Related
Suppose I want to calculate the proportion of different values within each group. For example, using the mtcars data, how do I calculate the relative frequency of number of gears by am (automatic/manual) in one go with dplyr?
library(dplyr)
data(mtcars)
mtcars <- tbl_df(mtcars)
# count frequency
mtcars %>%
group_by(am, gear) %>%
summarise(n = n())
# am gear n
# 0 3 15
# 0 4 4
# 1 4 8
# 1 5 5
What I would like to achieve:
am gear n rel.freq
0 3 15 0.7894737
0 4 4 0.2105263
1 4 8 0.6153846
1 5 5 0.3846154
Try this:
mtcars %>%
group_by(am, gear) %>%
summarise(n = n()) %>%
mutate(freq = n / sum(n))
# am gear n freq
# 1 0 3 15 0.7894737
# 2 0 4 4 0.2105263
# 3 1 4 8 0.6153846
# 4 1 5 5 0.3846154
From the dplyr vignette:
When you group by multiple variables, each summary peels off one level of the grouping. That makes it easy to progressively roll-up a dataset.
Thus, after the summarise, the last grouping variable specified in group_by, 'gear', is peeled off. In the mutate step, the data is grouped by the remaining grouping variable(s), here 'am'. You may check grouping in each step with groups.
The outcome of the peeling is of course dependent of the order of the grouping variables in the group_by call. You may wish to do a subsequent group_by(am), to make your code more explicit.
For rounding and prettification, please refer to the nice answer by #Tyler Rinker.
You can use count() function, which has however a different behaviour depending on the version of dplyr:
dplyr 0.7.1: returns an ungrouped table: you need to group again by am
dplyr < 0.7.1: returns a grouped table, so no need to group again, although you might want to ungroup() for later manipulations
dplyr 0.7.1
mtcars %>%
count(am, gear) %>%
group_by(am) %>%
mutate(freq = n / sum(n))
dplyr < 0.7.1
mtcars %>%
count(am, gear) %>%
mutate(freq = n / sum(n))
This results into a grouped table, if you want to use it for further analysis, it might be useful to remove the grouped attribute with ungroup().
#Henrik's is better for usability as this will make the column character and no longer numeric but matches what you asked for...
mtcars %>%
group_by (am, gear) %>%
summarise (n=n()) %>%
mutate(rel.freq = paste0(round(100 * n/sum(n), 0), "%"))
## am gear n rel.freq
## 1 0 3 15 79%
## 2 0 4 4 21%
## 3 1 4 8 62%
## 4 1 5 5 38%
EDIT Because Spacedman asked for it :-)
as.rel_freq <- function(x, rel_freq_col = "rel.freq", ...) {
class(x) <- c("rel_freq", class(x))
attributes(x)[["rel_freq_col"]] <- rel_freq_col
x
}
print.rel_freq <- function(x, ...) {
freq_col <- attributes(x)[["rel_freq_col"]]
x[[freq_col]] <- paste0(round(100 * x[[freq_col]], 0), "%")
class(x) <- class(x)[!class(x)%in% "rel_freq"]
print(x)
}
mtcars %>%
group_by (am, gear) %>%
summarise (n=n()) %>%
mutate(rel.freq = n/sum(n)) %>%
as.rel_freq()
## Source: local data frame [4 x 4]
## Groups: am
##
## am gear n rel.freq
## 1 0 3 15 79%
## 2 0 4 4 21%
## 3 1 4 8 62%
## 4 1 5 5 38%
Despite the many answers, one more approach which uses prop.table in combination with dplyr or data.table.
