R Loop code over several lists of dataframes

Question

I have several lists of dataframes and I want to format the date in each single dataframe within all lists of dataframes. Here is an example code:

v1 = c("2000-05-01", "2000-05-02", "2000-05-03", "2000-05-04", "2000-05-05")
v2 = seq(2,20, length = 5)
v3 = seq(-2,7, length = 5)
v4 = seq(-6,3, length = 5)

df1 = data.frame(Date = v1, df1_Tmax = v2, df1_Tmean = v3, df1_Tmin = v4)
dfl1 <- list(df1, df1, df1, df1)
names(dfl1) = c("ABC_1", "DEF_1", "GHI_1", "JKL_1")

v1 = c("2000-05-01", "2000-05-02", "2000-05-03", "2000-05-04", "2000-05-05")
v2 = seq(3,21, length = 5)
v3 = seq(-3,8, length = 5)
v4 = seq(-7,4, length = 5)

df2 = data.frame(Date = v1, df2_Tmax = v2, df2_Tmean = v3, df2_Tmin = v4)
dfl2 <- list(df2, df2, df2, df2)
names(dfl2) = c("ABC_2", "DEF_2", "GHI_2", "JKL_2")

v1 = c("2000-05-01", "2000-05-02", "2000-05-03", "2000-05-04", "2000-05-05")
v2 = seq(4,22, length = 5)
v3 = seq(-4,9, length = 5)
v4 = seq(-8,5, length = 5)

df3 = data.frame(Date = v1, df3_Tmax = v2, df3_Tmean = v3, df3_Tmin = v4)
dfl3 <- list(df3, df3, df3, df3)
names(dfl3) = c("ABC_3", "DEF_3", "GHI_3", "JKL_3")

v1 = c("2000-05-01", "2000-05-02", "2000-05-03", "2000-05-04", "2000-05-05")
v2 = seq(2,20, length = 5)
v3 = seq(-2,8, length = 5)
v4 = seq(-6,3, length = 5)

abc = data.frame(Date = v1, ABC_Tmax = v2, ABC_Tmean = v3, ABC_Tmin = v4)

abclist <-list(abc, abc, abc, abc)

names(abclist) = c("ABC_abc", "DEF_abc", "GHI_abc", "JKL_abc")

I know how to change the date-column manually:

dfl1$ABC_1$Date = as.Date(dfl1$ABC_1$Date,format="%Y-%m-%d")
class(dfl1$ABC_1$Date)

But how can I do that for each single Date-Column in all of my lists of dataframes?

A. Suliman · Accepted Answer · 2019-05-31 14:17:22Z

2

Here is one option using get and assign

nms <- c('dfl1', 'dfl2', 'dfl3', 'abclist')
lapply(nms, function(x) assign(x,lapply(get(x), 
                                 function(y) {y$Date1 <- as.Date(y$Date, format="%Y-%m-%d") 
                                 return(y)}), 
                        envir = .GlobalEnv))

PS: Be careful with assign since it will change your global environment .GlobalEnv. Many R users will suggest the list solution over assign.

edited May 31, 2019 at 14:17

answered May 31, 2019 at 13:36

A. Suliman

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4 Comments

Essi Over a year ago

Why is that so much of a problem? What are the risks?

kath Over a year ago

Have a look at stackoverflow.com/questions/17559390/why-is-using-assign-bad

Essi Over a year ago

Oh I see the problem now. It adds another columns with the updated Date. That is actually not that good.

A. Suliman Over a year ago

@Essi I just add Date1 to see the difference you can do y$Date <- as.Date(y$Date, format="%Y-%m-%d").

kath · Accepted Answer · 2019-05-31 13:16:08Z

2

This can be done with lapply:

lapply(dfl1, function(x) {
  x$Date <- as.Date(x$Date, format="%Y-%m-%d") 
  return(x)})

If you want to do this for all of you df-lists you need to store them in a list and then you can use a slightly modified version of the above call:

df_list <- list(dfl1, dfl2, dfl3, abclist)

lapply(df_list, function(x) {
  x[[1]]$Date <- as.Date(x[[1]]$Date, format="%Y-%m-%d") 
  return(x)})

This assumes that the Date-column has always the same name "Date".

edited May 31, 2019 at 13:16

answered May 31, 2019 at 13:14

kath

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8 Comments

Essi Over a year ago

Can you please show me how to do that for all of my dataframe lists? That was actually my question.

kath Over a year ago

Yes, sorry the last part got lost when I first wrote the answer.

Essi Over a year ago

Thank you! I see that this command stores all lists in one big list. What do I have to do to get my original files again? I mean all lists seperated with the same names as before, but the updated date-columns?

kath Over a year ago

You could apply the first version to each df list separately, i.e. use dfl2 instead of dfl1 in the first call. Having the same column name should not be a problem as the dfs are stored in different list entries...

kath Over a year ago

You can reassign them: dfl1 <- df_list$dfl1 and so on, but then I guess you are more looking for the solution from A. Suliman...

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