purrr rbind list of data frames per groupCombine every ith element of a list of lists together using dplyr, purrrDrop factor levels in a subsetted data frameHow to join (merge) data frames (inner, outer, left, right)Convert a list of data frames into one data frameR - list to data frameDrop data frame columns by nameChanging column names of a data frameChange and save only one row in a filepurrr list evaluation strangenessError for neuralnet package in RUsing flextable in r markdown loop not producing tables
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purrr rbind list of data frames per group
Combine every ith element of a list of lists together using dplyr, purrrDrop factor levels in a subsetted data frameHow to join (merge) data frames (inner, outer, left, right)Convert a list of data frames into one data frameR - list to data frameDrop data frame columns by nameChanging column names of a data frameChange and save only one row in a filepurrr list evaluation strangenessError for neuralnet package in RUsing flextable in r markdown loop not producing tables
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
2
After using purrr and friends to read in a load of csvs I have ended up with a tibble that looks something like this:
library(tidyverse)
df <-
tibble(
df_name = c("A", "B", "A", "A", "B"),
data = list(iris)
)
df
# A tibble: 5 x 2
df_name data
<chr> <list>
1 A <data.frame [150 × 5]>
2 B <data.frame [150 × 5]>
3 A <data.frame [150 × 5]>
4 A <data.frame [150 × 5]>
5 B <data.frame [150 × 5]>
I want to rbind (or equivalent) all data with a common df_name. I'd like the output to be a named list. I can do this with tapply:
desired = tapply(df$data, df$df_name, function(y) do.call(rbind,y))
List of 2
$ A:'data.frame': 450 obs. of 5 variables:
..$ Sepal.Length: num [1:450] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:450] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:450] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:450] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
$ B:'data.frame': 300 obs. of 5 variables:
..$ Sepal.Length: num [1:300] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:300] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:300] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:300] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
- attr(*, "dim")= int 2
- attr(*, "dimnames")=List of 1
..$ : chr [1:2] "A" "B"
I can't figure out how to do the same with purrr verbs. I think perhaps I need to start by setting the list names:
df_p <-
df %>%
mutate(data = setNames(data, df_name))
I found this question but I can't figure out how to apply in this situation.
r purrr
asked Mar 27 at 9:21
Pete900Pete900
8951 gold badge10 silver badges26 bronze badges
add a comment |
2
After using purrr and friends to read in a load of csvs I have ended up with a tibble that looks something like this:
library(tidyverse)
df <-
tibble(
df_name = c("A", "B", "A", "A", "B"),
data = list(iris)
)
df
# A tibble: 5 x 2
df_name data
<chr> <list>
1 A <data.frame [150 × 5]>
2 B <data.frame [150 × 5]>
3 A <data.frame [150 × 5]>
4 A <data.frame [150 × 5]>
5 B <data.frame [150 × 5]>
I want to rbind (or equivalent) all data with a common df_name. I'd like the output to be a named list. I can do this with tapply:
desired = tapply(df$data, df$df_name, function(y) do.call(rbind,y))
List of 2
$ A:'data.frame': 450 obs. of 5 variables:
..$ Sepal.Length: num [1:450] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:450] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:450] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:450] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
$ B:'data.frame': 300 obs. of 5 variables:
..$ Sepal.Length: num [1:300] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:300] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:300] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:300] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
- attr(*, "dim")= int 2
- attr(*, "dimnames")=List of 1
..$ : chr [1:2] "A" "B"
I can't figure out how to do the same with purrr verbs. I think perhaps I need to start by setting the list names:
df_p <-
df %>%
mutate(data = setNames(data, df_name))
I found this question but I can't figure out how to apply in this situation.