library(dplyr)
mtcars %>%
group_by(am, gear) %>%
tally() %>%
mutate(freq = prop.table(n))
#> # A tibble: 4 × 4
#> # Groups: am [2]
#> am gear n freq
#> <dbl> <dbl> <int> <dbl>
#> 1 0 3 15 0.789
#> 2 0 4 4 0.211
#> 3 1 4 8 0.615
#> 4 1 5 5 0.385
library(data.table)
cars_dt <- as.data.table(mtcars)
cars_dt[, .(n = .N), keyby = .(am, gear)][, freq := prop.table(n), by = "am"][]
#> am gear n freq
#> 1: 0 3 15 0.7894737
#> 2: 0 4 4 0.2105263
#> 3: 1 4 8 0.6153846
#> 4: 1 5 5 0.3846154
Created on 2022-10-22 with reprex v2.0.2
I wrote a small function for this repeating task:
count_pct <- function(df) {
return(
df %>%
tally %>%
mutate(n_pct = 100*n/sum(n))
)
}
I can then use it like:
mtcars %>%
group_by(cyl) %>%
count_pct
It returns:
# A tibble: 3 x 3
cyl n n_pct
<dbl> <int> <dbl>
1 4 11 34.4
2 6 7 21.9
3 8 14 43.8
For the sake of completeness of this popular question, since version 1.0.0 of dplyr, parameter .groups controls the grouping structure of the summarise function after group_by summarise help.
With .groups = "drop_last", summarise drops the last level of grouping. This was the only result obtained before version 1.0.0.
library(dplyr)
library(scales)
original <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n()) %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
#> `summarise()` regrouping output by 'am' (override with `.groups` argument)
original
#> # A tibble: 4 x 4
#> # Groups: am [2]
#> am gear n rel.freq
#> <dbl> <dbl> <int> <chr>
#> 1 0 3 15 78.9%
#> 2 0 4 4 21.1%
#> 3 1 4 8 61.5%
#> 4 1 5 5 38.5%
new_drop_last <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n(), .groups = "drop_last") %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
dplyr::all_equal(original, new_drop_last)
#> [1] TRUE
With .groups = "drop", all levels of grouping are dropped. The result is turned into an independent tibble with no trace of the previous group_by
# .groups = "drop"
new_drop <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n(), .groups = "drop") %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
new_drop
#> # A tibble: 4 x 4
#> am gear n rel.freq
#> <dbl> <dbl> <int> <chr>
#> 1 0 3 15 46.9%
#> 2 0 4 4 12.5%
#> 3 1 4 8 25.0%
#> 4 1 5 5 15.6%
If .groups = "keep", same grouping structure as .data (mtcars, in this case). summarise does not peel off any variable used in the group_by.
Finally, with .groups = "rowwise", each row is it's own group. It is equivalent to "keep" in this situation
# .groups = "keep"
new_keep <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n(), .groups = "keep") %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
new_keep
#> # A tibble: 4 x 4
#> # Groups: am, gear [4]
#> am gear n rel.freq
#> <dbl> <dbl> <int> <chr>
#> 1 0 3 15 100.0%
#> 2 0 4 4 100.0%
#> 3 1 4 8 100.0%
#> 4 1 5 5 100.0%
# .groups = "rowwise"
new_rowwise <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n(), .groups = "rowwise") %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
dplyr::all_equal(new_keep, new_rowwise)
#> [1] TRUE
Another point that can be of interest is that sometimes, after applying group_by and summarise, a summary line can help.
# create a subtotal line to help readability
subtotal_am <- mtcars %>%
group_by (am) %>%
summarise (n=n()) %>%
mutate(gear = NA, rel.freq = 1)
#> `summarise()` ungrouping output (override with `.groups` argument)
mtcars %>% group_by (am, gear) %>%
summarise (n=n()) %>%
mutate(rel.freq = n/sum(n)) %>%
bind_rows(subtotal_am) %>%
arrange(am, gear) %>%
mutate(rel.freq = scales::percent(rel.freq, accuracy = 0.1))
#> `summarise()` regrouping output by 'am' (override with `.groups` argument)
#> # A tibble: 6 x 4
#> # Groups: am [2]
#> am gear n rel.freq
#> <dbl> <dbl> <int> <chr>
#> 1 0 3 15 78.9%
#> 2 0 4 4 21.1%
#> 3 0 NA 19 100.0%
#> 4 1 4 8 61.5%
#> 5 1 5 5 38.5%
#> 6 1 NA 13 100.0%
Created on 2020-11-09 by the reprex package (v0.3.0)
Hope you find this answer useful.
