I want to count per country
the number of times the status
is open
and the number of times the status
is closed
. Then calculate the closerate
per country
.
Data:
customer <- c(1,2,3,4,5,6,7,8,9)
country <- c('BE', 'NL', 'NL','NL','BE','NL','BE','BE','NL')
closeday <- c('2017-08-23', '2017-08-05', '2017-08-22', '2017-08-26',
'2017-08-25', '2017-08-13', '2017-08-30', '2017-08-05', '2017-08-23')
closeday <- as.Date(closeday)
df <- data.frame(customer,country,closeday)
Adding status
:
df$status <- ifelse(df$closeday < '2017-08-20', 'open', 'closed')
customer country closeday status
1 1 BE 2017-08-23 closed
2 2 NL 2017-08-05 open
3 3 NL 2017-08-22 closed
4 4 NL 2017-08-26 closed
5 5 BE 2017-08-25 closed
6 6 NL 2017-08-13 open
7 7 BE 2017-08-30 closed
8 8 BE 2017-08-05 open
9 9 NL 2017-08-23 closed
Calculation closerate
closerate <- length(which(df$status == 'closed')) /
(length(which(df$status == 'closed')) + length(which(df$status == 'open')))
[1] 0.6666667
Obviously, this is the closerate
for the total. The challenge is to get the closerate
per country
. I tried adding the closerate
calculation to df
by:
df$closerate <- length(which(df$status == 'closed')) /
(length(which(df$status == 'closed')) + length(which(df$status == 'open')))
But it gives all lines a closerate
of 0.66 because I'm not grouping. I believe I should not use the length function because counting can be done by grouping. I read some information about using dplyr
to count logical outputs per group but this didn't work out.
This is the desired output:
aggregate(list(output = df$status == "closed"),
list(country = df$country),
function(x)
c(close = sum(x),
open = length(x) - sum(x),
rate = mean(x)))
# country output.close output.open output.rate
#1 BE 3.00 1.00 0.75
#2 NL 3.00 2.00 0.60
There was a solution using table
in the comments which appears to have been deleted. Anyway, you could also use table
output = as.data.frame.matrix(table(df$country, df$status))
output$closerate = output$closed/(output$closed + output$open)
output
# closed open closerate
#BE 3 1 0.75
#NL 3 2 0.60
You can use tapply
:
data.frame(open=tapply(df$status=="open", df$country, sum),
closed=tapply(df$status=="closed", df$country, sum)
closerate=tapply(df$status=="closed", df$country, mean))`
A data.table
method would be.
library(data.table)
setDT(df)[, {temp <- status=="closed"; # store temporary logical variable
.(closed=sum(temp), open=sum(!temp), closeRate=mean(temp))}, # calculate stuff
by=country] # by country
which returns
country closed open closeRate
1: BE 3 1 0.75
2: NL 3 2 0.60
Here is a dplyr
solution.
output <- df %>%
count(country, status) %>%
group_by(country) %>%
mutate(total = sum(n)) %>%
mutate(percent = n/total)
Returns...
output
country status n total percent
BE closed 3 4 0.75
BE open 1 4 0.25
NL closed 3 5 0.60
NL open 2 5 0.40
Here's a quick solution with tidyverse
:
library(dplyr)
df %>% group_by(country) %>%
mutate(status =ifelse(closeday < '2017-08-20', 'open', 'closed'),
closerate=mean(status=="closed"))
Returning:
# A tibble: 9 x 5
# Groups: country [2]
customer country closeday status closerate
<dbl> <fctr> <date> <chr> <dbl>
1 1 BE 2017-08-23 closed 0.75
2 2 NL 2017-08-05 open 0.60
3 3 NL 2017-08-22 closed 0.60
4 4 NL 2017-08-26 closed 0.60
5 5 BE 2017-08-25 closed 0.75
6 6 NL 2017-08-13 open 0.60
7 7 BE 2017-08-30 closed 0.75
8 8 BE 2017-08-05 open 0.75
9 9 NL 2017-08-23 closed 0.60
Here I am utilizing the coercion of logicals into integer when the vector of TRUE/FALSE is put into the mean()
function.
Alternatively, with data.table
:
library(data.table)
setDT(df)[,status:=ifelse(closeday < '2017-08-20', 'open', 'closed')]
df[, .(closerate=mean(status=="closed")), by=country]