R: Replacing values of one list element with values of a second list element

Question

I want to replace the values of one element of a list with the values of a second element of a list. Specifically,

I have a list containing multiple data sets.

Each data set has 2 variables

The variables are factors

The n'th element of the second variable of each data set needs to be replaced with the n'th element of the first variable in each data set

Also, the replaced value should be called "replaced"

dat1 <- data.frame(names1 =c("a", "b", "c", "f", "x"),values= c("val1_1", "val2_1", "val3_1", "val4_1", "val5_1"))
   dat1$values <- as.factor(dat1$values)
dat2 <- data.frame(names1 =c("a", "b", "f2", "s5", "h"),values= c("val1_2", "val2_2", "val3_2", "val4_2", "val5_2"))
   dat2$values <- as.factor(dat2$values)
list1 <- list(dat1, dat2)

The result should be the same list, but just with the 5th value replaced.

[[1]]
     names1  values
1         a  val1_1
2         b  val2_1
3         c  val3_1
4         f  val4_1
5  replaced       x
[[2]]
     names1  values
1         a  val1_2
2         b  val2_2
3        f2  val3_2
4        s5  val4_2
5  replaced       h

This is a simplified example. I have more than 4500 data sets. — tobias sch
– tobias sch, Commented Feb 3, 2019 at 14:55

Ronak Shah · Accepted Answer · 2019-02-03 15:13:43Z

A base R approach using lapply, since both the columns are factors we need to add new levels first before replacing them with new values otherwise those value would turn as NAs.

n <- 5

lapply(list1, function(x) {
   levels(x$values) <- c(levels(x$values), as.character(x$names1[n]))
   x$values[n] <- x$names1[n]
   levels(x$names1) <- c(levels(x$names1), "replaced")
   x$names1[n] <- "replaced"
   x
})

#[[1]]
#    names1 values
#1        a val1_1
#2        b val2_1
#3        c val3_1
#4        f val4_1
#5 replaced      x

#[[2]]
#    names1 values
#1        a val1_2
#2        b val2_2
#3       f2 val3_2
#4       s5 val4_2
#5 replaced      h

There is also another approach where we can convert both the columns to characters, then replace the values at required position and again convert them back to factors but since every dataframe in the list can be huge we do not want to convert all the values to characters and then back to factor just to change one value which could be computationally very expensive.

akrun · Accepted Answer · 2019-02-03 16:05:13Z

3

Here is one option with tidyverse. Loop through the list with map, slice the row of interest (in this case, it is the last row, so n() can be used), mutate the column value and bind with the original data without the last row

library(tidyverse)
map(list1, ~ .x %>% 
               slice(n()) %>%
               mutate(values = names1, names1 = 'replaced') %>% 
               bind_rows(.x %>% slice(-n()), .))
#[[1]]
#    names1 values
#1        a val1_1
#2        b val2_1
#3        c val3_1
#4        f val4_1
#5 replaced      x

#[[2]]
#    names1 values
#1        a val1_2
#2        b val2_2
#3       f2 val3_2
#4       s5 val4_2
#5 replaced      h

Or it can be made more compact with fct_c from forcats. Different factor levels can be combined together with fct_c for the 'values' and 'names1' column

library(forcats)
map(list1, ~ .x %>% 
        mutate(values = fct_c(values[-n()], names1[n()]), 
               names1 = fct_c(names1[-n()], factor('replaced'))))

Or using similar approach with base R where we loop through the list with lapply, then convert the data.frame to matrix, rbind the subset of matrix i.e. the last row removed with the values of interest, and convert to data.frame (by default, stringsAsFactors = TRUE - so it gets converted to factor)

lapply(list1,  function(x)  as.data.frame(rbind(as.matrix(x)[-5, ], 
              c('replaced',  as.character(x$names1[5])))))

edited Feb 3, 2019 at 16:05

answered Feb 3, 2019 at 15:45

akrun

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3 Comments

tobias sch Over a year ago

How do you get to the row number, Akrun? is n <- 5 used from the above answer?

akrun Over a year ago

@tobiassch No, Here, you have only 5 rows, and n() is the last row. If you have a custom n, then use that in slice(n)

tobias sch Over a year ago

Yes! I see, will try!

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