Here is a general function implementing Henrik's solution on dplyr 0.7.1.
freq_table <- function(x,
group_var,
prop_var) {
group_var <- enquo(group_var)
prop_var <- enquo(prop_var)
x %>%
group_by(!!group_var, !!prop_var) %>%
summarise(n = n()) %>%
mutate(freq = n /sum(n)) %>%
ungroup
}
Also, try add_count() (to get around pesky group_by .groups).
mtcars %>%
count(am, gear) %>%
add_count(am, wt = n, name = "nn") %>%
mutate(proportion = n / nn)
Here is a base R answer using aggregate and ave :
df1 <- with(mtcars, aggregate(list(n = mpg), list(am = am, gear = gear), length))
df1$prop <- with(df1, n/ave(n, am, FUN = sum))
#Also with prop.table
#df1$prop <- with(df1, ave(n, am, FUN = prop.table))
df1
# am gear n prop
#1 0 3 15 0.7894737
#2 0 4 4 0.2105263
#3 1 4 8 0.6153846
#4 1 5 5 0.3846154
We can also use prop.table but the output displays differently.
prop.table(table(mtcars$am, mtcars$gear), 1)
# 3 4 5
# 0 0.7894737 0.2105263 0.0000000
# 1 0.0000000 0.6153846 0.3846154
This answer is based upon Matifou's answer.
First I modified it to ensure that I don't get the freq column returned as a scientific notation column by using the scipen option.
Then I multiple the answer by 100 to get a percent rather than decimal to make the freq column easier to read as a percentage.
getOption("scipen")
options("scipen"=10)
mtcars %>%
count(am, gear) %>%
mutate(freq = (n / sum(n)) * 100)
Suppose I want to calculate the proportion of different values within each group. For example, using the mtcars data, how do I calculate the relative frequency of number of gears by am (automatic/manual) in one go with dplyr?
library(dplyr)
data(mtcars)
mtcars <- tbl_df(mtcars)
# count frequency
mtcars %>%
group_by(am, gear) %>%
summarise(n = n())
# am gear n
# 0 3 15
# 0 4 4
# 1 4 8
# 1 5 5
What I would like to achieve:
am gear n rel.freq
0 3 15 0.7894737
0 4 4 0.2105263
1 4 8 0.6153846
1 5 5 0.3846154
Try this:
mtcars %>%
group_by(am, gear) %>%
summarise(n = n()) %>%
mutate(freq = n / sum(n))
# am gear n freq
# 1 0 3 15 0.7894737
# 2 0 4 4 0.2105263
# 3 1 4 8 0.6153846
# 4 1 5 5 0.3846154
From the dplyr vignette:
When you group by multiple variables, each summary peels off one level of the grouping. That makes it easy to progressively roll-up a dataset.
Thus, after the summarise, the last grouping variable specified in group_by, 'gear', is peeled off. In the mutate step, the data is grouped by the remaining grouping variable(s), here 'am'. You may check grouping in each step with groups.
The outcome of the peeling is of course dependent of the order of the grouping variables in the group_by call. You may wish to do a subsequent group_by(am), to make your code more explicit.
For rounding and prettification, please refer to the nice answer by #Tyler Rinker.
You can use count() function, which has however a different behaviour depending on the version of dplyr:
dplyr 0.7.1: returns an ungrouped table: you need to group again by am
dplyr < 0.7.1: returns a grouped table, so no need to group again, although you might want to ungroup() for later manipulations
dplyr 0.7.1
mtcars %>%
count(am, gear) %>%
group_by(am) %>%
mutate(freq = n / sum(n))
dplyr < 0.7.1
mtcars %>%
count(am, gear) %>%
mutate(freq = n / sum(n))
This results into a grouped table, if you want to use it for further analysis, it might be useful to remove the grouped attribute with ungroup().
