Assign values in a for loop using R

Question

all:

The problem came when I tried to use for loop to create 5 dataframes and assign these five dataframes to a list. Please see an example:

library(data.table)
set.seed(123)    
df <- as.data.table(list(rnorm(10,1,1), rnorm(10,1,1)))
list <- list() 
for(i in 2011:2015){
             list[[paste0("A_",i)]] <- df[, year := as.numeric(i)]
             }

So, as I expected, the value of year variable should be as same as i in each element. For example, value of year in list[1] should be 2011. However, the above code returns 2015 for year in all elements:

    > list[1]
$A_2011
            V1         V2 year
 1:  0.4395244  2.2240818 2015
 2:  0.7698225  1.3598138 2015
 3:  2.5587083  1.4007715 2015
 4:  1.0705084  1.1106827 2015
 5:  1.1292877  0.4441589 2015
 6:  2.7150650  2.7869131 2015
 7:  1.4609162  1.4978505 2015
 8: -0.2650612 -0.9666172 2015
 9:  0.3131471  1.7013559 2015
10:  0.5543380  0.5272086 2015

I can't figure out what's wrong with my code. I would appreciate if anyone could point out the problem here. I would like to see any other solutions using lapply or so, if any. Many thanks!

Quick fix: Add list[[paste0("A_",i)]]$year <- i in your loop. — Majo
– Majo, Commented Apr 20, 2017 at 15:11
Generally a lot less painful to keep it in one table, like yrs = 2011:2015; res <- df[rep(1:.N, length(yrs))][, year := rep(yrs, each = nrow(df))][] or df[, rbindlist(Map(cbind, .(.SD), year = yrs))] — Frank
– Frank, Commented Apr 20, 2017 at 15:14
@Majo The fix works! However, I really want to know why my code does not work as expected. I think the logic is correct there. — Chuan
– Chuan, Commented Apr 20, 2017 at 15:18

Frank · Accepted Answer · 2017-04-20 15:22:51Z

2

for(i in 2011:2015){
  list[[paste0("A_",i)]] <- df[, year := as.numeric(i)]
} 
I would appreciate if anyone could point out the problem here.

<- makes a pointer to the same data.table, df, instead of making a copy. Wrapping this in copy() should fix that. However, it's cleaner to work with a single big table:

yrs = 2011:2015
res <- df[, rbindlist(Map(cbind, .(.SD), year = yrs))]

This has a couple advantages:

Lists of data.tables run into weird issues with pointers.
A big data.table allows one to use by= to iterative over subtables, which can be a lot more efficient.

answered Apr 20, 2017 at 15:22

Frank

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1 Comment

Chuan Over a year ago

Clear enough! Thank you @Frank

Carl Boneri · Accepted Answer · 2017-04-20 15:10:14Z

1

This worked: Using dplyr to append the new column as opposed to above:

setNames(lapply(2011:2015, function(i){
    as.data.table(list(rnorm(10,1,1), rnorm(10,1,1))) %>% 
        mutate(year = i)
}), sprintf("A_%s", 2011:2015))

Edit for seed handling :

setNames(lapply(2011:2015, function(i){
    set.seed(123)
    as.data.table(list(rnorm(10,1,1), rnorm(10,1,1))) %>% 
        mutate(year = i)
}), sprintf("A_%s", 2011:2015))

Output:

List of 5
 $ A_2011:'data.frame': 10 obs. of  3 variables:
  ..$ V1  : num [1:10] 0.44 0.77 2.56 1.07 1.13 ...
  ..$ V2  : num [1:10] 2.224 1.36 1.401 1.111 0.444 ...
  ..$ year: int [1:10] 2011 2011 2011 2011 2011 2011 2011 2011 2011 2011
 $ A_2012:'data.frame': 10 obs. of  3 variables:
  ..$ V1  : num [1:10] 0.44 0.77 2.56 1.07 1.13 ...
  ..$ V2  : num [1:10] 2.224 1.36 1.401 1.111 0.444 ...
  ..$ year: int [1:10] 2012 2012 2012 2012 2012 2012 2012 2012 2012 2012
 $ A_2013:'data.frame': 10 obs. of  3 variables:
  ..$ V1  : num [1:10] 0.44 0.77 2.56 1.07 1.13 ...
  ..$ V2  : num [1:10] 2.224 1.36 1.401 1.111 0.444 ...
  ..$ year: int [1:10] 2013 2013 2013 2013 2013 2013 2013 2013 2013 2013
 $ A_2014:'data.frame': 10 obs. of  3 variables:
  ..$ V1  : num [1:10] 0.44 0.77 2.56 1.07 1.13 ...
  ..$ V2  : num [1:10] 2.224 1.36 1.401 1.111 0.444 ...
  ..$ year: int [1:10] 2014 2014 2014 2014 2014 2014 2014 2014 2014 2014
 $ A_2015:'data.frame': 10 obs. of  3 variables:
  ..$ V1  : num [1:10] 0.44 0.77 2.56 1.07 1.13 ...
  ..$ V2  : num [1:10] 2.224 1.36 1.401 1.111 0.444 ...
  ..$ year: int [1:10] 2015 2015 2015 2015 2015 2015 2015 2015 2015 2015

answered Apr 20, 2017 at 15:10

Carl Boneri

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2 Comments

Chuan Over a year ago

Thank you! Using dplyr is definitely another good way to go. But do you know why my code does not work as expected?

Carl Boneri Over a year ago

Sadly I'm not as well versed as I would like to be with data.table to explain more clearly

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Assign values in a for loop using R

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