I would like to reassign a given records to a single group if the records are duplicated. In the below dataset I would like to to have 12-4 all being assigned to group A or B but not both. Is there a way to go abou it?
library(tidyverse)
dat <- tibble(
group = c("A", "A", "A", "A", "B", "B", "B", "B", "B"),
assigned = c("12-1", "12-2", "12-3", "12-4", "12-4", "12-5", "12-6",
"12-7", "12-8")
)
# Attempts to tease out records for each group
dat %>% pivot_wider(names_from = group, values_from = assigned)
You can group by record and reassign all to the same group, chosen at random from the available groups:
dat %>%
group_by(assigned) %>%
mutate(group = nth(group, sample(n())[1])) %>%
ungroup()
#> # A tibble: 9 x 2
#> group assigned
#> <chr> <chr>
#> 1 A 12-1
#> 2 A 12-2
#> 3 A 12-3
#> 4 A 12-4
#> 5 A 12-4
#> 6 B 12-5
#> 7 B 12-6
#> 8 B 12-7
#> 9 B 12-8
library(tidyverse)
dat <- tibble(
group = c("A", "A", "A", "A", "B", "B", "B", "B", "B"),
assigned = c(
"12-1", "12-2", "12-3", "12-4", "12-4", "12-5", "12-6",
"12-7", "12-8"
)
)
dat %>%
select(-group) %>%
left_join(
dat %>%
left_join(dat %>% count(group)) %>%
# reassign to the smallest group
arrange(n) %>%
select(-n) %>%
distinct(assigned, .keep_all = TRUE)
)
#> Joining, by = "group"
#> Joining, by = "assigned"
#> # A tibble: 9 × 2
#> assigned group
#> <chr> <chr>
#> 1 12-1 A
#> 2 12-2 A
#> 3 12-3 A
#> 4 12-4 A
#> 5 12-4 A
#> 6 12-5 B
#> 7 12-6 B
#> 8 12-7 B
#> 9 12-8 B
Created on 2022-04-04 by the reprex package (v2.0.0)
Related
I have a tibble which resembles the following:
data<-tibble(ref=c("ABC", "ABC", "XYZ", "XYZ", "FGH", "FGH", "FGH"),
type=c("A", "B", "A", "A", "A", "A", "B"))
ref type
1 ABC A
2 ABC B
3 XYZ A
4 XYZ A
5 FGH A
6 FGH A
7 FGH B
I need to group by ref and if--within a group--type B is present, return that row, else default to return any row (but only 1 row) of type A.
Expected output:
ref type
1 ABC B
2 XYZ A
3 FGH B
with large amounts of data, it is better to do sorting before grouping
tidyverse
library(tidyverse)
df<-tibble(ref=c("ABC", "ABC", "XYZ", "XYZ", "FGH", "FGH", "FGH"),
type=c("A", "B", "A", "A", "A", "A", "B"))
distinct(df) %>%
arrange(ref, desc(type)) %>%
group_by(ref) %>%
slice_head(n = 1) %>%
ungroup()
#> # A tibble: 3 × 2
#> ref type
#> <chr> <chr>
#> 1 ABC B
#> 2 FGH B
#> 3 XYZ A
data.table
Created on 2022-04-27 by the reprex package (v2.0.1)
df<-data.frame(ref=c("ABC", "ABC", "XYZ", "XYZ", "FGH", "FGH", "FGH"),
type=c("A", "B", "A", "A", "A", "A", "B"))
library(data.table)
setDT(df)[order(ref, -type), .SD[1], by = ref]
#> ref type
#> 1: ABC B
#> 2: FGH B
#> 3: XYZ A
Created on 2022-04-27 by the reprex package (v2.0.1)
If you only have A and B, then you can arrange and simply get the first row, i.e.