#Henrik's is better for usability as this will make the column character and no longer numeric but matches what you asked for...
mtcars %>%
group_by (am, gear) %>%
summarise (n=n()) %>%
mutate(rel.freq = paste0(round(100 * n/sum(n), 0), "%"))
## am gear n rel.freq
## 1 0 3 15 79%
## 2 0 4 4 21%
## 3 1 4 8 62%
## 4 1 5 5 38%
EDIT Because Spacedman asked for it :-)
as.rel_freq <- function(x, rel_freq_col = "rel.freq", ...) {
class(x) <- c("rel_freq", class(x))
attributes(x)[["rel_freq_col"]] <- rel_freq_col
x
}
print.rel_freq <- function(x, ...) {
freq_col <- attributes(x)[["rel_freq_col"]]
x[[freq_col]] <- paste0(round(100 * x[[freq_col]], 0), "%")
class(x) <- class(x)[!class(x)%in% "rel_freq"]
print(x)
}
mtcars %>%
group_by (am, gear) %>%
summarise (n=n()) %>%
mutate(rel.freq = n/sum(n)) %>%
as.rel_freq()
## Source: local data frame [4 x 4]
## Groups: am
##
## am gear n rel.freq
## 1 0 3 15 79%
## 2 0 4 4 21%
## 3 1 4 8 62%
## 4 1 5 5 38%
Despite the many answers, one more approach which uses prop.table in combination with dplyr or data.table.
library(dplyr)
mtcars %>%
group_by(am, gear) %>%
tally() %>%
mutate(freq = prop.table(n))
#> # A tibble: 4 × 4
#> # Groups: am [2]
#> am gear n freq
#> <dbl> <dbl> <int> <dbl>
#> 1 0 3 15 0.789
#> 2 0 4 4 0.211
#> 3 1 4 8 0.615
#> 4 1 5 5 0.385
library(data.table)
cars_dt <- as.data.table(mtcars)
cars_dt[, .(n = .N), keyby = .(am, gear)][, freq := prop.table(n), by = "am"][]
#> am gear n freq
#> 1: 0 3 15 0.7894737
#> 2: 0 4 4 0.2105263
#> 3: 1 4 8 0.6153846
#> 4: 1 5 5 0.3846154
Created on 2022-10-22 with reprex v2.0.2
I wrote a small function for this repeating task:
count_pct <- function(df) {
return(
df %>%
tally %>%
mutate(n_pct = 100*n/sum(n))
)
}
I can then use it like:
mtcars %>%
group_by(cyl) %>%
count_pct
It returns:
# A tibble: 3 x 3
cyl n n_pct
<dbl> <int> <dbl>
1 4 11 34.4
2 6 7 21.9
3 8 14 43.8
For the sake of completeness of this popular question, since version 1.0.0 of dplyr, parameter .groups controls the grouping structure of the summarise function after group_by summarise help.
With .groups = "drop_last", summarise drops the last level of grouping. This was the only result obtained before version 1.0.0.
library(dplyr)
library(scales)
original <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n()) %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
#> `summarise()` regrouping output by 'am' (override with `.groups` argument)
original
#> # A tibble: 4 x 4
#> # Groups: am [2]
#> am gear n rel.freq
#> <dbl> <dbl> <int> <chr>
#> 1 0 3 15 78.9%
#> 2 0 4 4 21.1%
#> 3 1 4 8 61.5%
#> 4 1 5 5 38.5%
new_drop_last <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n(), .groups = "drop_last") %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
dplyr::all_equal(original, new_drop_last)
#> [1] TRUE
With .groups = "drop", all levels of grouping are dropped. The result is turned into an independent tibble with no trace of the previous group_by
# .groups = "drop"
new_drop <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n(), .groups = "drop") %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
new_drop
#> # A tibble: 4 x 4
#> am gear n rel.freq
#> <dbl> <dbl> <int> <chr>
#> 1 0 3 15 46.9%
#> 2 0 4 4 12.5%
#> 3 1 4 8 25.0%
#> 4 1 5 5 15.6%
If .groups = "keep", same grouping structure as .data (mtcars, in this case). summarise does not peel off any variable used in the group_by.
