使用R计算相同的速率
calculate the rate under the same in using R
我有一个问题要计算相同身份证号码下的费率。
这是示例数据集 d:
id answer
1 1
1 0
1 0
1 1
1 1
1 1
1 0
2 0
2 0
2 0
3 1
3 0
理想的输出是
id rate freq
1 4/7 (=0.5714) 7
2 0 3
3 1/2 (=0.5) 2
谢谢。
尝试
library(data.table)
setDT(df1)[,list(rate= mean(answer), freq=.N) ,id]
# id rate freq
#1: 1 0.5714286 7
#2: 2 0.0000000 3
#3: 3 0.5000000 2
或者
library(dplyr)
df1 %>%
group_by(id) %>%
summarise(rate=mean(answer), freq=n())
数据
df1 <- structure(list(id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
3L, 3L), answer = c(1L, 0L, 0L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 1L,
0L)), .Names = c("id", "answer"), class = "data.frame",
row.names = c(NA, -12L))
为了好玩，您可以使用aggregate
> aggregate(answer~id, function(x) c(rate=mean(x), freq=length(x)), data=df1)
id answer.rate answer.freq
1 1 0.5714286 7.0000000
2 2 0.0000000 3.0000000
3 3 0.5000000 2.0000000
我有一个问题要计算相同身份证号码下的费率。 这是示例数据集 d:
id answer
1 1
1 0
1 0
1 1
1 1
1 1
1 0
2 0
2 0
2 0
3 1
3 0
理想的输出是
id rate freq
1 4/7 (=0.5714) 7
2 0 3
3 1/2 (=0.5) 2
谢谢。
尝试
library(data.table)
setDT(df1)[,list(rate= mean(answer), freq=.N) ,id]
# id rate freq
#1: 1 0.5714286 7
#2: 2 0.0000000 3
#3: 3 0.5000000 2
或者
library(dplyr)
df1 %>%
group_by(id) %>%
summarise(rate=mean(answer), freq=n())
数据
df1 <- structure(list(id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
3L, 3L), answer = c(1L, 0L, 0L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 1L,
0L)), .Names = c("id", "answer"), class = "data.frame",
row.names = c(NA, -12L))
为了好玩，您可以使用aggregate
> aggregate(answer~id, function(x) c(rate=mean(x), freq=length(x)), data=df1)
id answer.rate answer.freq
1 1 0.5714286 7.0000000
2 2 0.0000000 3.0000000
3 3 0.5000000 2.0000000