Function not working the same way when using lapply

Question

First off - I apologise for the badly phrased question, if you have any more appropriate suggestions for it I'm all ears!

I have written a function that adds a new variable into my df based on what the object is called. For example, I have x_data, and when I put it into my function, a new variable called x_or_y is created, with all entries showing "x". You guessed it - there is also a y_data, that when plugged into the function it creates the same variable x_or_y and fills it with "y"

I'm having some issues however when I have a list of dfs (x_data and y_data) and I use lapply() to the list - it just returns the same letter for all. See below my example.

Example:

set.seed(123)
x_data <- data.frame(A = rnorm(20),
                 B = rnorm(20))
y_data <- data.frame(A = rnorm(20),
                 B = rnorm(20))

Function for "x or y":

add_x_or_y  <- function(z) {
  z$x_or_y <- ifelse(grepl(pattern = "x", 
                                 deparse(substitute(z)), fixed = TRUE), 
                           "x", "y")
  z
}

So trying the function out by itself:

head(add_x_or_y(x_data),3)

            A          B x_or_y
1 -0.56047565 -1.0678237      x
2 -0.23017749 -0.2179749      x
3  1.5587083 -1.0260044       x

head(add_x_or_y(y_data),3)

           A          B x_or_y
1 -0.6947070  0.3796395      y
2 -0.2079173 -0.5023235      y
3 -1.2653964 -0.3332074      y

Okay, so that works - but in real life I have about 20 different dfs that I need to put through the function. That's a lot of wasted space writing it each time, so let's make a list and use lapply().

x_y_list <- list(x_data = x_data,
                 y_data = y_data)
x_y_list <- lapply(x_y_list, add_x_or_y)

head(x_y_list$x_data,3)
           A          B x_or_y
1 -0.5604756 -1.0678237      y
2 -0.2301775 -0.2179749      y
3  1.5587083 -1.0260044      y

head(x_y_list$y_data,3)
           A          B x_or_y
1 -0.6947070  0.3796395      y
2 -0.2079173 -0.5023235      y
3 -1.2653964 -0.3332074      y

It didn't work! :( The first one should have "x"'s in the x_or_y variable. I'm sure it's a really really simple solution here, but I cannot seem to find it - please help Stack Overflow!

As far as I can tell, you'll have to pass names(x_y_list) to a modified function to achieve the same behavior — CPak
– CPak, Commented May 21, 2018 at 17:48
Put print(deparse(substitute(z))) as the first instruction of the function and see what is happening. — Rui Barradas
– Rui Barradas, Commented May 21, 2018 at 17:50
Names objects != Named elements within a list. And please show the actual function (or give background to actual problem) as I highly doubt deparse(substitute(z)) is even needed. — Parfait
– Parfait, Commented May 21, 2018 at 19:45
@RuiBarradas great suggestion using print(), this is helping to see what's going on. The function is returning "X[[i]]" "X[[i]]" which means it's not taking the name of each element in the list, just the subscript. Confused why it shows a capital "X" though? — TheNerdyCat
– TheNerdyCat, Commented May 22, 2018 at 7:06

r.user.05apr · Accepted Answer · 2018-05-22 07:09:48Z

2

If you are using lapply, you are just handing over the data frames without their names. Perhaps if you change the approach alittle:

set.seed(123)
x_data <- data.frame(A = rnorm(20),
                     B = rnorm(20))
y_data <- data.frame(A = rnorm(20),
                     B = rnorm(20))

add_x_or_y <- function(df, df_name) {
  df$x_or_y <- ifelse(grepl("x", df_name) == TRUE, "x", "y")
  df
}

x_y_list <- list(x_data = x_data,
                 y_data = y_data)

x_y_list <- mapply(add_x_or_y, 
                   x_y_list, 
                   names(x_y_list), SIMPLIFY = FALSE)

head(x_y_list$x_data, 3)
head(x_y_list$y_data, 3)

answered May 22, 2018 at 7:09

r.user.05apr

5,4783 gold badges27 silver badges43 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

TheNerdyCat Over a year ago

Brilliant! This does the job :) I didn't even think to add another argument, but it makes a lot of sense. I've learnt something today - thank you!

Collectives™ on Stack Overflow

Function not working the same way when using lapply

Example:

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

Example:

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related