I have the following nested list called l (dput below):
> l
$A
$A$`1`
[1] 1 2 3
$A$`2`
[1] 3 2 1
$B
$B$`1`
[1] 2 2 2
$B$`2`
[1] 3 4 3
I would like to convert this to a grouped dataframe where A and B are the first group column and 1 and 2 are the subgroups with respective values. The desired output should look like this:
group subgroup values
1 A 1 1
2 A 1 2
3 A 1 3
4 A 2 3
5 A 2 2
6 A 2 1
7 B 1 2
8 B 1 2
9 B 1 2
10 B 2 3
11 B 2 4
12 B 2 3
As you can see A and B are the main group and 1 and 2 are the subgroups. Using purrr::flatten(l) or unnest doesn't work. So I was wondering if anyone knows how to convert a nested list to a grouped row dataframe?
dput of l:
l <- list(A = list(`1` = c(1, 2, 3), `2` = c(3, 2, 1)), B = list(`1` = c(2,
2, 2), `2` = c(3, 4, 3)))
Using stack and rowbind with id:
data.table::rbindlist(lapply(l, stack), idcol = "id")
# id values ind
# 1: A 1 1
# 2: A 2 1
# 3: A 3 1
# 4: A 3 2
# 5: A 2 2
# 6: A 1 2
# 7: B 2 1
# 8: B 2 1
# 9: B 2 1
# 10: B 3 2
# 11: B 4 2
# 12: B 3 2
You can use enframe() to convert the list into a data.frame, and unnest the value column twice.
library(tidyr)
tibble::enframe(l, name = "group") %>%
unnest_longer(value, indices_to = "subgroup") %>%
unnest(value)
# A tibble: 12 × 3
group value subgroup
<chr> <dbl> <chr>
1 A 1 1
2 A 2 1
3 A 3 1
4 A 3 2
5 A 2 2
6 A 1 2
7 B 2 1
8 B 2 1
9 B 2 1
10 B 3 2
11 B 4 2
12 B 3 2
Turn the list directly into a data frame, then pivot it into a long format and arrange to your desired order.
library(tidyverse)
lst %>%
as.data.frame() %>%
pivot_longer(everything(), names_to = c("group", "subgroup"),
values_to = "values",
names_pattern = "(.+?)\\.(.+?)") %>%
arrange(group, subgroup)
# A tibble: 12 × 3
group subgroup values
<chr> <chr> <dbl>
1 A 1 1
2 A 1 2
3 A 1 3
4 A 2 3
5 A 2 2
6 A 2 1
7 B 1 2
8 B 1 2
9 B 1 2
10 B 2 3
11 B 2 4
12 B 2 3
You can combine rrapply with unnest, which has the benefit to work in lists of arbitrary lengths:
library(rrapply)
library(tidyr)
rrapply(l, how = "melt") |>
unnest(value)
# A tibble: 12 × 3
L1 L2 value
<chr> <chr> <dbl>
1 A 1 1
2 A 1 2
3 A 1 3
4 A 2 3
5 A 2 2
6 A 2 1
7 B 1 2
8 B 1 2
9 B 1 2
10 B 2 3
11 B 2 4
12 B 2 3
Related
I have a data frame that looks like this :
a
b
c
1
2
10
2
2
10
3
2
10
4
2
10
5
2
10
I want to create a column with mutate function of something else under the dplyr framework of functions (or base) that will be sequence from b to c (i.e from 2 to 10 with length the number of rows of this tibble or data frame)
Ideally my new data frame I want to like like this :
a
b
c
c
1
2
10
2
2
2
10
4
3
2
10
6
4
2
10
8
5
2
10
10
How can I do this with R using dplyr ?
