R：根据另一列以0填充行

Question

I have a df with 3 columns: day, hour, sessions.

I need to complete hours that are not present in data frame with 0, based on weekdays. Hours of days goes from 1 to 23.

For example, for Sunday there is data from 6 am, but not for < 6 am (1,2,3,4,5).

Is there a function for these? pkg?

Original data example:

  day   hour  sessions
  <fct> <chr>    <dbl>
1 Sun   06        1.00
2 Sun   10        2.00
3 Sun   11        2.00
4 Sun   12        1.00
5 Sun   18        1.00
6 Sun   20        2.00

Desired output: (after 6 am, there are more hours to be added with 0, but you get the idea)

  day   hour  sessions
  <fct> <chr>    <dbl>
1 Sun   01        0.00
2 Sun   02        0.00
3 Sun   03        0.00
4 Sun   04        0.00
5 Sun   05        0.00
6 Sun   06        1.00
7 Sun   10        2.00
8 Sun   11        2.00
9 Sun   12        1.00
10 Sun   18        1.00
11 Sun   20        2.00

Data:

df <- structure(list(day = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 
4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 6L, 
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 
7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L), .Label = c("Sun", "Sat", 
"Fri", "Thu", "Wed", "Tue", "Mon"), class = "factor"), hour = c("06", 
"10", "11", "12", "18", "20", "21", "23", "01", "08", "09", "11", 
"13", "14", "16", "20", "21", "23", "07", "08", "09", "10", "11", 
"12", "14", "15", "16", "17", "18", "19", "20", "21", "23", "08", 
"10", "11", "12", "13", "14", "15", "16", "18", "19", "20", "04", 
"09", "10", "11", "13", "15", "16", "17", "18", "19", "06", "07", 
"09", "10", "11", "12", "15", "16", "17", "18", "21", "23", "04", 
"06", "09", "10", "11", "12", "14", "15", "16", "17", "19", "20", 
"21", "22"), sessions = c(1, 2, 2, 1, 1, 2, 1, 2, 1, 1, 1, 1, 
2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 3, 2, 3, 1, 1, 1, 2, 2, 1, 2, 2, 
1, 2, 1, 1, 1, 1, 3, 1, 1, 2, 3, 1, 1, 1, 3, 1, 2, 4, 4, 3, 1, 
1, 2, 1, 1, 3, 4, 2, 2, 1, 1, 3, 1, 1, 2, 2, 1, 1, 4, 1, 1, 2, 
1, 1, 1, 2, 3)), .Names = c("day", "hour", "sessions"), row.names = c(NA, 
-80L), totals = list(structure(list(sessions = "134"), .Names = "sessions")), minimums = list(
    structure(list(sessions = "1"), .Names = "sessions")), maximums = list(
    structure(list(sessions = "4"), .Names = "sessions")), rowCount = 116L, vars = "day", drop = TRUE, class = c("grouped_df", 
"tbl_df", "tbl", "data.frame"))

Answer 1

Here is a simple solution with the dplyr package. I coerced the hour variable to a numeric first.

library(dplyr)
allHours <- function(data) {
  data.frame(hour = 1:23, 
             sessions = sapply(1:23, function(x) sum(data$sessions[data$hour == x])))
}
df %>% 
  mutate(hour = as.numeric(hour)) %>%
  group_by(day) %>%
  do(allHours(.))

Output:

# A tibble: 161 x 3
# Groups:   day [7]
      day  hour sessions
   <fctr> <int>    <dbl>
 1    Sun     1        0
 2    Sun     2        0
 3    Sun     3        0
 4    Sun     4        0
 5    Sun     5        0
 6    Sun     6        1
 7    Sun     7        0
 8    Sun     8        0
 9    Sun     9        0
10    Sun    10        2
# ... with 151 more rows