r purrr
asked Mar 27 at 9:21
Pete900Pete900
8951 gold badge10 silver badges26 bronze badges
add a comment |
2
2
2
After using purrr and friends to read in a load of csvs I have ended up with a tibble that looks something like this:
library(tidyverse)
df <-
tibble(
df_name = c("A", "B", "A", "A", "B"),
data = list(iris)
)
df
# A tibble: 5 x 2
df_name data
<chr> <list>
1 A <data.frame [150 × 5]>
2 B <data.frame [150 × 5]>
3 A <data.frame [150 × 5]>
4 A <data.frame [150 × 5]>
5 B <data.frame [150 × 5]>
I want to rbind (or equivalent) all data with a common df_name. I'd like the output to be a named list. I can do this with tapply:
desired = tapply(df$data, df$df_name, function(y) do.call(rbind,y))
List of 2
$ A:'data.frame': 450 obs. of 5 variables:
..$ Sepal.Length: num [1:450] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:450] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:450] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:450] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
$ B:'data.frame': 300 obs. of 5 variables:
..$ Sepal.Length: num [1:300] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:300] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:300] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:300] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
- attr(*, "dim")= int 2
- attr(*, "dimnames")=List of 1
..$ : chr [1:2] "A" "B"
I can't figure out how to do the same with purrr verbs. I think perhaps I need to start by setting the list names:
df_p <-
df %>%
mutate(data = setNames(data, df_name))
I found this question but I can't figure out how to apply in this situation.
r purrr
asked Mar 27 at 9:21
Pete900Pete900
8951 gold badge10 silver badges26 bronze badges
After using purrr and friends to read in a load of csvs I have ended up with a tibble that looks something like this:
library(tidyverse)
df <-
tibble(
df_name = c("A", "B", "A", "A", "B"),
data = list(iris)
)
df
# A tibble: 5 x 2
df_name data
<chr> <list>
1 A <data.frame [150 × 5]>
2 B <data.frame [150 × 5]>
3 A <data.frame [150 × 5]>
4 A <data.frame [150 × 5]>
5 B <data.frame [150 × 5]>
I want to rbind (or equivalent) all data with a common df_name. I'd like the output to be a named list. I can do this with tapply:
desired = tapply(df$data, df$df_name, function(y) do.call(rbind,y))
List of 2
$ A:'data.frame': 450 obs. of 5 variables:
..$ Sepal.Length: num [1:450] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:450] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:450] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:450] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
$ B:'data.frame': 300 obs. of 5 variables:
..$ Sepal.Length: num [1:300] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:300] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:300] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:300] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
- attr(*, "dim")= int 2
- attr(*, "dimnames")=List of 1
..$ : chr [1:2] "A" "B"
I can't figure out how to do the same with purrr verbs. I think perhaps I need to start by setting the list names:
df_p <-
df %>%
mutate(data = setNames(data, df_name))
I found this question but I can't figure out how to apply in this situation.
r purrr
r purrr
asked Mar 27 at 9:21
Pete900Pete900
8951 gold badge10 silver badges26 bronze badges
asked Mar 27 at 9:21
Pete900Pete900
8951 gold badge10 silver badges26 bronze badges
asked Mar 27 at 9:21
Pete900Pete900
8951 gold badge10 silver badges26 bronze badges
asked Mar 27 at 9:21
Pete900Pete900
8951 gold badge10 silver badges26 bronze badges
asked Mar 27 at 9:21
Pete900Pete900
8951 gold badge10 silver badges26 bronze badges
8951 gold badge10 silver badges26 bronze badges
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
3
We can use tidyr::unnest
library(tidyverse)
df %>% split(.$df_name) %>% map(.%>%unnest() %>% select(-df_name))
#OR
df %>% split(.$df_name) %>% map(~unnest(.) %>% select(-df_name))
df %>% unnest(data) %>% split(.$df_name)
As @kath pointed out that we can use unnest directly
df %>% split(.$df_name) %>% map(unnest)
answered Mar 27 at 9:36
A. SulimanA. Suliman
7,4734 gold badges13 silver badges26 bronze badges
2
map(unnest)would also work (to add another option ;) )
– kath
Mar 27 at 9:47
add a comment |
2
You can use reduce from purrr and bind_rows (similar to rbind) from dplyr.
df_list <- df %>%
group_by(df_name) %>%
summarize(data = list(reduce(data, bind_rows)))
df_list
# A tibble: 2 x 2
# df_name data
# <chr> <list>
# 1 A <data.frame [450 x 5]>
# 2 B <data.frame [300 x 5]>
For the exact same structure as in your tapply-version we would need to add the following:
df_list2 <- df_list %>%
split(.$df_name) %>%
map(~ .x$data[[1]])
str(df_list2)
List of 2
$ A:'data.frame': 450 obs. of 5 variables:
..$ Sepal.Length: num [1:450] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:450] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:450] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:450] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
$ B:'data.frame': 300 obs. of 5 variables:
..$ Sepal.Length: num [1:300] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:300] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:300] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:300] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
answered Mar 27 at 9:40
kathkath
5,70411 silver badges27 bronze badges
Both great answers so thank you!
– Pete900
Mar 27 at 9:43
add a comment |
1
I would use unnest and group_split :
df %>% unnest(data) %>% group_split(df_name)
# [[1]]
# # A tibble: 450 x 6
# df_name Sepal.Length Sepal.Width Petal.Length Petal.Width Species
# <chr> <dbl> <dbl> <dbl> <dbl> <fct>
# 1 A 5.1 3.5 1.4 0.2 setosa
# 2 A 4.9 3 1.4 0.2 setosa
# 3 A 4.7 3.2 1.3 0.2 setosa
# 4 A 4.6 3.1 1.5 0.2 setosa
# 5 A 5 3.6 1.4 0.2 setosa
# 6 A 5.4 3.9 1.7 0.4 setosa
# 7 A 4.6 3.4 1.4 0.3 setosa
# 8 A 5 3.4 1.5 0.2 setosa
# 9 A 4.4 2.9 1.4 0.2 setosa
# 10 A 4.9 3.1 1.5 0.1 setosa
# # ... with 440 more rows
#
# [[2]]
# # A tibble: 300 x 6
# df_name Sepal.Length Sepal.Width Petal.Length Petal.Width Species
# <chr> <dbl> <dbl> <dbl> <dbl> <fct>
# 1 B 5.1 3.5 1.4 0.2 setosa
# 2 B 4.9 3 1.4 0.2 setosa
# 3 B 4.7 3.2 1.3 0.2 setosa
# 4 B 4.6 3.1 1.5 0.2 setosa
# 5 B 5 3.6 1.4 0.2 setosa
# 6 B 5.4 3.9 1.7 0.4 setosa
# 7 B 4.6 3.4 1.4 0.3 setosa
# 8 B 5 3.4 1.5 0.2 setosa
# 9 B 4.4 2.9 1.4 0.2 setosa
# 10 B 4.9 3.1 1.5 0.1 setosa
# # ... with 290 more rows
answered Mar 27 at 10:31
Moody_MudskipperMoody_Mudskipper
27.1k4 gold badges48 silver badges81 bronze badges
Thanks, I hadn't heard of group_split before.
– Pete900
Mar 27 at 11:09
It's fairly new :) tidyverse.org/articles/2019/02/dplyr-0-8-0
– Moody_Mudskipper
Mar 27 at 11:15
Although as-is the resulting list elements are not named but I can see how to set them.
– Pete900
Mar 27 at 11:17
1
fair point, actually the help says "it does not name the elements of the list based on the grouping as this typically loses information and is confusing.", which I think is a mistake and some users complained about it in the github issues already. The function is still marked as experimental so hopefully they'll change it.
