I love the new tidyr pivot_wider function but since it hasn't been officially added to the CRAN package I was wondering how to convert the following code into the older spread() function (I do not have access to the server to DL tidyr from github)
test <- data.frame(x = c(1,1,2,2,2,2,3,3,3,4),
y = c(rep("a", 5), rep("b", 5)))
test %>%
count(x, y) %>%
group_by(x) %>%
mutate(prop = prop.table(n)) %>%
mutate(v1 = paste0(n, ' (', round(prop, 2), ')')) %>%
pivot_wider(id_cols = x, names_from = y, values_from = v1)
Desired Output:
# A tibble: 4 x 3
# Groups: x [4]
x a b
<dbl> <chr> <chr>
1 1 2 (1) NA
2 2 3 (0.75) 1 (0.25)
3 3 NA 3 (1)
4 4 NA 1 (1)
I tried (but is not quite right):
test %>%
count(x, y) %>%
group_by(x) %>%
mutate(prop = prop.table(n)) %>%
mutate(v1 = paste0(n, ' (', round(prop, 2), ')')) %>%
spread(y, v1) %>%
select(-n, -prop)
Any help appreciated!
One option is to remove the columns 'n', 'prop' before the spread statement as including them would create unique rows with that column values as well
library(dplyr)
library(tidyr)
test %>%
count(x, y) %>%
group_by(x) %>%
mutate(prop = prop.table(n)) %>%
mutate(v1 = paste0(n, ' (', round(prop, 2), ')')) %>%
select(-n, -prop) %>%
spread(y, v1)
# A tibble: 4 x 3
# Groups: x [4]
# x a b
# <dbl> <chr> <chr>
#1 1 2 (1) <NA>
#2 2 3 (0.75) 1 (0.25)
#3 3 <NA> 3 (1)
#4 4 <NA> 1 (1)
Or using base R
tbl <- table(test)
tbl[] <- paste0(tbl, "(", prop.table(tbl, 1), ")")
You can use data.table package:
> library(data.table)
> setDT(test)[,.(n=.N),by=.(x,y)][,.(y=y,n=n,final=gsub('\\(1\\)','',paste0(n,'(',round(prop.table(n),2), ')'))),by=x]
x y n final
1: 1 a 2 2
2: 2 a 3 3(0.75)
3: 2 b 1 1(0.25)
4: 3 b 3 3
5: 4 b 1 1
Related
I can concatenate one column of data.frame, following the code as below if the column name is available.
However, How about the "column" name saved in the variable?
Further question is, how can I specify the columns by the value of a variable? (!!sym() ?)
Here are test code:
> library(dplyr)
> packageVersion("dplyr")
[1] ‘1.0.7’
> df <- data.frame(x = 1:3, y = c("A", "B", "A"))
> df %>%
group_by(y) %>%
summarise(z = paste(x, collapse = ","))
# A tibble: 2 x 2
y z
<chr> <chr>
1 A 1,3
2 B 2
I have a variable a, with the value x, How can I do above summarize?
> a <- "x"
> df %>%
group_by(y) %>%
summarise(z = paste(a, collapse = ","))
# A tibble: 2 x 2
y z
<chr> <chr>
1 A x
2 B x
Solution-1: use !!sym()
> a <- "x"
> df %>%
group_by(y) %>%
summarise(z = paste(!!sym(a), collapse = ","))
# A tibble: 2 x 2
y z
<chr> <chr>
1 A 1,3
2 B 2
Solution-2: Assign the column to new variable
> df %>%
group_by(y) %>%
rename(new_col = a) %>%
summarise(z = paste(new_col, collapse = ","))
# A tibble: 2 x 2
y z
<chr> <chr>
1 A 1,3
2 B 2
Are there any other ways to do the job?
similar questions could be found: https://stackoverflow.com/a/15935166/2530783 ,https://stackoverflow.com/a/50537209/2530783,
Here are some other options -
Use .data -
library(dplyr)
a <- "x"
df %>% group_by(y) %>% summarise(z = toString(.data[[a]]))
# y z
# <chr> <chr>
#1 A 1, 3
#2 B 2
get
df %>% group_by(y) %>% summarise(z = toString(get(a)))
as.name
df %>% group_by(y) %>% summarise(z = toString(!!as.name(a)))
paste(..., collapse = ',') is equivalent to toString.
