Group by function query - r

Question

Hi guys i am new to R,
While i have attached screenshot of the df i am working with (https://i.stack.imgur.com/CUz4l.png), here is a short description
I have a data frame with a total of 7 columns, one of which is a month column, rest of the 6 columns are (integer) values and these also have empty rows
Need to summarise by count of all the 6 columns and group them by month
tried the following code: group_by(Month) %>% summarise(count=n(),na.omit())
get the following error:
Error: Problem with summarise() input ..2.
x argument "object" is missing, with no default
i Input ..2 is na.omit().
i The error occurred in group 1: Month = "1".
Run rlang::last_error() to see where the error occurred.
Can someone please assist?
[head of data][1] (https://i.stack.imgur.com/stfoG.png)
> dput(head(Dropoff))
structure(list(Start.Date = c("01-11-2019 06:07", "01-11-2019 06:07",
"01-11-2019 06:08", "01-11-2019 06:08", "02-11-2019 06:08", "02-11-2019 06:07"
), End.Date = c("01-11-2019 06:12", "01-11-2019 09:28", "01-11-2019 10:02",
"01-11-2019 13:05", "02-11-2019 06:13", "02-11-2019 06:16"),
Month = structure(c(3L, 3L, 3L, 3L, 3L, 3L), .Label = c("1",
"2", "11"), class = "factor"), nps = c(9L, 10L, 9L, 8L, 9L,
9L), effort = c(9L, 10L, 9L, 9L, 9L, 8L), knowledge = c(NA,
NA, 5L, NA, NA, 5L), confidence = c(5L, 5L, NA, NA, 5L, NA
), listening = c(NA, NA, NA, 5L, NA, NA), fcr = c(1L, 1L,
1L, 1L, 1L, 1L), fixing.issues = c(NA, NA, NA, NA, NA, NA
)), row.names = c(NA, 6L), class = "data.frame")
id like the output to look something like this
Month
count of nps
count of effort
1
xxx
xxx
2
xxx
xxx
11
6
6
....so on (count)for all the variables
the following
df%>% group_by(Month) %>% summarise(count=n())
provides this output
[1]: https://i.stack.imgur.com/u3nxv.png
this is not what i am hoping for

Answer 1

looks like the na.omit() causes problems in this case. Given that you want to count NA but not have them in any following sum, you might use
df[is.na(df)] = 0
and then
df %>% group_by(Month) %>% summarise(count=n())

Answer 2

thanks for the clarifications. The semi-manual solution
df %>% group_by(Month) %>% summarize(
c_nps= sum(!is.na(nps)),
c_effort= sum(!is.na(effort)),
c_knowledge= sum(!is.na(knowledge)),
c_confidence= sum(!is.na(confidence)),
c_listening= sum(!is.na(listening)),
c_fcr= sum(!is.na(fcr))
)
should do the trick. Since it's only 6 columns to be summarized, I would use the manual specification over an automated implementation (i.e. count non-NA in all other columns).
It results in
# A tibble: 1 x 7
Month c_nps c_effort c_knowledge c_confidence c_listening c_fcr
<fct> <int> <int> <int> <int> <int> <int>
1 11 6 6 2 3 1 6
Cheers and good luck!

Answer 3

From you example I understand, that you want to count the non-NA values in every column.
Dropoff %>% group_by(Month) %>%
summarise_at(vars(nps:fixing.issues), list(count=~sum(!is.na(.x))))
summarize_at: The term performs a summarize at every column given in the vars() expression. Here I chose all columns from nps to fixing.issues.
As summarizing function (which describes how the data is summarized), I defined to count all non-NA values. The syntax is to give all functions as named list. Here the ~ does the same as function(x). A more lengthy way to write it would be: function(x) sum(!is.na(x))
The "count" expression works as follows: check the vector of the column (x) if those are NA values is.na. The ! negates this expression. As this is a vector with only true/false values, you can just count the true values with sum.
The expression works for all kind of column types (text, numbers, ...)
Giving the result:
# A tibble: 1 x 8
Month nps_count effort_count knowledge_count confidence_count listening_count fcr_count fixing.issues_count
<fct> <int> <int> <int> <int> <int> <int> <int>
1 11 6 6 2 3 1 6 0
If that is not what you are aiming at, please precise your question.