library(tidyverse)
n=5
a = seq(1,n,length.out=n)
b = rep(2,n)
c = rep(10,n)
data = tibble(a,b,c)
We may do
library(dplyr)
data %>%
rowwise %>%
mutate(new = seq(b, c, length.out = n)[a]) %>%
ungroup
-output
# A tibble: 5 × 4
a b c new
<dbl> <dbl> <dbl> <dbl>
1 1 2 10 2
2 2 2 10 4
3 3 2 10 6
4 4 2 10 8
5 5 2 10 10
If you want this done "by group" for each a value (creating many new rows), we can create the sequence as a list column and then unnest it:
data %>%
mutate(result = map2(b, c, seq, length.out = n)) %>%
unnest(result)
# # A tibble: 25 × 4
# a b c result
# <dbl> <dbl> <dbl> <dbl>
# 1 1 2 10 2
# 2 1 2 10 4
# 3 1 2 10 6
# 4 1 2 10 8
# 5 1 2 10 10
# 6 2 2 10 2
# 7 2 2 10 4
# 8 2 2 10 6
# 9 2 2 10 8
# 10 2 2 10 10
# # … with 15 more rows
# # ℹ Use `print(n = ...)` to see more rows
If you want to keep the same number of rows and go from the first b value to the last c value, we can use seq directly in mutate:
data %>%
mutate(result = seq(from = first(b), to = last(c), length.out = n()))
# # A tibble: 5 × 4
# a b c result
# <dbl> <dbl> <dbl> <dbl>
# 1 1 2 10 2
# 2 2 2 10 4
# 3 3 2 10 6
# 4 4 2 10 8
# 5 5 2 10 10
This one?
library(dplyr)
df %>%
mutate(c1 = a*b)
a b c c1
1 1 2 10 2
2 2 2 10 4
3 3 2 10 6
4 4 2 10 8
5 5 2 10 10
i've got some data in two columns:
# A tibble: 16 x 2
code niveau
<chr> <dbl>
1 A 1
2 1 2
3 2 2
4 3 2
5 4 2
6 5 2
7 B 1
8 6 2
9 7 2
My desired output is:
A tibble: 16 x 3
code niveau cat
<chr> <dbl> <chr>
1 A 1 A
2 1 2 A
3 2 2 A
4 3 2 A
5 4 2 A
6 5 2 A
7 B 1 B
8 6 2 B
I there a tidy way to convert these data without looping through it?
Here some dummy data:
data<-tibble(code=c('A', 1,2,3,4,5,'B', 6,7,8,9,'C',10,11,12,13), niveau=c(1, 2,2,2,2,2,1,2,2,2,2,1,2,2,2,2))
desired_output<-tibble(code=c('A', 1,2,3,4,5,'B', 6,7,8,9,'C',10,11,12,13), niveau=c(1, 2,2,2,2,2,1,2,2,2,2,1,2,2,2,2),
cat=c(rep('A', 6),rep('B', 5), rep('C', 5)))
Nicolas
Probably, you can create a new column cat and replace code values with NA where there is a number. We can then use fill to replace missing values with previous non-NA value.
library(dplyr)
data %>% mutate(cat = replace(code, grepl('\\d', code), NA)) %>% tidyr::fill(cat)
# A tibble: 16 x 3
# code niveau cat
# <chr> <dbl> <chr>
# 1 A 1 A
# 2 1 2 A
# 3 2 2 A
# 4 3 2 A
# 5 4 2 A
# 6 5 2 A
# 7 B 1 B
# 8 6 2 B
# 9 7 2 B
#10 8 2 B
#11 9 2 B
#12 C 1 C
#13 10 2 C
#14 11 2 C
#15 12 2 C
#16 13 2 C
We can use str_detect from stringr
library(dplyr)
library(stringr)
library(tidyr)
data %>%
mutate(cat = replace(code, str_detect(code, '\\d'), NA)) %>%
fill(cat)
Suppose a df with some missing values like this:
ID col_A_1 col_A_2 col_B_1 col_B_2
1 1 1 NA NA a
2 1 2 NA 1 b
3 1 3 1 2 c
4 1 4 2 3 d
5 1 NA 3 4 e
6 2 NA 1 5 f
7 2 NA 2 6 g
8 2 1 3 7 h
9 2 2 4 8 <NA>
10 2 3 5 NA <NA>
I want to fill the missing values using tidyr fill(), however, only the missing values in columns containing A.
