Selectively modify single element of a list in a list-column (tidy solution)

I want to replace a single element of a list within a list-column with the value from another column:

library(tidyverse)

df <- tibble(a = list(list(number = 1, letter = "a"), list(number = 2, letter = "b")), b = c("c", "d"))

print(df)

#> # A tibble: 2 x 2

#>   a          b    

#>   <list>     <chr>

#> 1 <list [2]> c    

#> 2 <list [2]> d

This is what the output should look like:

df2 <- tibble(a = list(list(number = 1, letter = "c"), list(number = 2, letter = "d")), b = c("c", "d"))

print(df2)

#> # A tibble: 2 x 2

#>   a          b    

#>   <list>     <chr>

#> 1 <list [2]> c    

#> 2 <list [2]> d

I was able to use a for loop and confusing subsetting syntax, but I am wondering if there is a tidy function or more elegant way to accomplish my goal:

for (i in 1:nrow(df)) {

  df$a[[i]]$letter <- df$b[[i]]

}

asked Jan 20 at 3:51

hlinee

1339

add a comment |

I want to replace a single element of a list within a list-column with the value from another column:

library(tidyverse)

df <- tibble(a = list(list(number = 1, letter = "a"), list(number = 2, letter = "b")), b = c("c", "d"))

print(df)

#> # A tibble: 2 x 2

#>   a          b    

#>   <list>     <chr>

#> 1 <list [2]> c    

#> 2 <list [2]> d

This is what the output should look like:

df2 <- tibble(a = list(list(number = 1, letter = "c"), list(number = 2, letter = "d")), b = c("c", "d"))

print(df2)

#> # A tibble: 2 x 2

#>   a          b    

#>   <list>     <chr>

#> 1 <list [2]> c    

#> 2 <list [2]> d

I was able to use a for loop and confusing subsetting syntax, but I am wondering if there is a tidy function or more elegant way to accomplish my goal:

for (i in 1:nrow(df)) {

  df$a[[i]]$letter <- df$b[[i]]

}

asked Jan 20 at 3:51

hlinee

1339

add a comment |

I want to replace a single element of a list within a list-column with the value from another column:

library(tidyverse)

df <- tibble(a = list(list(number = 1, letter = "a"), list(number = 2, letter = "b")), b = c("c", "d"))

print(df)

#> # A tibble: 2 x 2

#>   a          b    

#>   <list>     <chr>

#> 1 <list [2]> c    

#> 2 <list [2]> d

This is what the output should look like:

df2 <- tibble(a = list(list(number = 1, letter = "c"), list(number = 2, letter = "d")), b = c("c", "d"))

print(df2)

#> # A tibble: 2 x 2

#>   a          b    

#>   <list>     <chr>

#> 1 <list [2]> c    

#> 2 <list [2]> d

I was able to use a for loop and confusing subsetting syntax, but I am wondering if there is a tidy function or more elegant way to accomplish my goal:

for (i in 1:nrow(df)) {

  df$a[[i]]$letter <- df$b[[i]]

}

asked Jan 20 at 3:51

hlinee

1339

I want to replace a single element of a list within a list-column with the value from another column:

library(tidyverse)

df <- tibble(a = list(list(number = 1, letter = "a"), list(number = 2, letter = "b")), b = c("c", "d"))

print(df)

#> # A tibble: 2 x 2

#>   a          b    

#>   <list>     <chr>

#> 1 <list [2]> c    

#> 2 <list [2]> d

This is what the output should look like:

df2 <- tibble(a = list(list(number = 1, letter = "c"), list(number = 2, letter = "d")), b = c("c", "d"))

print(df2)

#> # A tibble: 2 x 2

#>   a          b    

#>   <list>     <chr>

#> 1 <list [2]> c    

#> 2 <list [2]> d

I was able to use a for loop and confusing subsetting syntax, but I am wondering if there is a tidy function or more elegant way to accomplish my goal:

for (i in 1:nrow(df)) {

  df$a[[i]]$letter <- df$b[[i]]

}

r tidyverse

asked Jan 20 at 3:51

hlinee

1339

asked Jan 20 at 3:51

hlinee

1339

asked Jan 20 at 3:51

hlinee

1339

asked Jan 20 at 3:51

hlinee

1339

asked Jan 20 at 3:51

hlinee

1339

add a comment |

1 Answer
1

active

oldest

votes

We could use a base R approach using mapply by changing values in letter element of every list from the corresponding b element.

df$a <- mapply(function(x, y) {x['letter'] <- y;list(x)}, df$a, df$b)



df$a

#[[1]]

#[[1]]$number

#[1] 1



#[[1]]$letter

#[1] "c"





#[[2]]

#[[2]]$number

#[1] 2



#[[2]]$letter

#[1] "d"

If you are interested in the tidyverse approach we could use the same logic using map2

library(tidyverse)



df$a <- map2(df$a, df$b, function(x, y) {x['letter'] <- y;x})

Or using a different approach using lapply

do.call("rbind", lapply(seq_len(nrow(df)), function(x) {

        df$a[[x]]['letter'] <- df$b[x]

        df[x, ]

}))

edited Jan 20 at 4:54

answered Jan 20 at 4:46

Ronak Shah

38.2k104161

Could you explain why it is necessary to add list(x) after the assignment statement? I tried it without the list(x) and apparently it replaces the whole list with the variable, but I am not sure how adding list(x) corrects this problem.

– hlinee
Jan 23 at 23:51

1

@hlinee it will return the object which is last in mapply call. So if you don't return list(x) it will return only the modified x['letter'] part and will ignore the remaining part of list which is number in this case.

– Ronak Shah
Jan 24 at 8:32

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

We could use a base R approach using mapply by changing values in letter element of every list from the corresponding b element.

df$a <- mapply(function(x, y) {x['letter'] <- y;list(x)}, df$a, df$b)



df$a

#[[1]]

#[[1]]$number

#[1] 1



#[[1]]$letter

#[1] "c"





#[[2]]

#[[2]]$number

#[1] 2



#[[2]]$letter

#[1] "d"

If you are interested in the tidyverse approach we could use the same logic using map2

library(tidyverse)



df$a <- map2(df$a, df$b, function(x, y) {x['letter'] <- y;x})

Or using a different approach using lapply

do.call("rbind", lapply(seq_len(nrow(df)), function(x) {

        df$a[[x]]['letter'] <- df$b[x]

        df[x, ]

}))

edited Jan 20 at 4:54

answered Jan 20 at 4:46

Ronak Shah

38.2k104161

Could you explain why it is necessary to add list(x) after the assignment statement? I tried it without the list(x) and apparently it replaces the whole list with the variable, but I am not sure how adding list(x) corrects this problem.

– hlinee
Jan 23 at 23:51

1

@hlinee it will return the object which is last in mapply call. So if you don't return list(x) it will return only the modified x['letter'] part and will ignore the remaining part of list which is number in this case.

– Ronak Shah
Jan 24 at 8:32

add a comment |

We could use a base R approach using mapply by changing values in letter element of every list from the corresponding b element.

df$a <- mapply(function(x, y) {x['letter'] <- y;list(x)}, df$a, df$b)



df$a

#[[1]]

#[[1]]$number

#[1] 1



#[[1]]$letter

#[1] "c"





#[[2]]

#[[2]]$number

#[1] 2



#[[2]]$letter

#[1] "d"

If you are interested in the tidyverse approach we could use the same logic using map2

library(tidyverse)



df$a <- map2(df$a, df$b, function(x, y) {x['letter'] <- y;x})

Or using a different approach using lapply

do.call("rbind", lapply(seq_len(nrow(df)), function(x) {

        df$a[[x]]['letter'] <- df$b[x]

        df[x, ]

}))

edited Jan 20 at 4:54

answered Jan 20 at 4:46

Ronak Shah

38.2k104161

Could you explain why it is necessary to add list(x) after the assignment statement? I tried it without the list(x) and apparently it replaces the whole list with the variable, but I am not sure how adding list(x) corrects this problem.

– hlinee
Jan 23 at 23:51

1

@hlinee it will return the object which is last in mapply call. So if you don't return list(x) it will return only the modified x['letter'] part and will ignore the remaining part of list which is number in this case.

– Ronak Shah
Jan 24 at 8:32

add a comment |

We could use a base R approach using mapply by changing values in letter element of every list from the corresponding b element.

df$a <- mapply(function(x, y) {x['letter'] <- y;list(x)}, df$a, df$b)



df$a

#[[1]]

#[[1]]$number

#[1] 1



#[[1]]$letter

#[1] "c"





#[[2]]

#[[2]]$number

#[1] 2



#[[2]]$letter

#[1] "d"

If you are interested in the tidyverse approach we could use the same logic using map2

library(tidyverse)



df$a <- map2(df$a, df$b, function(x, y) {x['letter'] <- y;x})

Or using a different approach using lapply

do.call("rbind", lapply(seq_len(nrow(df)), function(x) {

        df$a[[x]]['letter'] <- df$b[x]

        df[x, ]

}))

edited Jan 20 at 4:54

answered Jan 20 at 4:46

Ronak Shah

38.2k104161

We could use a base R approach using mapply by changing values in letter element of every list from the corresponding b element.

df$a <- mapply(function(x, y) {x['letter'] <- y;list(x)}, df$a, df$b)



df$a

#[[1]]

#[[1]]$number

#[1] 1



#[[1]]$letter

#[1] "c"





#[[2]]

#[[2]]$number

#[1] 2



#[[2]]$letter

#[1] "d"

If you are interested in the tidyverse approach we could use the same logic using map2

library(tidyverse)



df$a <- map2(df$a, df$b, function(x, y) {x['letter'] <- y;x})

Or using a different approach using lapply

do.call("rbind", lapply(seq_len(nrow(df)), function(x) {

        df$a[[x]]['letter'] <- df$b[x]

        df[x, ]

}))

edited Jan 20 at 4:54

answered Jan 20 at 4:46

Ronak Shah

38.2k104161

edited Jan 20 at 4:54

answered Jan 20 at 4:46

Ronak Shah

38.2k104161

answered Jan 20 at 4:46

Ronak Shah

38.2k104161

answered Jan 20 at 4:46

Ronak Shah

38.2k104161

Could you explain why it is necessary to add list(x) after the assignment statement? I tried it without the list(x) and apparently it replaces the whole list with the variable, but I am not sure how adding list(x) corrects this problem.

– hlinee
Jan 23 at 23:51

1

@hlinee it will return the object which is last in mapply call. So if you don't return list(x) it will return only the modified x['letter'] part and will ignore the remaining part of list which is number in this case.

– Ronak Shah
Jan 24 at 8:32

add a comment |

Could you explain why it is necessary to add list(x) after the assignment statement? I tried it without the list(x) and apparently it replaces the whole list with the variable, but I am not sure how adding list(x) corrects this problem.

– hlinee
Jan 23 at 23:51

1

@hlinee it will return the object which is last in mapply call. So if you don't return list(x) it will return only the modified x['letter'] part and will ignore the remaining part of list which is number in this case.

– Ronak Shah
Jan 24 at 8:32

Could you explain why it is necessary to add list(x) after the assignment statement? I tried it without the list(x) and apparently it replaces the whole list with the variable, but I am not sure how adding list(x) corrects this problem.

– hlinee
Jan 23 at 23:51

@hlinee it will return the object which is last in mapply call. So if you don't return list(x) it will return only the modified x['letter'] part and will ignore the remaining part of list which is number in this case.

– Ronak Shah
Jan 24 at 8:32

add a comment |

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