As an example, I have the following data frame:
df <- data.frame(a1=1,a2=2,a3=3,b1=1,b2=2,b3=3)
I have a function:
fn <- function(x,y,z) x^y+(z-x)^(y-x)
I want the following:
df <- df %>% mutate(a=fn(a1,a2,a3),b=fn(b1,b2,b3))
The problem is, I have tons of triplets in my dataset, so it is not ideal to write them out one by one.
Here are base R options using:
split.default + lapply + do.call
cbind(
df,
lapply(
split.default(df, gsub("\\d+", "", names(df))),
function(x) do.call(fn, unname(x))
)
)
reshape + lapply + do.call
cbind(
df,
lapply(
subset(
reshape(
setNames(df, gsub("(\\d+)$", "\\.\\1", names(df))),
direction = "long",
varying = 1:length(df)
),
select = -c(time, id)
),
function(x) do.call(fn, as.list(x))
)
)
Output
a1 a2 a3 b1 b2 b3 a b
1 1 2 3 1 2 3 3 3
I would convert df to long format then use lag to create 3 columns then apply fn() on them
library(tidyverse)
df_long <- df %>%
pivot_longer(everything(),
names_to = c(".value", "set"),
names_pattern = "(.)(.)")
df_longer <- df_long %>%
pivot_longer(-c(set),
names_to = "key",
values_to = "val") %>%
arrange(key)
df_longer
#> # A tibble: 6 x 3
#> set key val
#> <chr> <chr> <dbl>
#> 1 1 a 1
#> 2 2 a 2
#> 3 3 a 3
#> 4 1 b 1
#> 5 2 b 2
#> 6 3 b 3
lag then apply fn(), keep only non-NA val_fn
df_longer <- df_longer %>%
group_by(key) %>%
mutate(val_lag1 = lag(val, n = 1),
val_lag2 = lag(val, n = 2)) %>%
mutate(val_fn = fn(val_lag2, val_lag1, val)) %>%
filter(!is.na(val_fn))
df_longer
#> # A tibble: 2 x 6
#> # Groups: key [2]
#> set key val val_lag1 val_lag2 val_fn
#> <chr> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 3 a 3 2 1 3
#> 2 3 b 3 2 1 3
Created on 2020-12-03 by the reprex package (v0.3.0)
I think it would be easier/shorter to combine columns into their separate group and apply the function to each column.
library(dplyr)
library(tidyr)
df %>%
pivot_longer(cols = everything(),
names_to = '.value',
names_pattern = '([a-z]+)') %>%
summarise(across(.fns = ~do.call(fn, as.list(.)))) -> result
result
# a b
# <dbl> <dbl>
#1 3 3
You can bind the result to your original dataset if needed.
bind_cols(df, result)
# a1 a2 a3 b1 b2 b3 a b
#1 1 2 3 1 2 3 3 3
My code is dirty.
if condition smaller than two, names = unpopular.
df <- data.frame(vote=c("A","A","A","B","B","B","B","B","B","C","D"),
val=c(rep(1,11))
)
df %>% group_by(vote) %>% summarise(val=sum(val))
out
vote val
<fct> <dbl>
1 A 3
2 B 6
3 C 1
4 D 1
but I need
vote val
<fct> <dbl>
1 A 3
2 B 6
3 unpopular 2
my idea is
df2 <- df %>% group_by(vote) %>% summarise(val=sum(val))
df2$vote[df2$val < 2] <- "unpop"
df2 %>% group_by....
it's not cool.
do you know any cool & helpful function ?
