are the final output.

www
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0

Let's say you have data.frames

Plant1987 <- data.frame(plantID=1:4, x=rnorm(4))
Plant1988 <- data.frame(plantID=1:4, x=rnorm(4))
Plant1989 <- data.frame(plantID=1:4, x=rnorm(4))

You could put a $year column in each with

year <- 1987:1989
for(yeari in year) {
  eval(parse(text=paste0("Plant",yeari,"$year<-",yeari)))
}

Plant1987
#   plantID           x year
# 1       1  0.67724230 1987
# 2       2 -1.74773250 1987
# 3       3  0.67982621 1987
# 4       4  0.04731677 1987
# ...etc for other years...

...and either bind them together into one data.frame with

df <- Plant1987
for(yeari in year[-1]) {
  df <- rbind(df, eval(parse(text=paste0("Plant",yeari))))
}

df
#    plantID            x year
# 1        1  0.677242300 1987
# 2        2 -1.747732498 1987
# 3        3  0.679826213 1987
# 4        4  0.047316768 1987
# 5        1  1.043299473 1988
# 6        2  0.003758675 1988
# 7        3  0.601255190 1988
# 8        4  0.904374498 1988
# 9        1  0.082030356 1989
# 10       2 -1.409670456 1989
# 11       3 -0.064881722 1989
# 12       4  1.312507736 1989

...or in a list as 

itsalist <- list()
for(yeari in year) {
  eval(parse(text=paste0("itsalist$Plant",yeari,"<-Plant",yeari)))
}

itsalist
# $Plant1987
#   plantID           x year
# 1       1  0.67724230 1987
# 2       2 -1.74773250 1987
# 3       3  0.67982621 1987
# 4       4  0.04731677 1987
# 
# $Plant1988
#   plantID           x year
# 1       1 1.043299473 1988
# 2       2 0.003758675 1988
# 3       3 0.601255190 1988
# 4       4 0.904374498 1988
# 
# $Plant1989
#   plantID           x year
# 1       1  0.08203036 1989
# 2       2 -1.40967046 1989
# 3       3 -0.06488172 1989
# 4       4  1.31250774 1989
Matt Tyers
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