Calculate the number of rows in a data frame above a threshold as a function or other column factors

Question

I would like to find the number for rows for each object for every day when the value is greater than 11 and displays them in a data frame for analysis. The data set is large (5000 rows), so a function is needed for this.

subject = c(rep("A", 12), rep("B", 12))        
day = c(1,1,1,1,2,2,2,2,3,3,3,3,1,1,1,1,2,2,2,2,3,3,3,3)
value = c(13,14,15,5,12,9,6,14,4,2,1,2,13,14,15,5,12,9,6,14,2,2,2,3)
df = data.frame(subject, day, value)
df

   subject day value
1        A   1    13
2        A   1    14
3        A   1    15
4        A   1     5
5        A   2    12
6        A   2     9
7        A   2     6
8        A   2    14
9        A   3     4
10       A   3     2
11       A   3     1
12       A   3     2
13       B   1    13
14       B   1    14
15       B   1    15
16       B   1     5
17       B   2    12
18       B   2     9
19       B   2     6
20       B   2    14
21       B   3     2
22       B   3     2
23       B   3     2
24       B   3     3

The output I would like would be

subject.agg = c(rep("A", 3), rep("B", 3)) 
day.agg = as.factor(c(1,2,3,1,2,3))
highvalues = (c(3,2,0,3,2,0))
df.agg = data.frame(subject.agg,day.agg,highvalues)
df.agg

  subject.agg day.agg highvalues
1           A       1          3
2           A       2          2
3           A       3          0
4           B       1          3
5           B       2          2
6           B       3          0

Any help really enjoyed.

Answer 1

One parameter aggregatefrombase R

aggregate(cbind(highvalues=value>11)~., df,  sum)

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Or using data.table

library(data.table)
setDT(df)[value>11, .(highvalues=.N), by = .(subject, day)]
#     subject day highvalues
#1:       A   1          3
#2:       A   2          2
#3:       A   3          3
#4:       B   1          3
#5:       B   2          2
#6:       B   3          3

Answer 2

You can choose a method tidyverse:

df %>%
  filter(value > 11) %>%
  group_by(subject,day) %>%
  mutate(highvalue = n()) %>%
  select(subject, day, highvalue) %>%
  unique()

Answer 3

library(data.table)
dt = setDT(df)
dt[, sum(value>11),by = .(subject,day)]
   subject day V1
1:       A   1  3
2:       A   2  2
3:       A   3  3
4:       B   1  3
5:       B   2  2
6:       B   3  3