– Moody_Mudskipper
Mar 27 at 11:27
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
We can use tidyr::unnest
library(tidyverse)
df %>% split(.$df_name) %>% map(.%>%unnest() %>% select(-df_name))
#OR
df %>% split(.$df_name) %>% map(~unnest(.) %>% select(-df_name))
df %>% unnest(data) %>% split(.$df_name)
As @kath pointed out that we can use unnest directly
df %>% split(.$df_name) %>% map(unnest)
answered Mar 27 at 9:36
A. SulimanA. Suliman
7,4734 gold badges13 silver badges26 bronze badges
2
map(unnest)would also work (to add another option ;) )
– kath
Mar 27 at 9:47
add a comment |
3
We can use tidyr::unnest
library(tidyverse)
df %>% split(.$df_name) %>% map(.%>%unnest() %>% select(-df_name))
#OR
df %>% split(.$df_name) %>% map(~unnest(.) %>% select(-df_name))
df %>% unnest(data) %>% split(.$df_name)
As @kath pointed out that we can use unnest directly
df %>% split(.$df_name) %>% map(unnest)
answered Mar 27 at 9:36
A. SulimanA. Suliman
7,4734 gold badges13 silver badges26 bronze badges
2
map(unnest)would also work (to add another option ;) )
– kath
Mar 27 at 9:47
add a comment |
3
3
3
We can use tidyr::unnest
library(tidyverse)
df %>% split(.$df_name) %>% map(.%>%unnest() %>% select(-df_name))
#OR
df %>% split(.$df_name) %>% map(~unnest(.) %>% select(-df_name))
df %>% unnest(data) %>% split(.$df_name)
As @kath pointed out that we can use unnest directly
df %>% split(.$df_name) %>% map(unnest)
answered Mar 27 at 9:36
A. SulimanA. Suliman
7,4734 gold badges13 silver badges26 bronze badges
We can use tidyr::unnest
library(tidyverse)
df %>% split(.$df_name) %>% map(.%>%unnest() %>% select(-df_name))
#OR
df %>% split(.$df_name) %>% map(~unnest(.) %>% select(-df_name))
df %>% unnest(data) %>% split(.$df_name)
As @kath pointed out that we can use unnest directly
df %>% split(.$df_name) %>% map(unnest)
answered Mar 27 at 9:36
A. SulimanA. Suliman
7,4734 gold badges13 silver badges26 bronze badges
edited Mar 27 at 9:53
answered Mar 27 at 9:36
A. SulimanA. Suliman
7,4734 gold badges13 silver badges26 bronze badges
answered Mar 27 at 9:36
A. SulimanA. Suliman
7,4734 gold badges13 silver badges26 bronze badges
answered Mar 27 at 9:36
A. SulimanA. Suliman
7,4734 gold badges13 silver badges26 bronze badges
7,4734 gold badges13 silver badges26 bronze badges
2
map(unnest)would also work (to add another option ;) )
– kath
Mar 27 at 9:47
add a comment |
2
map(unnest)would also work (to add another option ;) )
– kath
Mar 27 at 9:47
2
2
map(unnest) would also work (to add another option ;) )– kath
Mar 27 at 9:47
map(unnest) would also work (to add another option ;) )– kath
Mar 27 at 9:47
add a comment |
2
You can use reduce from purrr and bind_rows (similar to rbind) from dplyr.
df_list <- df %>%
group_by(df_name) %>%
summarize(data = list(reduce(data, bind_rows)))
df_list
# A tibble: 2 x 2
# df_name data
# <chr> <list>
# 1 A <data.frame [450 x 5]>
# 2 B <data.frame [300 x 5]>
For the exact same structure as in your tapply-version we would need to add the following:
df_list2 <- df_list %>%
split(.$df_name) %>%
map(~ .x$data[[1]])
str(df_list2)
List of 2
$ A:'data.frame': 450 obs. of 5 variables:
..$ Sepal.Length: num [1:450] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:450] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:450] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:450] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
$ B:'data.frame': 300 obs. of 5 variables:
..$ Sepal.Length: num [1:300] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:300] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:300] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:300] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
answered Mar 27 at 9:40
kathkath
5,70411 silver badges27 bronze badges
Both great answers so thank you!