I was able to achieve it using:
library(dplyr)
library(tidyr)
df %>%
group_by(ID) %>%
fill(names(.)[grepl("A", names(.))], .direction = "up") %>%
fill(names(.)[grepl("A", names(.))], .direction = "down") %>%
ungroup()
ID col_A_1 col_A_2 col_B_1 col_B_2
<dbl> <int> <int> <int> <chr>
1 1 1 1 NA a
2 1 2 1 1 b
3 1 3 1 2 c
4 1 4 2 3 d
5 1 4 3 4 e
6 2 1 1 5 f
7 2 1 2 6 g
8 2 1 3 7 h
9 2 2 4 8 <NA>
10 2 3 5 NA <NA>
however, I'm looking for other variable/column selection possibilities inside tidyr fill().
Sample data:
df <- data.frame(ID = c(rep(1, 5), rep(2, 5)),
col_A_1 = c(1:4, NA, NA, NA, 1:3),
col_A_2 = c(NA, NA, 1:3, 1:5),
col_B_1 = c(NA, 1:8, NA),
col_B_2 = c(letters[1:8], NA, NA),
stringsAsFactors = FALSE)
The fill can take select_helpers
library(tidyverse)
df %>%
group_by(ID) %>%
fill(matches('A'), .direction = 'up') %>%
fill(matches('A'), .direction = 'down')
# A tibble: 10 x 5
# Groups: ID [2]
# ID col_A_1 col_A_2 col_B_1 col_B_2
# <dbl> <int> <int> <int> <chr>
# 1 1 1 1 NA a
# 2 1 2 1 1 b
# 3 1 3 1 2 c
# 4 1 4 2 3 d
# 5 1 4 3 4 e
# 6 2 1 1 5 f
# 7 2 1 2 6 g
# 8 2 1 3 7 h
# 9 2 2 4 8 <NA>
#10 2 3 5 NA <NA>
I have two lists. Each of them with many vectors (around 500) of different lengths and I would like to get a tibble data frame with three columns.
My reproducible example is the following:
> a
[[1]]
[1] 1 3 6
[[2]]
[1] 5 4
> b
[[1]]
[1] 3 4
[[2]]
[1] 5 6 7
I would like to get the following tibble data frame:
name index value
a 1 1
a 1 3
a 1 6
a 2 5
a 2 4
b 1 3
b 1 4
b 2 5
b 2 6
b 2 7
I would be grateful if someone could help me with this issue
using Base R:
transform(stack(c(a=a,b=b)),name=substr(ind,1,1),ind=substr(ind,2,2))
values ind name
1 1 1 a
2 2 1 a
3 3 1 a
4 5 2 a
5 6 2 a
6 3 1 b
7 4 1 b
8 5 2 b
9 6 2 b
10 7 2 b
using tidyverse:
library(tidyverse)
list(a=a,b=b)%>%map(~stack(setNames(.x,1:length(.x))))%>%bind_rows(.id = "name")
name values ind
1 a 1 1
2 a 2 1
3 a 3 1
4 a 5 2
5 a 6 2
6 b 3 1
7 b 4 1
8 b 5 2
9 b 6 2
10 b 7 2
Here is one option with tidyverse
library(tidyverse)
list(a= a, b = b) %>%
map_df(enframe, name = "index", .id = 'name') %>%
unnest
# A tibble: 10 x 3
# name index value
# <chr> <int> <dbl>
# 1 a 1 1
# 2 a 1 3
# 3 a 1 6
# 4 a 2 5
# 5 a 2 4
# 6 b 1 3
# 7 b 1 4
# 8 b 2 5
# 9 b 2 6
#10 b 2 7
data
a <- list(c(1, 3, 6), c(5, 4))
b <- list(c(3, 4), c(5, 6, 7))
Suppose I have a data frame:
df <- data.frame(SID=sample(1:4,15,replace=T), Var1=c(rep("A",5),rep("B",5),rep("C",5)), Var2=sample(2:4,15,replace=T))
which comes out to something like this:
SID Var1 Var2
1 4 A 2
2 3 A 2
3 4 A 3
4 3 A 3
5 1 A 4
6 1 B 2
7 3 B 2
8 4 B 4
9 4 B 4
10 3 B 2
11 2 C 2
12 2 C 2
13 4 C 4
14 2 C 4
15 3 C 3