library(dplyr)
data %>%
group_by(ref) %>%
filter(type %in% c('A', 'B')) %>% #If other types exist
arrange(desc(type)) %>%
slice(1L)
# A tibble: 3 x 2
# Groups: ref [3]
ref type
<chr> <chr>
1 ABC B
2 FGH B
3 XYZ A
We can use which.max over boolean to extract the desired rows
data %>%
group_by(ref) %>%
slice(which.max(type == "B")) %>%
ungroup()
which gives
# A tibble: 3 x 2
ref type
<chr> <chr>
1 ABC B
2 FGH B
3 XYZ A
I have the following dataset:
df <- structure(list(var = c("a", "a", "a", "a", "a", "a", "a", "a",
"a", "a", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b"),
beta_2 = c(-0.0441739987111475, -0.237256549142376, -0.167105040977351,
-0.140660549127359, -0.0623609020878716, -0.279740636040755,
-0.0211523654970921, 0.135368375550385, -0.0612770247281429,
-0.13183964102725, 0.363736380163624, -0.0134490092107583,
-0.0179957210095045, -0.00897746346470879, -0.0588242539401108,
-0.0571976057977875, -0.0290052449275881, 0.263181562031473,
0.00398338217426211, 0.0945495450635497), beta_3 = c(8.54560737016843e-05,
-0.0375859675101865, -0.0334219898732454, 0.0332275634691021,
6.41499442849741e-05, -0.0200724300602369, 8.046644459034e-05,
0.0626880671346749, 0.066218613897726, 0.0101268565262127,
0.44671567722757, 0.180543425234781, 0.526177616390516, 0.281245231195401,
-0.0362628519010746, 0.0609803646123324, 0.104137160504616,
0.804375133555955, 0.211218123083386, 0.824756942938928),
beta_4 = c(-8.50289708803184e-06, 0.0376601781861706, 0.104418586040791,
-0.0949557776511923, 2.11896613386966e-05, 0.0969765824620132,
4.95280289930771e-06, -0.0967836292162074, -0.132623370126544,
0.0579395551175153, -0.140392004360494, 0.00950912868877355,
-0.388317615535003, -0.0282634228070272, 0.0547116932731301,
0.0119441792873249, -0.0413015877795695, -0.720387490330028,
-0.0321860166581817, -0.627489324697221)), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -20L))
df
# # A tibble: 20 × 4
# var beta_2 beta_3 beta_4
# <chr> <dbl> <dbl> <dbl>
# 1 a -0.0442 0.0000855 -0.00000850
# 2 a -0.237 -0.0376 0.0377
# 3 a -0.167 -0.0334 0.104
# 4 a -0.141 0.0332 -0.0950
# 5 a -0.0624 0.0000641 0.0000212
# ...
I would like to summarise each beta_ column grouped by var so that I have the mean of beta_2, beta_3 and beta_4 in a single cell.
I can do it with the following code:
df %>%
pivot_longer(!var) %>%
group_by(var, name) %>%
summarise(mean_beta = mean(value) %>% round(2), .groups = "drop") %>%
aggregate(mean_beta ~ var, ., function(x) paste0(x, collapse = ", ")) %>%
as_tibble()
# # A tibble: 2 × 2
# var mean_beta
# <chr> <chr>
# 1 a -0.1, 0.01, 0
# 2 b 0.05, 0.34, -0.19
I'm looking for a more straightforward, tidyverse-only solution. I have tried using map inside summarise but couldn't get what I wanted. Any idea?
You may do the following -
library(dplyr)
df %>%
group_by(var) %>%
summarise(mean_beta = cur_data() %>%
summarise(across(.fns =
~.x %>% mean(na.rm = TRUE) %>% round(2))) %>%
unlist() %>% toString())
# var mean_beta
# <chr> <chr>
#1 a -0.1, 0.01, 0
#2 b 0.05, 0.34, -0.19
cur_data() provides the sub-data within each group as dataframe that can be summarised for each column and concatenated together.
Another possible solution:
library(tidyverse)
df %>%
group_by(var) %>%
summarise(across(everything(), mean)) %>%
{bind_cols(var=.$var, mean_betas=apply(., 1, \(x) str_c(x[-1], collapse = ", ")))}
#> # A tibble: 2 × 2
#> var mean_betas
#> <chr> <chr>
#> 1 a "-0.10101983, 0.008141079, -0.002735024"
#> 2 b " 0.05400016, 0.340388682, -0.190217246"
This question already has answers here:
Getting the top values by group
(6 answers)
Closed 1 year ago.