– Pete900
Mar 27 at 9:43
add a comment |
2
You can use reduce from purrr and bind_rows (similar to rbind) from dplyr.
df_list <- df %>%
group_by(df_name) %>%
summarize(data = list(reduce(data, bind_rows)))
df_list
# A tibble: 2 x 2
# df_name data
# <chr> <list>
# 1 A <data.frame [450 x 5]>
# 2 B <data.frame [300 x 5]>
For the exact same structure as in your tapply-version we would need to add the following:
df_list2 <- df_list %>%
split(.$df_name) %>%
map(~ .x$data[[1]])
str(df_list2)
List of 2
$ A:'data.frame': 450 obs. of 5 variables:
..$ Sepal.Length: num [1:450] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:450] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:450] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:450] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
$ B:'data.frame': 300 obs. of 5 variables:
..$ Sepal.Length: num [1:300] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:300] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:300] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:300] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
answered Mar 27 at 9:40
kathkath
5,70411 silver badges27 bronze badges
Both great answers so thank you!
– Pete900
Mar 27 at 9:43
add a comment |
2
2
2
You can use reduce from purrr and bind_rows (similar to rbind) from dplyr.
df_list <- df %>%
group_by(df_name) %>%
summarize(data = list(reduce(data, bind_rows)))
df_list
# A tibble: 2 x 2
# df_name data
# <chr> <list>
# 1 A <data.frame [450 x 5]>
# 2 B <data.frame [300 x 5]>
For the exact same structure as in your tapply-version we would need to add the following:
df_list2 <- df_list %>%
split(.$df_name) %>%
map(~ .x$data[[1]])
str(df_list2)
List of 2
$ A:'data.frame': 450 obs. of 5 variables:
..$ Sepal.Length: num [1:450] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:450] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:450] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:450] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
$ B:'data.frame': 300 obs. of 5 variables:
..$ Sepal.Length: num [1:300] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:300] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:300] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:300] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
answered Mar 27 at 9:40
kathkath
5,70411 silver badges27 bronze badges
You can use reduce from purrr and bind_rows (similar to rbind) from dplyr.
df_list <- df %>%
group_by(df_name) %>%
summarize(data = list(reduce(data, bind_rows)))
df_list
# A tibble: 2 x 2
# df_name data
# <chr> <list>
# 1 A <data.frame [450 x 5]>
# 2 B <data.frame [300 x 5]>
For the exact same structure as in your tapply-version we would need to add the following:
df_list2 <- df_list %>%
split(.$df_name) %>%
map(~ .x$data[[1]])
str(df_list2)
List of 2
$ A:'data.frame': 450 obs. of 5 variables:
..$ Sepal.Length: num [1:450] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:450] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:450] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:450] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
$ B:'data.frame': 300 obs. of 5 variables:
..$ Sepal.Length: num [1:300] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
..$ Sepal.Width : num [1:300] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
..$ Petal.Length: num [1:300] 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
..$ Petal.Width : num [1:300] 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
..$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
answered Mar 27 at 9:40
kathkath
5,70411 silver badges27 bronze badges
edited Mar 27 at 9:57
answered Mar 27 at 9:40
kathkath
5,70411 silver badges27 bronze badges
answered Mar 27 at 9:40
kathkath
5,70411 silver badges27 bronze badges
answered Mar 27 at 9:40
kathkath
5,70411 silver badges27 bronze badges
5,70411 silver badges27 bronze badges
Both great answers so thank you!
– Pete900
Mar 27 at 9:43
add a comment |
Both great answers so thank you!
– Pete900
Mar 27 at 9:43
Both great answers so thank you!
– Pete900
Mar 27 at 9:43
Both great answers so thank you!
