Question

我得到了一个data.frame，看起来像下面那个：

OBJECT ID   TASK
1           A
1           C
1           D
1           E
2           A
2           B
2           C
2           D
2           F

现在我想计算data.frame中唯一的连续组合，以获得以下结果：

PREDECESSOR  SUCCESSOR  COUNT
A            C          1
C            D          2
D            E          1
A            B          1
B            C          1
D            F          1

我已经想出在两个for循环的帮助下提取连续值，但是我在新data.frame（或list中的分配和计数任务失败了}）。

Answer 1

使用data.table的解决方案：

代码：

library(data.table)
setDT(df)
df[, TASK0 := shift(TASK), OBJECT]
df[!is.na(TASK0), .N, .(TASK, TASK0)][, .(
    COUNT = sum(N)), .(PREDECESSOR = TASK0, SUCCESSOR = TASK)]

结果：

   PREDECESSOR SUCCESSOR COUNT
1:           A         C     1
2:           C         D     2
3:           D         E     1
4:           A         B     1
5:           B         C     1
6:           D         F     1

说明：

setDT(df)：将data.frame转换为data.table对象
[, TASK0 := shift(TASK), OBJECT]：获取每个OBJECT
!is.na(TASK0)：摆脱每个OBJECT的第一行（他们没有PREDECESSOR）
.N, .(TASK, TASK0)：计算TASK和TASK0（之前的字母组合）的出现次数
sum(N)：总计数

数据（df）：

structure(list(OBJECT = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L), 
    TASK = c("A", "C", "D", "E", "A", "B", "C", "D", "F")), .Names = c("OBJECT", 
"TASK"), row.names = c(NA, -9L), class = c("data.table", "data.frame"
))

Answer 2

aggregate(COUNT~.,
          data.frame(PREDECESSOR = head(df1$TASK, -1),
                     SUCCESSOR = tail(df1$TASK, -1),
                     COUNT = 1),
          length)
#  PREDECESSOR SUCCESSOR COUNT
#1           E         A     1
#2           A         B     1
#3           A         C     1
#4           B         C     1
#5           C         D     2
#6           D         E     1
#7           D         F     1

您可以使用类似的方法，即使您希望split OBJECT.ID之前temp = do.call(rbind, lapply(split(df1, df1$OBJECT.ID), function(X){ aggregate(COUNT~., data.frame(PREDECESSOR = head(X$TASK, -1), SUCCESSOR = tail(X$TASK, -1), COUNT = 1), length) })) aggregate(COUNT~., temp, length) # PREDECESSOR SUCCESSOR COUNT #1 A C 1 #2 B C 1 #3 C D 2 #4 D E 1 #5 A B 1 #6 D F 1

df1 = structure(list(OBJECT.ID = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
2L), TASK = c("A", "C", "D", "E", "A", "B", "C", "D", "F")), .Names = c("OBJECT.ID", 
"TASK"), class = "data.frame", row.names = c(NA, -9L))

数据

{{1}}

Answer 3

为了获得计数，你可以用以下两行来完成：

cc <- cbind(df$TASK,c(df$TASK[-1],"LAST"))
table(paste(cc[,1],cc[2],sep="-"))

结果是

A-B  A-C  B-C  C-D  D-E  D-F  E-A F-LAST 
1    1    1    2    1    1    1   1

计算data.frame中的后继组合

3 个答案: