我希望找到每个主题的行数,其中每个主题的值大于11,并将其输出到数据框中进行分析。数据集很大(5000行),因此需要一个函数。
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
我想要的输出是
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
非常感谢任何帮助。
答案 0 :(得分:6)
一个选项是来自aggregate
base R
aggregate(cbind(highvalues=value>11)~., df, sum)
或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
答案 1 :(得分:2)
您可以采用tidyverse
方式:
df %>%
filter(value > 11) %>%
group_by(subject,day) %>%
mutate(highvalue = n()) %>%
select(subject, day, highvalue) %>%
unique()
答案 2 :(得分:1)
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