– Pete900
Mar 27 at 9:43
add a comment |
1
I would use unnest and group_split :
df %>% unnest(data) %>% group_split(df_name)
# [[1]]
# # A tibble: 450 x 6
# df_name Sepal.Length Sepal.Width Petal.Length Petal.Width Species
# <chr> <dbl> <dbl> <dbl> <dbl> <fct>
# 1 A 5.1 3.5 1.4 0.2 setosa
# 2 A 4.9 3 1.4 0.2 setosa
# 3 A 4.7 3.2 1.3 0.2 setosa
# 4 A 4.6 3.1 1.5 0.2 setosa
# 5 A 5 3.6 1.4 0.2 setosa
# 6 A 5.4 3.9 1.7 0.4 setosa
# 7 A 4.6 3.4 1.4 0.3 setosa
# 8 A 5 3.4 1.5 0.2 setosa
# 9 A 4.4 2.9 1.4 0.2 setosa
# 10 A 4.9 3.1 1.5 0.1 setosa
# # ... with 440 more rows
#
# [[2]]
# # A tibble: 300 x 6
# df_name Sepal.Length Sepal.Width Petal.Length Petal.Width Species
# <chr> <dbl> <dbl> <dbl> <dbl> <fct>
# 1 B 5.1 3.5 1.4 0.2 setosa
# 2 B 4.9 3 1.4 0.2 setosa
# 3 B 4.7 3.2 1.3 0.2 setosa
# 4 B 4.6 3.1 1.5 0.2 setosa
# 5 B 5 3.6 1.4 0.2 setosa
# 6 B 5.4 3.9 1.7 0.4 setosa
# 7 B 4.6 3.4 1.4 0.3 setosa
# 8 B 5 3.4 1.5 0.2 setosa
# 9 B 4.4 2.9 1.4 0.2 setosa
# 10 B 4.9 3.1 1.5 0.1 setosa
# # ... with 290 more rows
answered Mar 27 at 10:31
Moody_MudskipperMoody_Mudskipper
27.1k4 gold badges48 silver badges81 bronze badges
Thanks, I hadn't heard of group_split before.
– Pete900
Mar 27 at 11:09
It's fairly new :) tidyverse.org/articles/2019/02/dplyr-0-8-0
– Moody_Mudskipper
Mar 27 at 11:15
Although as-is the resulting list elements are not named but I can see how to set them.
– Pete900
Mar 27 at 11:17
1
fair point, actually the help says "it does not name the elements of the list based on the grouping as this typically loses information and is confusing.", which I think is a mistake and some users complained about it in the github issues already. The function is still marked as experimental so hopefully they'll change it.
– Moody_Mudskipper
Mar 27 at 11:27
add a comment |
1
I would use unnest and group_split :
df %>% unnest(data) %>% group_split(df_name)
# [[1]]
# # A tibble: 450 x 6
# df_name Sepal.Length Sepal.Width Petal.Length Petal.Width Species
# <chr> <dbl> <dbl> <dbl> <dbl> <fct>
# 1 A 5.1 3.5 1.4 0.2 setosa
# 2 A 4.9 3 1.4 0.2 setosa
# 3 A 4.7 3.2 1.3 0.2 setosa
# 4 A 4.6 3.1 1.5 0.2 setosa
# 5 A 5 3.6 1.4 0.2 setosa
# 6 A 5.4 3.9 1.7 0.4 setosa
# 7 A 4.6 3.4 1.4 0.3 setosa
# 8 A 5 3.4 1.5 0.2 setosa
# 9 A 4.4 2.9 1.4 0.2 setosa
# 10 A 4.9 3.1 1.5 0.1 setosa
# # ... with 440 more rows
#
# [[2]]
# # A tibble: 300 x 6
# df_name Sepal.Length Sepal.Width Petal.Length Petal.Width Species
# <chr> <dbl> <dbl> <dbl> <dbl> <fct>
# 1 B 5.1 3.5 1.4 0.2 setosa
# 2 B 4.9 3 1.4 0.2 setosa
# 3 B 4.7 3.2 1.3 0.2 setosa
# 4 B 4.6 3.1 1.5 0.2 setosa
# 5 B 5 3.6 1.4 0.2 setosa
# 6 B 5.4 3.9 1.7 0.4 setosa
# 7 B 4.6 3.4 1.4 0.3 setosa
# 8 B 5 3.4 1.5 0.2 setosa
# 9 B 4.4 2.9 1.4 0.2 setosa
# 10 B 4.9 3.1 1.5 0.1 setosa
# # ... with 290 more rows
answered Mar 27 at 10:31
Moody_MudskipperMoody_Mudskipper
27.1k4 gold badges48 silver badges81 bronze badges
Thanks, I hadn't heard of group_split before.