I have this data:
df <- data.frame(
node = c("A", "B", "A", "A", "A", "B", "A", "A", "A", "B", "B", "B", "B"),
left = c("ab", "ab", "ab", "ab", "cc", "xx", "cc", "ab", "zz", "xx", "xx", "zz", "zz")
)
I want to count grouped frequencies and proportions and slice/filter out a sequence of grouped rows. Say, given the small dataset, I want to have the rows with the two highest Freq_left values per group. How can that be done? I can only extract the rows with the maximum Freq_left values but not the desired sequence of rows:
df %>%
group_by(node, left) %>%
# summarise
summarise(
Freq_left = n(),
Prop_left = round(Freq_left/nrow(.)*100, 4)
) %>%
slice_max(Freq_left)
# A tibble: 2 × 4
# Groups: node [2]
node left Freq_left Prop_left
<chr> <chr> <int> <dbl>
1 A ab 4 30.8
2 B xx 3 23.1
Expected output:
node left Freq_left Prop_left
<chr> <chr> <int> <dbl>
A ab 4 30.8
A cc 2 15.4
B xx 3 23.1
B zz 2 15.4
You could use dplyr::top_n or dplyr::slice_max:
Thanks to #PaulSmith for pointing out that dplyr::top_n is superseded in favor of dplyr::slice_max:
library(dplyr)
df %>%
group_by(node, left) %>%
# summarise
summarise(
Freq_left = n(),
Prop_left = round(Freq_left/nrow(.)*100, 4)
) %>%
slice_max(order_by = Prop_left, n = 2)
#> `summarise()` has grouped output by 'node'. You can override using the `.groups` argument.
#> # A tibble: 4 × 4
#> # Groups: node [2]
#> node left Freq_left Prop_left
#> <chr> <chr> <int> <dbl>
#> 1 A ab 4 30.8
#> 2 A cc 2 15.4
#> 3 B xx 3 23.1
#> 4 B zz 2 15.4
I'm taking the mean, 3 by 3, by grouping. For that, I'm using the summarise function. In this context I would like to select the last date from the four that make up the average.
I tried to select the maximum, but this way I'm just selecting the highest date for the whole group.
test = data.frame(my_groups = c("A", "A", "A", "B", "B", "C", "C", "C", "A", "A", "A"),
measure = c(10, 20, 5, 2, 62 ,2, 5, 4, 6, 7, 25),
time= c("20-09-2020", "25-09-2020", "19-09-2020", "20-05-2020", "20-06-2021",
"11-01-2021", "13-01-2021", "13-01-2021", "15-01-2021", "15-01-2021", "19-01-2021"))
# > test
# my_groups measure time
# 1 A 10 20-09-2020
# 2 A 20 25-09-2020
# 3 A 5 19-09-2020
# 4 B 2 20-05-2020
# 5 B 62 20-06-2021
# 6 C 2 11-01-2021
# 7 C 5 13-01-2021
# 8 C 4 13-01-2021
# 9 A 6 15-01-2021
# 10 A 7 15-01-2021
# 11 A 25 19-01-2021
test %>%
arrange(time) %>%
group_by(my_groups) %>%
summarise(mean_3 = rollapply(measure, 3, mean, by = 3, align = "left", partial = F),
final_data = max(time))
# my_groups mean_3 final_data
# <chr> <dbl> <chr>
# 1 A 12.7 25-09-2020
# 2 A 11.7 25-09-2020
# 3 C 3.67 13-01-2021
In the second line I wish the date was 19-01-2021, and not the global maximum of group A, (25-09-2020).
Any hint on how I could do that?
I have 2 dplyr ways for you. Not happy with it because when the rollapply with max and dates doesn't find anything it in group B it uses a double by default which doesn't match the characters from group A and C.
Mutate:
test %>%
arrange(time) %>%
group_by(my_groups) %>%
mutate(final = rollapply(time, 3, max, by = 3, fill = NA, align = "left", partial = F),
mean_3 = rollapply(measure, 3, mean, by = 3, fill = NA, align = "left", partial = F)) %>%
filter(!is.na(final)) %>%
select(my_groups, final, mean_3) %>%
arrange(my_groups)
# A tibble: 3 x 3
# Groups: my_groups [2]
my_groups final mean_3
<chr> <chr> <dbl>
1 A 19-01-2021 12.7
2 A 25-09-2020 11.7
3 C 13-01-2021 3.67
Summarise that doesn't summarise, but is a bit cleaner in code:
test %>%
arrange(time) %>%
group_by(my_groups) %>%
summarise(final = rollapply(time, 3, max, by = 3, fill = NA, align = "left", partial = F),
mean_3 = rollapply(measure, 3, mean, by = 3, fill = NA, align = "left", partial = F)) %>%
filter(!is.na(final))
`summarise()` has grouped output by 'my_groups'. You can override using the `.groups` argument.
