R:具有ID的组的频率

时间:2016-11-02 11:00:33

标签: r frequency

我有一个这样的数据框:

ID Cont
1   a
1   a
1   b
2   a
2   c
2   d

我需要通过ID报告“Cont”的频率。输出应为

ID Cont Freq
1   a    2
1   b    1
2   a    1
2   c    1
2   d    1

3 个答案:

答案 0 :(得分:5)

使用dplyrgroup_byID以及Cont summarise使用n()获取Freq:< / p>

library(dplyr)
res <- df %>% group_by(ID,Cont) %>% summarise(Freq=n())
##Source: local data frame [5 x 3]
##Groups: ID [?]
##
##     ID   Cont  Freq
##  <int> <fctr> <int>
##1     1      a     2
##2     1      b     1
##3     2      a     1
##4     2      c     1
##5     2      d     1

数据:

df <- structure(list(ID = c(1L, 1L, 1L, 2L, 2L, 2L), Cont = structure(c(1L, 
1L, 2L, 1L, 3L, 4L), .Label = c("a", "b", "c", "d"), class = "factor")), .Names = c("ID", 
"Cont"), class = "data.frame", row.names = c(NA, -6L))
##  ID Cont
##1  1    a
##2  1    a
##3  1    b
##4  2    a
##5  2    c
##6  2    d

答案 1 :(得分:3)

library(data.table)
setDT(x)[, .(Freq = .N), by = .(ID, Cont)]

#     ID Cont Freq
# 1:  1    a    2
# 2:  1    b    1
# 3:  2    a    1
# 4:  2    c    1
# 5:  2    d    1

答案 2 :(得分:3)

以基地R:

var response = JSON.parse(responseBody);
response["#!#"].forEach( function(entry) {
   test["foo"] = entry._type === "application";
   ... 
});

如果您想按ID排序,请添加以下行:

df1 <- subset(as.data.frame(table(df)), Freq != 0)