– Pete900
Mar 27 at 11:09
It's fairly new :) tidyverse.org/articles/2019/02/dplyr-0-8-0
– Moody_Mudskipper
Mar 27 at 11:15
Although as-is the resulting list elements are not named but I can see how to set them.
– Pete900
Mar 27 at 11:17
1
fair point, actually the help says "it does not name the elements of the list based on the grouping as this typically loses information and is confusing.", which I think is a mistake and some users complained about it in the github issues already. The function is still marked as experimental so hopefully they'll change it.
– Moody_Mudskipper
Mar 27 at 11:27
add a comment |
1
1
1
I would use unnest and group_split :
df %>% unnest(data) %>% group_split(df_name)
# [[1]]
# # A tibble: 450 x 6
# df_name Sepal.Length Sepal.Width Petal.Length Petal.Width Species
# <chr> <dbl> <dbl> <dbl> <dbl> <fct>
# 1 A 5.1 3.5 1.4 0.2 setosa
# 2 A 4.9 3 1.4 0.2 setosa
# 3 A 4.7 3.2 1.3 0.2 setosa
# 4 A 4.6 3.1 1.5 0.2 setosa
# 5 A 5 3.6 1.4 0.2 setosa
# 6 A 5.4 3.9 1.7 0.4 setosa
# 7 A 4.6 3.4 1.4 0.3 setosa
# 8 A 5 3.4 1.5 0.2 setosa
# 9 A 4.4 2.9 1.4 0.2 setosa
# 10 A 4.9 3.1 1.5 0.1 setosa
# # ... with 440 more rows
#
# [[2]]
# # A tibble: 300 x 6
# df_name Sepal.Length Sepal.Width Petal.Length Petal.Width Species
# <chr> <dbl> <dbl> <dbl> <dbl> <fct>
# 1 B 5.1 3.5 1.4 0.2 setosa
# 2 B 4.9 3 1.4 0.2 setosa
# 3 B 4.7 3.2 1.3 0.2 setosa
# 4 B 4.6 3.1 1.5 0.2 setosa
# 5 B 5 3.6 1.4 0.2 setosa
# 6 B 5.4 3.9 1.7 0.4 setosa
# 7 B 4.6 3.4 1.4 0.3 setosa
# 8 B 5 3.4 1.5 0.2 setosa
# 9 B 4.4 2.9 1.4 0.2 setosa
# 10 B 4.9 3.1 1.5 0.1 setosa
# # ... with 290 more rows
answered Mar 27 at 10:31
Moody_MudskipperMoody_Mudskipper
27.1k4 gold badges48 silver badges81 bronze badges
I would use unnest and group_split :
df %>% unnest(data) %>% group_split(df_name)
# [[1]]
# # A tibble: 450 x 6
# df_name Sepal.Length Sepal.Width Petal.Length Petal.Width Species
# <chr> <dbl> <dbl> <dbl> <dbl> <fct>
# 1 A 5.1 3.5 1.4 0.2 setosa
# 2 A 4.9 3 1.4 0.2 setosa
# 3 A 4.7 3.2 1.3 0.2 setosa
# 4 A 4.6 3.1 1.5 0.2 setosa
# 5 A 5 3.6 1.4 0.2 setosa
# 6 A 5.4 3.9 1.7 0.4 setosa
# 7 A 4.6 3.4 1.4 0.3 setosa
# 8 A 5 3.4 1.5 0.2 setosa
# 9 A 4.4 2.9 1.4 0.2 setosa
# 10 A 4.9 3.1 1.5 0.1 setosa
# # ... with 440 more rows
#
# [[2]]
# # A tibble: 300 x 6
# df_name Sepal.Length Sepal.Width Petal.Length Petal.Width Species
# <chr> <dbl> <dbl> <dbl> <dbl> <fct>
# 1 B 5.1 3.5 1.4 0.2 setosa
# 2 B 4.9 3 1.4 0.2 setosa
# 3 B 4.7 3.2 1.3 0.2 setosa
# 4 B 4.6 3.1 1.5 0.2 setosa
# 5 B 5 3.6 1.4 0.2 setosa
# 6 B 5.4 3.9 1.7 0.4 setosa
# 7 B 4.6 3.4 1.4 0.3 setosa
# 8 B 5 3.4 1.5 0.2 setosa
# 9 B 4.4 2.9 1.4 0.2 setosa
# 10 B 4.9 3.1 1.5 0.1 setosa
# # ... with 290 more rows
answered Mar 27 at 10:31
Moody_MudskipperMoody_Mudskipper
27.1k4 gold badges48 silver badges81 bronze badges
answered Mar 27 at 10:31
Moody_MudskipperMoody_Mudskipper
27.1k4 gold badges48 silver badges81 bronze badges
answered Mar 27 at 10:31