Finally, with .groups = "rowwise", each row is it's own group. It is equivalent to "keep" in this situation
# .groups = "keep"
new_keep <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n(), .groups = "keep") %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
new_keep
#> # A tibble: 4 x 4
#> # Groups: am, gear [4]
#> am gear n rel.freq
#> <dbl> <dbl> <int> <chr>
#> 1 0 3 15 100.0%
#> 2 0 4 4 100.0%
#> 3 1 4 8 100.0%
#> 4 1 5 5 100.0%
# .groups = "rowwise"
new_rowwise <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n(), .groups = "rowwise") %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
dplyr::all_equal(new_keep, new_rowwise)
#> [1] TRUE
Another point that can be of interest is that sometimes, after applying group_by and summarise, a summary line can help.
# create a subtotal line to help readability
subtotal_am <- mtcars %>%
group_by (am) %>%
summarise (n=n()) %>%
mutate(gear = NA, rel.freq = 1)
#> `summarise()` ungrouping output (override with `.groups` argument)
mtcars %>% group_by (am, gear) %>%
summarise (n=n()) %>%
mutate(rel.freq = n/sum(n)) %>%
bind_rows(subtotal_am) %>%
arrange(am, gear) %>%
mutate(rel.freq = scales::percent(rel.freq, accuracy = 0.1))
#> `summarise()` regrouping output by 'am' (override with `.groups` argument)
#> # A tibble: 6 x 4
#> # Groups: am [2]
#> am gear n rel.freq
#> <dbl> <dbl> <int> <chr>
#> 1 0 3 15 78.9%
#> 2 0 4 4 21.1%
#> 3 0 NA 19 100.0%
#> 4 1 4 8 61.5%
#> 5 1 5 5 38.5%
#> 6 1 NA 13 100.0%
Created on 2020-11-09 by the reprex package (v0.3.0)
Hope you find this answer useful.
Here is a general function implementing Henrik's solution on dplyr 0.7.1.
freq_table <- function(x,
group_var,
prop_var) {
group_var <- enquo(group_var)
prop_var <- enquo(prop_var)
x %>%
group_by(!!group_var, !!prop_var) %>%
summarise(n = n()) %>%
mutate(freq = n /sum(n)) %>%
ungroup
}
Also, try add_count() (to get around pesky group_by .groups).
mtcars %>%
count(am, gear) %>%
add_count(am, wt = n, name = "nn") %>%
mutate(proportion = n / nn)
Here is a base R answer using aggregate and ave :
df1 <- with(mtcars, aggregate(list(n = mpg), list(am = am, gear = gear), length))
df1$prop <- with(df1, n/ave(n, am, FUN = sum))
#Also with prop.table
#df1$prop <- with(df1, ave(n, am, FUN = prop.table))
df1
# am gear n prop
#1 0 3 15 0.7894737
#2 0 4 4 0.2105263
#3 1 4 8 0.6153846
#4 1 5 5 0.3846154
We can also use prop.table but the output displays differently.
prop.table(table(mtcars$am, mtcars$gear), 1)
# 3 4 5
# 0 0.7894737 0.2105263 0.0000000
# 1 0.0000000 0.6153846 0.3846154
This answer is based upon Matifou's answer.
First I modified it to ensure that I don't get the freq column returned as a scientific notation column by using the scipen option.
Then I multiple the answer by 100 to get a percent rather than decimal to make the freq column easier to read as a percentage.
getOption("scipen")
options("scipen"=10)
mtcars %>%
count(am, gear) %>%
mutate(freq = (n / sum(n)) * 100)
I've been trying a few ways to achieve (do, row_number) this but still stuck.
I have 3 groups: month, city, and gender.
I would like to get only the top 5 count of these 3 group bys.
This code works fine only with 2 groups:
df_top5_2grp <- df %>%
group_by(month, city) %>%
tally() %>%
top_n(n = 5, wt = n) %>%
arrange(retention_month, desc(n))
However, it won't return the top 5 count if I add an additional group:
df_top5_3grp <- df %>%
group_by(month, city, gender) %>%
tally() %>%
top_n(n = 5, wt = n) %>%
arrange(retention_month, gender, desc(n))
It returns all rows instead. The only difference is I added gender.
Any help is appreciated. Thanks!
You probably need an ungroup() in there.
