CODEWITHSUNDEEP
rlistdata.tablenames
Tom
Tom

Reputation: 2341

Getting a column name from each dataframe in a list

My data is as follows:

library(data.table)   
df <- fread(
    "A   B  C  D  E  F  iso   year   
     0   A   NA  1  NA  NA  NLD   2009   
     1   Y   NA  2  NA  NA  NLD   2009   
     0   Q   NA  3  NA  NA  AUS   2011   
     1   NA  NA  4  NA  NA  AUS   2011   
     0   0   NA  7  NA  NA  NLD   2008   
     1   1   NA  1  NA  NA  NLD   2008   
     0   1   NA  3  NA  NA  AUS   2012   
     0   NA  1   NA  1  NA  ECU   2009   
     1   NA  0   NA  2  0   ECU   2009   
     0   NA  0   NA  3  0   BRA   2011   
     1   NA  0   NA  4  0   BRA   2011   
     0   NA  1   NA  7  NA  ECU   2008   
     1   NA  0   NA  1  0   ECU   2008   
     0   NA  0   NA  3  2   BRA   2012   
     1   NA  0   NA  4  NA  BRA   2012",
   header = TRUE
)

# Creates a list of dataframes
df_iso <- split(df, df$iso) # Creates a list of dataframes

I would now like to extract the column name of column 8 in each dataset of the list.

Obviously in this case they are all "year", but in my actual data they are different.

If I do colnames(df_iso[[1]])[8] I get "year", so I tried:

I tried:

names <- list()
for (i in length(df_iso)) { 
     names <- as.vector(append(names , colnames(df_iso[[i]])[8]))
 }

Surprisingly that does not work. I would like either a list or a vector which for each data.frame in df_iso gives me `"year"``, can anyone help?

Upvotes: 1

Views: 41

Answers (3)

akrun
akrun

Reputation: 886938

An option with purrr

library(purrr)
map(df_iso, ~ names(.x)[8])

Upvotes: 0

Henry Cyranka
Henry Cyranka

Reputation: 3060

Solution using sapply:

sapply(df_iso, function(i){
colnames(i)[8]
})

   AUS    BRA    ECU    NLD 
"year" "year" "year" "year"

Upvotes: 0

tmfmnk
tmfmnk

Reputation: 39858

One possibility could be:

lapply(df_iso, function(x) names(x)[8])

$AUS
[1] "year"

$BRA
[1] "year"

$ECU
[1] "year"

$NLD
[1] "year"

Upvotes: 2

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