We can do a double grouping
library(dplyr)
df %>%
group_by(vote) %>%
summarise(val=sum(val)) %>%
group_by(vote = replace(vote, val <2, 'unpop')) %>%
summarise(val = sum(val))
-output
# A tibble: 3 x 2
# vote val
# <chr> <dbl>
#1 A 3
#2 B 6
#3 unpop 2
Or another option with rowsum
df %>%
group_by(vote = replace(vote, vote %in%
names(which((rowsum(val, vote) < 2)[,1])), 'unpopular')) %>%
summarise(val = sum(val))
Or using fct_lump_n from forcats
library(forcats)
df %>%
group_by(vote = fct_lump_n(vote, 2, other_level = "unpop")) %>%
summarise(val = sum(val))
# A tibble: 3 x 2
# vote val
# <fct> <dbl>
#1 A 3
#2 B 6
#3 unpop 2
Or using table
df %>%
group_by(vote = replace(vote,
vote %in% names(which(table(vote) < 2)), 'unpop')) %>%
summarise(val = sum(val))
If you want to vote based on sum of val in base R you can do this as :
aggregate(val~vote, transform(aggregate(val~vote, df, sum),
vote = replace(vote, val < 2, 'unpop')), sum)
# vote val
#1 A 3
#2 B 6
#3 unpop 2
I have a dataset like so:
df<-data.frame(x=c("A","A","A","A", "B","B","B","B","B",
"C","C","C","C","C","D","D","D","D","D"),
y= as.factor(c(rep("Eoissp2",4),rep("Eoissp1",5),"Eoissp1","Eoisp4","Automerissp1","Automerissp2","Acharias",rep("Eoissp2",3),rep("Eoissp1",2))))
I want to identify, for each subset of x, the corresponding levels in y that are entirely duplicates containing the expression Eois. Therefore, A , B, and D will be returned in a vector because every level of A , B, and D contains the expression Eois , while level C consists of various unique levels (e.g. Eois, Automeris and Acharias). For this example the output would be:
output<- c("A", "B", "D")
Using new df:
> df %>% filter(str_detect(y,"Eois")) %>% group_by(x) %>% distinct(y) %>%
count() %>% filter(n==1) %>% select(x)
# A tibble: 2 x 1
# Groups: x [2]
x
<fct>
1 A
2 B
(Answer below uses the original df posted by the question author.)
Using the pipe function in magrittr & functions from dplyr:
> df %>% group_by(x) %>% distinct(y)
# A tibble: 7 x 2
# Groups: x [3]
x y
<fct> <fct>
1 A plant1a
2 B plant1b
3 C plant1a
4 C plant2a
5 C plant3a
6 C plant4a
7 C plant5a
Then you can roll up the results like this:
> results <- df %>% group_by(x) %>% distinct(y) %>%
count() %>% filter(n==1) %>% select(x)
> results
# A tibble: 2 x 1
# Groups: x [2]
x
<fct>
1 A
2 B
If you know your original data frame is always going to come with the x's in order, you can drop the group_by part.
A dplyr based solution could be as:
library(dplyr)
df %>% group_by(x) %>%
filter(grepl("Eoiss", y)) %>%
mutate(y = sub("\\d+", "", y)) %>%
filter(n() >1 & length(unique(y)) == 1) %>%
select(x) %>% unique(.)
# A tibble: 3 x 1
# Groups: x [3]
# x
# <fctr>
#1 A
#2 B
#3 D
Data
df<-data.frame(x=c("A","A","A","A", "B","B","B","B","B",
"C","C","C","C","C","D","D","D","D","D"),
y= as.factor(c(rep("Eoissp2",4),
rep("Eoissp1",5),"Eoissp1","Eoisp4","Automerissp1","Automerissp2",
"Acharias",rep("Eoissp2",3),rep("Eoissp1",2))))
In dplyr 0.5.0, calling summarise on a grouped data frame does not guarantee any resultant row order (Currently, it reorders the rows by group, not sure how it handles duplicate grouping levels).
To get around this, I would like to replace all summarise(x = ...) operations with mutate(x = ...) %>% filter(row_number() == 1). Are there any disadvantages or drawbacks to doing this?
Example of the two operations.
tmp_df <-
data.frame(group = rep(c(2L, 1L), each = 5), b = rep(c(-1, 1), each = 5)) %>%
group_by(group)
tmp_df %>%
summarise(b = sum(b))
tmp_df %>%
mutate(b = sum(b)) %>%
filter(row_number() == 1)
producing:
> tmp_df %>%
+ summarise(b = sum(b))
# A tibble: 2 × 2
group b
<int> <dbl>
1 1 5
2 2 -5
> tmp_df %>%
+ mutate(b = sum(b)) %>%
+ filter(row_number() == 1)
Source: local data frame [2 x 2]
Groups: group [2]
group b
<int> <dbl>
1 2 -5
2 1 5
EDIT: In response to a comment, for readability I can define the function:
summarise_o <- function (.data, ...) {
# order preserving summarise
mutate_(.data, .dots = lazyeval::lazy_dots(...)) %>%
filter(row_number() == 1) %>%
return
}
and simply call:
tmp_df %>%
summarise_o(b = sum(b))
One option is to create the 'group' as a factor
tmp_df <- data.frame(group = rep(c(2L, 1L), each = 5), b = rep(c(-1, 1), each = 5)) %>%
group_by(group = factor(group, levels = unique(group)))
tmp_df %>%
summarise(b = sum(b))
# A tibble: 2 x 2
# group b
# <fctr> <dbl>
#1 2 -5
#2 1 5