In the first example below, it returns all the rows, since there are 7 groups, each with one row. So returning the top 5 of each of the seven groups returns all rows.
mtcars %>%
group_by(cyl, vs, am) %>% # grouping across three variables
tally() %>% # tally is a summarization that removes the last grouping
top_n(n = 5, wt = n)
# A tibble: 7 x 4
# Groups: cyl, vs [5] # NOTE! This reminds us the data is still grouped
cyl vs am n
<dbl> <dbl> <dbl> <int>
1 4 0 1 1
2 4 1 0 3
3 4 1 1 7
4 6 0 1 3
5 6 1 0 4
6 8 0 0 12
7 8 0 1 2
Adding ungroup makes it so the top 5 filtering happens across all the summarized groups, not within each group.
mtcars %>%
group_by(cyl, vs, am) %>%
tally() %>%
ungroup() %>%
top_n(n = 5, wt = n)
# A tibble: 5 x 4
cyl vs am n
<dbl> <dbl> <dbl> <int>
1 4 1 0 3
2 4 1 1 7
3 6 0 1 3
4 6 1 0 4
5 8 0 0 12
Is there a way to do this in one line of code resulting in one dataframe instead of two as seen below:
df1 <- mtcars %>% group_by(gear, carb) %>%
distinct(gear, cyl, am) %>%
summarise(UniqCnt = n()
df2 <- mtcars %>% group_by(gear, carb) %>%
summarise(Cnt = n())
I attempted this
attempt1 <- mtcars %>% (group_by(gear, carb) %>%
distinct(gear, cyl,am) %>%
summarise(UniqCnt = n())) %>%
(group_by(gear, carb) %>%
summarise(Cnt = n()))
but it did not work. I can rbind the two but I would prefer not to.
Thank you in advance.
You can use the n_distinct() function in your summarize(). For example
mtcars %>% group_by(gear, carb) %>%
summarize(UniqCnt = n_distinct(am), Cnt=n())
# gear carb UniqCnt Cnt
# <dbl> <dbl> <int> <int>
# 1 3 1 1 3
# 2 3 2 1 4
# 3 3 3 1 3
# 4 3 4 1 5
# 5 4 1 1 4
# 6 4 2 2 4
# 7 4 4 2 4
# 8 5 2 1 2
# 9 5 4 1 1
# 10 5 6 1 1
# 11 5 8 1 1
How about this:
df1 <- mtcars %>%
group_by(gear, carb) %>%
summarise(UniqCnt = n_distinct(gear, cyl, am),
Cnt = n())
Suppose I want to calculate the proportion of different values within each group. For example, using the mtcars data, how do I calculate the relative frequency of number of gears by am (automatic/manual) in one go with dplyr?
library(dplyr)
data(mtcars)
mtcars <- tbl_df(mtcars)
# count frequency
mtcars %>%
group_by(am, gear) %>%
summarise(n = n())
# am gear n
# 0 3 15
# 0 4 4
# 1 4 8
# 1 5 5
What I would like to achieve:
am gear n rel.freq
0 3 15 0.7894737
0 4 4 0.2105263
1 4 8 0.6153846
1 5 5 0.3846154
Try this:
mtcars %>%
group_by(am, gear) %>%
summarise(n = n()) %>%
mutate(freq = n / sum(n))
# am gear n freq
# 1 0 3 15 0.7894737
# 2 0 4 4 0.2105263
# 3 1 4 8 0.6153846
# 4 1 5 5 0.3846154
From the dplyr vignette:
When you group by multiple variables, each summary peels off one level of the grouping. That makes it easy to progressively roll-up a dataset.
Thus, after the summarise, the last grouping variable specified in group_by, 'gear', is peeled off. In the mutate step, the data is grouped by the remaining grouping variable(s), here 'am'. You may check grouping in each step with groups.
The outcome of the peeling is of course dependent of the order of the grouping variables in the group_by call. You may wish to do a subsequent group_by(am), to make your code more explicit.
For rounding and prettification, please refer to the nice answer by #Tyler Rinker.
