CODEWITHSUNDEEP
r
Reuben L.
Reuben L.

Reputation: 2859

Updating columns using lapply

Assume I have a set of dataframes named N01 to N99. They are identical in structure and all have a column named DATE.

Assume this DATE column is of factor class and I want to change it to POSIXlt. If I intend to update the NXX$DATE column using a function like strptime(), how can I do so for all N01-N99 dataframes?

I've tried using lapply but have trouble using assign in the function to assign to a specific column.

For example, this didn't work:

lapply( ls(pattern="N[0-9][0-9]"), function(x){ assign(paste(get(x),'$DATE',sep=''), strptime(get(x)$DATE),"%y/%m/%d), envir=.GlobalEnv)})

I also tried assigning to paste(x,"['TIME']") too, but it also created a new data.frame instead of updating the column.

How can I achieve what I want?

Upvotes: 0

Views: 1206

Answers (2)

Ramnath
Ramnath

Reputation: 55685

Here is one solution

# collect data frames into a list using mget
dats = mget(sprintf('N%02d', 1:99))

# loop through list of data frames and convert factor to date
dats = lapply(dats, function(x){
  x$DATE = strptime(x$DATE, "%y/%m/%d")
  return(x)
})

Upvotes: 3

akrun
akrun

Reputation: 886938

using lapply with list2env. But, you could do all the necessary analysis within the lapply and save each list element to separate file without even changing the original dataset using list2env. 

lst <- mget(ls(pattern="^N[0-9]+"))
list2env(
        lapply(lst, function(x) {x$DATE <- strptime(x$DATE, "%Y-%m-%d") #change here
                                          x}),
                                          envir=.GlobalEnv)    
<environment: R_GlobalEnv>

 str(N01)
#'data.frame':  5 obs. of  2 variables:
#$ DATE: POSIXlt, format: "2012-01-01" "2012-01-02" ...
#$ val : num  -0.212 -1.042 -1.153 0.322 -1.5

data

set.seed(25)
N01 <- data.frame(DATE = factor(seq(as.Date("2012-01-01"), length.out=5, by=1)), val=rnorm(5))
N02 <- data.frame(DATE = factor(seq(as.Date("2012-01-01"), length.out=5, by=1)), val=rnorm(5))
N25 <- data.frame(DATE = factor(seq(as.Date("2012-01-01"), length.out=5, by=1)), val=rnorm(5))

Upvotes: 2

Related Questions