# A tibble: 3 x 3
# Groups: my_groups [2]
my_groups final mean_3
<chr> <chr> <dbl>
1 A 19-01-2021 12.7
2 A 25-09-2020 11.7
3 C 13-01-2021 3.67
Edit:
Added isa's solution from comment. Partial = TRUE does the trick:
test %>%
arrange(time) %>%
group_by(my_groups) %>%
summarise(mean_3 = rollapply(measure, 3, mean, by = 3, align = "left", partial = F),
final_data = rollapply(time, 3, max, by = 3, align = "left", partial = T))
`summarise()` has grouped output by 'my_groups'. You can override using the `.groups` argument.
# A tibble: 3 x 3
# Groups: my_groups [2]
my_groups mean_3 final_data
<chr> <dbl> <chr>
1 A 12.7 19-01-2021
2 A 11.7 25-09-2020
3 C 3.67 13-01-2021
Another possible solution:
library(tidyverse)
test = data.frame(my_groups = c("A", "A", "A", "B", "B", "C", "C", "C", "A", "A", "A"),
measure = c(10, 20, 5, 2, 62 ,2, 5, 4, 6, 7, 25),
time= c("20-09-2020", "25-09-2020", "19-09-2020", "20-05-2020", "20-06-2021",
"11-01-2021", "13-01-2021", "13-01-2021", "15-01-2021", "15-01-2021", "19-01-2021"))
test %>%
group_by(data.table::rleid(my_groups)) %>%
filter(n() == 3) %>%
summarise(
groups = unique(my_groups),
mean_3 = mean(measure), final_data = max(time), .groups = "drop") %>%
select(-1)
#> # A tibble: 3 × 3
#> groups mean_3 final_data
#> <chr> <dbl> <chr>
#> 1 A 11.7 25-09-2020
#> 2 C 3.67 13-01-2021
#> 3 A 12.7 19-01-2021
EDIT
To allow for calculation of mean of 2 values, as asked for in a comment below by the OP, I revised my code, using data.table::frollmean and data.table::frollapply:
library(tidyverse)
library(lubridate)
library(data.table)
n <- 2 # choose the number with which to calculate the mean
test %>%
group_by(rleid(my_groups)) %>%
summarise(
groups = unique(my_groups),
mean_n = frollmean(measure, n), final_data = frollapply(dmy(time), n, max) %>%
as_date(origin = lubridate::origin), .groups = "drop") %>%
drop_na(mean_n) %>% select(-1)
#> # A tibble: 7 × 3
#> groups mean_n final_data
#> <chr> <dbl> <date>
#> 1 A 15 2020-09-25
#> 2 A 12.5 2020-09-25
#> 3 B 32 2021-06-20
#> 4 C 3.5 2021-01-13
#> 5 C 4.5 2021-01-13
#> 6 A 6.5 2021-01-15
#> 7 A 16 2021-01-19
The code below creates a simplified version of the dataframe and illustrates my desired end result (df_wider) based on the unnested version. My question is: How can I achieve the same end result (df_wider) from the nested version (nested_df), using purrr?
library(tidyverse)
df <- tibble(id_01 = c(rep("01", 3), rep("02", 3)),
a = (c("a", "a", "b", "c", "c", "d")),
b = letters[7:12],
id_02 = rep(c(1, 2, 1), 2)
)
df_wider <- pivot_wider(df,
id_cols = c(id_01, a),
names_from = id_02,
values_from = b,
names_sep = "_"
)
nested_df <- nest(df, data = -id_01)
To be clear, I am trying to pivot while the dataframes are nested (i.e., before unnesting).
We can use purrr::map() within dplyr::mutate():
library(tidyverse)
df <- tibble(
id_01 = c(rep("01", 3), rep("02", 3)),
a = (c("a", "a", "b", "c", "c", "d")),
b = letters[7:12],
id_02 = rep(c(1, 2, 1), 2)
)
nested_df <- df %>%
nest(data = -id_01) %>%
mutate(data = map(data, ~ .x %>%
pivot_wider(
id_cols = a,
names_from = id_02,
values_from = b
)))
nested_df
#> # A tibble: 2 x 2
#> id_01 data
#> <chr> <list>
#> 1 01 <tibble [2 x 3]>
#> 2 02 <tibble [2 x 3]>
nested_df %>%
unnest(data)
#> # A tibble: 4 x 4
#> id_01 a `1` `2`
#> <chr> <chr> <chr> <chr>
#> 1 01 a g h
#> 2 01 b i <NA>
#> 3 02 c j k
#> 4 02 d l <NA>
Created on 2021-03-26 by the reprex package (v1.0.0)