You can use count() function, which has however a different behaviour depending on the version of dplyr:
dplyr 0.7.1: returns an ungrouped table: you need to group again by am
dplyr < 0.7.1: returns a grouped table, so no need to group again, although you might want to ungroup() for later manipulations
dplyr 0.7.1
mtcars %>%
count(am, gear) %>%
group_by(am) %>%
mutate(freq = n / sum(n))
dplyr < 0.7.1
mtcars %>%
count(am, gear) %>%
mutate(freq = n / sum(n))
This results into a grouped table, if you want to use it for further analysis, it might be useful to remove the grouped attribute with ungroup().
#Henrik's is better for usability as this will make the column character and no longer numeric but matches what you asked for...
mtcars %>%
group_by (am, gear) %>%
summarise (n=n()) %>%
mutate(rel.freq = paste0(round(100 * n/sum(n), 0), "%"))
## am gear n rel.freq
## 1 0 3 15 79%
## 2 0 4 4 21%
## 3 1 4 8 62%
## 4 1 5 5 38%
EDIT Because Spacedman asked for it :-)
as.rel_freq <- function(x, rel_freq_col = "rel.freq", ...) {
class(x) <- c("rel_freq", class(x))
attributes(x)[["rel_freq_col"]] <- rel_freq_col
x
}
print.rel_freq <- function(x, ...) {
freq_col <- attributes(x)[["rel_freq_col"]]
x[[freq_col]] <- paste0(round(100 * x[[freq_col]], 0), "%")
class(x) <- class(x)[!class(x)%in% "rel_freq"]
print(x)
}
mtcars %>%
group_by (am, gear) %>%
summarise (n=n()) %>%
mutate(rel.freq = n/sum(n)) %>%
as.rel_freq()
## Source: local data frame [4 x 4]
## Groups: am
##
## am gear n rel.freq
## 1 0 3 15 79%
## 2 0 4 4 21%
## 3 1 4 8 62%
## 4 1 5 5 38%
Despite the many answers, one more approach which uses prop.table in combination with dplyr or data.table.
library(dplyr)
mtcars %>%
group_by(am, gear) %>%
tally() %>%
mutate(freq = prop.table(n))
#> # A tibble: 4 × 4
#> # Groups: am [2]
#> am gear n freq
#> <dbl> <dbl> <int> <dbl>
#> 1 0 3 15 0.789
#> 2 0 4 4 0.211
#> 3 1 4 8 0.615
#> 4 1 5 5 0.385
library(data.table)
cars_dt <- as.data.table(mtcars)
cars_dt[, .(n = .N), keyby = .(am, gear)][, freq := prop.table(n), by = "am"][]
#> am gear n freq
#> 1: 0 3 15 0.7894737
#> 2: 0 4 4 0.2105263
#> 3: 1 4 8 0.6153846
#> 4: 1 5 5 0.3846154
Created on 2022-10-22 with reprex v2.0.2
I wrote a small function for this repeating task:
count_pct <- function(df) {
return(
df %>%
tally %>%
mutate(n_pct = 100*n/sum(n))
)
}
I can then use it like:
mtcars %>%
group_by(cyl) %>%
count_pct
It returns:
# A tibble: 3 x 3
cyl n n_pct
<dbl> <int> <dbl>
1 4 11 34.4
2 6 7 21.9
3 8 14 43.8
For the sake of completeness of this popular question, since version 1.0.0 of dplyr, parameter .groups controls the grouping structure of the summarise function after group_by summarise help.
With .groups = "drop_last", summarise drops the last level of grouping. This was the only result obtained before version 1.0.0.
library(dplyr)
library(scales)
original <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n()) %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
#> `summarise()` regrouping output by 'am' (override with `.groups` argument)
original
#> # A tibble: 4 x 4
#> # Groups: am [2]
#> am gear n rel.freq
#> <dbl> <dbl> <int> <chr>
#> 1 0 3 15 78.9%
#> 2 0 4 4 21.1%
#> 3 1 4 8 61.5%
#> 4 1 5 5 38.5%
new_drop_last <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n(), .groups = "drop_last") %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
dplyr::all_equal(original, new_drop_last)
#> [1] TRUE
With .groups = "drop", all levels of grouping are dropped. The result is turned into an independent tibble with no trace of the previous group_by
# .groups = "drop"
new_drop <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n(), .groups = "drop") %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
new_drop
#> # A tibble: 4 x 4
#> am gear n rel.freq
#> <dbl> <dbl> <int> <chr>
#> 1 0 3 15 46.9%
#> 2 0 4 4 12.5%
#> 3 1 4 8 25.0%
#> 4 1 5 5 15.6%
If .groups = "keep", same grouping structure as .data (mtcars, in this case). summarise does not peel off any variable used in the group_by.