What I hope to accomplish is to find the count of unique SIDs (see below under update, this should have said count of unique (SID, Var1) combinations) where the given row's Var1 is excluded from this count and the count is grouped on Var2. So for the example above, I would like to output:
SID Var1 Var2 Count.Excluding.Var1
1 4 A 2 3
2 3 A 2 3
3 4 A 3 1
4 3 A 3 1
5 1 A 4 3
6 1 B 2 3
7 3 B 2 3
8 4 B 4 3
9 4 B 4 3
10 3 B 2 3
11 2 C 2 4
12 2 C 2 4
13 4 C 4 2
14 2 C 4 2
15 3 C 3 2
For the 1st observation, we have a count of 3 because there are 3 unique combinations of (SID, Var1) for the given Var2 value (2, in this case) where Var1 != A (Var1 value of 1st observation) -- specifically, the count includes observation 6, 7 and 11, but not 12 because we already accounted for a (SID, Var1)=(2,C) and not row 2 because we do not want Var1 to be "A". All of these rows have the same Var2 value.
I'd preferably like to use dplyr functions and the %>% operator.
&
UPDATE
I apologize for the confusion and my incorrect explanation above. I have corrected what I intended on asking for in the paranthesis, but I am leaving my original phrasing as well because majority of answers seem to interpret it this way.
As for the example, I apologize for not setting the seed. There seems to have been some confusion with regards to the Count.Excluding.Var1 for rows 11 and 12. With unique (SID, Var1) combinations, rows 11 and 12 should make sense as these count rows 1,2,6, and 7 xor 8.
A simple mapply can do the trick. But as OP requested for %>% based solution, an option could be as:
df %>% mutate(Count.Excluding.Var1 =
mapply(function(x,y)nrow(unique(df[df$Var1 != x & df$Var2 == y,1:2])),.$Var1,.$Var2))
# SID Var1 Var2 Count.Excluding.Var1
# 1 4 A 2 3
# 2 2 A 3 3
# 3 4 A 4 3
# 4 4 A 4 3
# 5 3 A 4 3
# 6 4 B 3 1
# 7 3 B 3 1
# 8 3 B 3 1
# 9 4 B 2 3
# 10 2 B 3 1
# 11 2 C 2 2
# 12 4 C 4 2
# 13 1 C 4 2
# 14 1 C 2 2
# 15 3 C 4 2
Data:
The above results are based on origional data provided by OP.
df <- data.frame(SID=sample(1:4,15,replace=T), Var1=c(rep("A",5),rep("B",5),rep("C",5)), Var2=sample(2:4,15,replace=T))
could not think of a dplyr solution, but here's one with apply
df$Count <- apply(df, 1, function(x) length(unique(df$SID[(df$Var1 != x['Var1']) & (df$Var2 == x['Var2'])])))
# SID Var1 Var2 Count
# 1 4 A 2 3
# 2 3 A 2 3
# 3 4 A 3 1
# 4 3 A 3 1
# 5 1 A 4 2
# 6 1 B 2 3
# 7 3 B 2 3
# 8 4 B 4 3
# 9 4 B 4 3
# 10 3 B 2 3
# 11 2 C 2 3
# 12 2 C 2 3
# 13 4 C 4 2
# 14 2 C 4 2
# 15 3 C 3 2
Here is a dplyr solution, as requested. For future reference, please use set.seed so we can reproduce your desired output with sample, else I have to enter data by hand...
