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my question is two-fold..

  1. I have a list of dataframes, and using lapply in R, I would like to add a column to each dataframe in the list.
  2. The added column should actually take values sequentially from a list, if possible. I have a list that is the same length as the list of dataframes, and each value in that list should be the added column value.

The reason I'm doing this is because the file name of each data set I'm importing has date info, e.g. the file name contains Jun12_2003. So I want to import each data set, and then assign a column for year and date, taking the info from the file name (so far doing that part with regexp).

Thanks for any help!

setophaga
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1 Answers1

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Use Map. It is short for mapply(..., SIMPLIFY = FALSE) as suggested by Ari.

df1 <- data.frame(x = runif(3), y = runif(3))
df2 <- data.frame(x = runif(3), y = runif(3))
dfs <- list(df1, df2)
years <- list(2013, 2014)

Map(cbind, dfs, year = years)
# [[1]]
#           x         y year
# 1 0.8843945 0.6285246 2013
# 2 0.8400041 0.1369520 2013
# 3 0.4398870 0.4660476 2013
# 
# [[2]]
#           x         y year
# 1 0.4153315 0.5831114 2014
# 2 0.9685105 0.2398060 2014
# 3 0.9507591 0.7585670 2014
flodel
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    What if the data frame list was in the thousands and you had a corresponding list to name those data frames in the list. I know that some dfs are empty, but those empty dfs should still have an added year from the years list that correspond to the data frame. Would an empty data frame result in a misorder? I have a similar problem, but the years are not aligning to correct columns. – sammyramz Jun 29 '16 at 22:01
  • Thank you for suggesting Map() instead of mapply() - it is a simpler implementation and thus this is the best possible answer to this question that I have come across – setophaga Apr 13 '17 at 04:28
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    This is the best solution - and in my case (at the time of writing), `Map(cbind,dfs,var=values)` worked as expected, but `mapply(cbind,dfs,var=values,simplify=FALSE)` did not! – Remko Duursma Jan 27 '19 at 20:35