Finally, with .groups = "rowwise", each row is it's own group. It is equivalent to "keep" in this situation
# .groups = "keep"
new_keep <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n(), .groups = "keep") %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
new_keep
#> # A tibble: 4 x 4
#> # Groups: am, gear [4]
#> am gear n rel.freq
#> <dbl> <dbl> <int> <chr>
#> 1 0 3 15 100.0%
#> 2 0 4 4 100.0%
#> 3 1 4 8 100.0%
#> 4 1 5 5 100.0%
# .groups = "rowwise"
new_rowwise <- mtcars %>%
group_by (am, gear) %>%
summarise (n=n(), .groups = "rowwise") %>%
mutate(rel.freq = scales::percent(n/sum(n), accuracy = 0.1))
dplyr::all_equal(new_keep, new_rowwise)
#> [1] TRUE
Another point that can be of interest is that sometimes, after applying group_by and summarise, a summary line can help.
# create a subtotal line to help readability
subtotal_am <- mtcars %>%
group_by (am) %>%
summarise (n=n()) %>%
mutate(gear = NA, rel.freq = 1)
#> `summarise()` ungrouping output (override with `.groups` argument)
mtcars %>% group_by (am, gear) %>%
summarise (n=n()) %>%
mutate(rel.freq = n/sum(n)) %>%
bind_rows(subtotal_am) %>%
arrange(am, gear) %>%
mutate(rel.freq = scales::percent(rel.freq, accuracy = 0.1))
#> `summarise()` regrouping output by 'am' (override with `.groups` argument)
#> # A tibble: 6 x 4
#> # Groups: am [2]
#> am gear n rel.freq
#> <dbl> <dbl> <int> <chr>
#> 1 0 3 15 78.9%
#> 2 0 4 4 21.1%
#> 3 0 NA 19 100.0%
#> 4 1 4 8 61.5%
#> 5 1 5 5 38.5%
#> 6 1 NA 13 100.0%
Created on 2020-11-09 by the reprex package (v0.3.0)
Hope you find this answer useful.
Here is a general function implementing Henrik's solution on dplyr 0.7.1.
freq_table <- function(x,
group_var,
prop_var) {
group_var <- enquo(group_var)
prop_var <- enquo(prop_var)
x %>%
group_by(!!group_var, !!prop_var) %>%
summarise(n = n()) %>%
mutate(freq = n /sum(n)) %>%
ungroup
}
Also, try add_count() (to get around pesky group_by .groups).
mtcars %>%
count(am, gear) %>%
add_count(am, wt = n, name = "nn") %>%
mutate(proportion = n / nn)
Here is a base R answer using aggregate and ave :
df1 <- with(mtcars, aggregate(list(n = mpg), list(am = am, gear = gear), length))
df1$prop <- with(df1, n/ave(n, am, FUN = sum))
#Also with prop.table
#df1$prop <- with(df1, ave(n, am, FUN = prop.table))
df1
# am gear n prop
#1 0 3 15 0.7894737
#2 0 4 4 0.2105263
#3 1 4 8 0.6153846
#4 1 5 5 0.3846154
We can also use prop.table but the output displays differently.
prop.table(table(mtcars$am, mtcars$gear), 1)
# 3 4 5
# 0 0.7894737 0.2105263 0.0000000
# 1 0.0000000 0.6153846 0.3846154
This answer is based upon Matifou's answer.
First I modified it to ensure that I don't get the freq column returned as a scientific notation column by using the scipen option.
Then I multiple the answer by 100 to get a percent rather than decimal to make the freq column easier to read as a percentage.
getOption("scipen")
options("scipen"=10)
mtcars %>%
count(am, gear) %>%
mutate(freq = (n / sum(n)) * 100)