I think this is your logic? You want the n_distinct(SID) for each Var2, but for each row, you want to exclude rows which have the same Var1 as the current row. So a key observation here is row 3, where a simple grouped summarise would yield a count of 2. Of the rows with Var2 = 3, row 3 has SID = 4, row 4 has SID = 3, row 15 has SID = 3, but we don't count row 3 or row 4, so final count is one unique SID.
Here we get first the count of unique SID for each Var2, then the count of unique SID for each Var1, Var2 combo. First count is too large by the amount of additional unique SID for each combo, so we subtract it and add one. There is an edge case where for a Var1, there is only one corresponding Var2. This should return 0 since you exclude all the possible values of SID. I added two rows to illustrate this.
library(tidyverse)
df <- read_table2(
"SID Var1 Var2
4 A 2
3 A 2
4 A 3
3 A 3
1 A 4
1 B 2
3 B 2
4 B 4
4 B 4
3 B 2
2 C 2
2 C 2
4 C 4
2 C 4
3 C 3
1 D 5
2 D 5"
)
df %>%
group_by(Var2) %>%
mutate(SID_per_Var2 = n_distinct(SID)) %>%
group_by(Var1, Var2) %>%
mutate(SID_per_Var1Var2 = n_distinct(SID)) %>%
ungroup() %>%
add_count(Var1) %>%
add_count(Var1, Var2) %>%
mutate(
Count.Excluding.Var1 = if_else(
n > nn,
SID_per_Var2 - SID_per_Var1Var2 + 1,
0
)
) %>%
select(SID, Var1, Var2, Count.Excluding.Var1)
#> # A tibble: 17 x 4
#> SID Var1 Var2 Count.Excluding.Var1
#> <int> <chr> <int> <dbl>
#> 1 4 A 2 3.
#> 2 3 A 2 3.
#> 3 4 A 3 1.
#> 4 3 A 3 1.
#> 5 1 A 4 3.
#> 6 1 B 2 3.
#> 7 3 B 2 3.
#> 8 4 B 4 3.
#> 9 4 B 4 3.
#> 10 3 B 2 3.
#> 11 2 C 2 4.
#> 12 2 C 2 4.
#> 13 4 C 4 2.
#> 14 2 C 4 2.
#> 15 3 C 3 2.
#> 16 1 D 5 0.
#> 17 2 D 5 0.
Created on 2018-04-12 by the reprex package (v0.2.0).
Here's a solution using purrr - you can wrap this in a mutate statement if you want, but I don't know that it adds much in this particular case.
library(purrr)
df$Count.Excluding.Var1 = map_int(1:nrow(df), function(n) {
df %>% filter(Var2 == Var2[n], Var1 != Var1[n]) %>% distinct() %>% nrow()
})
(Updated with input from comments by Calum You. Thanks!)
A 100% tidyverse solution:
library(tidyverse) # dplyr + purrr
df %>%
group_by(Var2) %>%
mutate(count = map_int(Var1,~n_distinct(SID[.x!=Var1],Var1[.x!=Var1])))
# # A tibble: 15 x 4
# # Groups: Var2 [3]
# SID Var1 Var2 count
# <int> <chr> <int> <int>
# 1 4 A 2 3
# 2 3 A 2 3
# 3 4 A 3 1
# 4 3 A 3 1
# 5 1 A 4 3
# 6 1 B 2 3
# 7 3 B 2 3
# 8 4 B 4 3
# 9 4 B 4 3
# 10 3 B 2 3
# 11 2 C 2 4
# 12 2 C 2 4
# 13 4 C 4 2
# 14 2 C 4 2
# 15 3 C 3 2