Moody_MudskipperMoody_Mudskipper
27.1k4 gold badges48 silver badges81 bronze badges
answered Mar 27 at 10:31
Moody_MudskipperMoody_Mudskipper
27.1k4 gold badges48 silver badges81 bronze badges
27.1k4 gold badges48 silver badges81 bronze badges
Thanks, I hadn't heard of group_split before.
– Pete900
Mar 27 at 11:09
It's fairly new :) tidyverse.org/articles/2019/02/dplyr-0-8-0
– Moody_Mudskipper
Mar 27 at 11:15
Although as-is the resulting list elements are not named but I can see how to set them.
– Pete900
Mar 27 at 11:17
1
fair point, actually the help says "it does not name the elements of the list based on the grouping as this typically loses information and is confusing.", which I think is a mistake and some users complained about it in the github issues already. The function is still marked as experimental so hopefully they'll change it.
– Moody_Mudskipper
Mar 27 at 11:27
add a comment |
Thanks, I hadn't heard of group_split before.
– Pete900
Mar 27 at 11:09
It's fairly new :) tidyverse.org/articles/2019/02/dplyr-0-8-0
– Moody_Mudskipper
Mar 27 at 11:15
Although as-is the resulting list elements are not named but I can see how to set them.
– Pete900
Mar 27 at 11:17
1
fair point, actually the help says "it does not name the elements of the list based on the grouping as this typically loses information and is confusing.", which I think is a mistake and some users complained about it in the github issues already. The function is still marked as experimental so hopefully they'll change it.
– Moody_Mudskipper
Mar 27 at 11:27
Thanks, I hadn't heard of group_split before.
– Pete900
Mar 27 at 11:09
Thanks, I hadn't heard of group_split before.
– Pete900
Mar 27 at 11:09
It's fairly new :) tidyverse.org/articles/2019/02/dplyr-0-8-0
– Moody_Mudskipper
Mar 27 at 11:15
It's fairly new :) tidyverse.org/articles/2019/02/dplyr-0-8-0
– Moody_Mudskipper
Mar 27 at 11:15
Although as-is the resulting list elements are not named but I can see how to set them.
– Pete900
Mar 27 at 11:17
Although as-is the resulting list elements are not named but I can see how to set them.
– Pete900
Mar 27 at 11:17
1
1
fair point, actually the help says "it does not name the elements of the list based on the grouping as this typically loses information and is confusing.", which I think is a mistake and some users complained about it in the github issues already. The function is still marked as experimental so hopefully they'll change it.
– Moody_Mudskipper
Mar 27 at 11:27
fair point, actually the help says "it does not name the elements of the list based on the grouping as this typically loses information and is confusing.", which I think is a mistake and some users complained about it in the github issues already. The function is still marked as experimental so hopefully they'll change it.
– Moody_Mudskipper
Mar 27 at 11:27
add a comment |
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