我在R中有两个数据集(下面这些表只是较小的版本),我想将它们组合成一个新的数据框。
> meetingtime2
#two columns of datetime that class=factor
ST ET
1 2014-12-22 07:00:00 2014-12-22 07:30:00
2 2014-12-22 07:30:00 2014-12-22 08:00:00
3 2014-12-22 08:00:00 2014-12-22 08:30:00
4 2014-12-22 08:30:00 2014-12-22 09:00:00
5 2014-12-22 09:00:00 2014-12-22 09:30:00
> roomdata2
#three columns; Room=factor, Capacity=integer, Video Conference=numeric
Room Capacity Video.Conference
1 0M02A 16 1
2 0M03A 8 0
3 0M03B 12 1
所需的输出是15行×5列矩阵。简单来说,每个房间的每个时段都有输出。
#the following is a MANUALLY created output of what the first few rows should look like
Room Capacity Video.Conference ST ET
1 0M02A 16 1 2014-12-22 07:00:00 2014-12-22 07:30:00
2 0M02A 16 1 2014-12-22 07:30:00 2014-12-22 08:00:00
3 0M02A 16 1 2014-12-22 08:00:00 2014-12-22 08:30:00
4 0M02A 16 1 2014-12-22 08:30:00 2014-12-22 09:00:00
5 0M02A 16 1 2014-12-22 09:00:00 2014-12-22 09:30:00
6 0M03A 16 1 2014-12-22 07:00:00 2014-12-22 07:30:00
7 0M03A 16 1 2014-12-22 07:30:00 2014-12-22 08:00:00
#and so forth to 15 rows.
我尝试过使用嵌套循环
#note, the code is written so I can apply to a bigger (1000's of rows) dataset
>mylist<-list()
>for(i in 1:(nrow(roomdata2)))
+{ for(j in 1:(nrow(meetingtime2)))
+mylist[[j]]<- data.frame(roomdata2[i,1],roomdata2[i,2],roomdata2[i,3],
+meetingtime2[j,1],meetingtime2[j,2])
}
>df<-do.call("rbind",mylist)
>df
我得到的输出。我得到了最后一个房间的所有时间段,而不是前面的房间
roomdata2.i..1. roomdata2.i..2. roomdata2.i..3. meetingtime2.j..1. meetingtime2.j..2.
1 0M03B 12 1 2014-12-22 07:00:00 2014-12-22 07:30:00
2 0M03B 12 1 2014-12-22 07:30:00 2014-12-22 08:00:00
3 0M03B 12 1 2014-12-22 08:00:00 2014-12-22 08:30:00
4 0M03B 12 1 2014-12-22 08:30:00 2014-12-22 09:00:00
5 0M03B 12 1 2014-12-22 09:00:00 2014-12-22 09:30:00
我知道我的代码远非正确,并且正在给我循环的最后一次迭代。
我看到的另一种方法是每次迭代的连续打印功能
>for(i in 1:(nrow(roomdata2)))
>for(j in 1:(nrow(meetingtime2)))
>print(paste(roomdata2[i,1],roomdata2[i,2],roomdata2[i,3],
+meetingtime2[j,1],meetingtime2[j,2]))
输出
[1] "0M02A 16 1 2014-12-22 07:00:00 2014-12-22 07:30:00"
[1] "0M02A 16 1 2014-12-22 07:30:00 2014-12-22 08:00:00"
[1] "0M02A 16 1 2014-12-22 08:00:00 2014-12-22 08:30:00"
[1] "0M02A 16 1 2014-12-22 08:30:00 2014-12-22 09:00:00"
[1] "0M02A 16 1 2014-12-22 09:00:00 2014-12-22 09:30:00"
[1] "0M03A 8 0 2014-12-22 07:00:00 2014-12-22 07:30:00"
[1] "0M03A 8 0 2014-12-22 07:30:00 2014-12-22 08:00:00"
[1] "0M03A 8 0 2014-12-22 08:00:00 2014-12-22 08:30:00"
[1] "0M03A 8 0 2014-12-22 08:30:00 2014-12-22 09:00:00"
[1] "0M03A 8 0 2014-12-22 09:00:00 2014-12-22 09:30:00"
[1] "0M03B 12 1 2014-12-22 07:00:00 2014-12-22 07:30:00"
[1] "0M03B 12 1 2014-12-22 07:30:00 2014-12-22 08:00:00"
[1] "0M03B 12 1 2014-12-22 08:00:00 2014-12-22 08:30:00"
[1] "0M03B 12 1 2014-12-22 08:30:00 2014-12-22 09:00:00"
[1] "0M03B 12 1 2014-12-22 09:00:00 2014-12-22 09:30:00"
#however the values are not separated, they are just in one set of string for each row.
所需的结果是直接在上面的表格,而是一个数据框,其中每个值都在一个单独的列中(每个日期和时间在一列中设置在一起)。
我已经查看了列表,lapply,foreach但是我无法绕过解决方案。 任何帮助将不胜感激,我是一个初学者,所以我很想学习。
干杯 * dputs
>dput(meetingtime2)
结构(列表(ST =结构(1:5,.Label = c(&#34; 22/12/2014 7:00&#34;, &#34; 22/12/2014 7:30&#34;,&#34; 22/12/2014 8:00&#34;,&#34; 22/12/2014 8:30&#34;,&# 34; 22/12/2014 9:00&#34; ),class =&#34; factor&#34;),ET = structure(1:5,.Label = c(&#34; 22/12/2014 7:30&#34;, &#34; 22/12/2014 8:00&#34;,&#34; 22/12/2014 8:30&#34;,&#34; 22/12/2014 9:00&#34;,&# 34; 22/12/2014 9:30&#34; ),class =&#34; factor&#34;)),. Name = c(&#34; ST&#34;,&#34; ET&#34;),row.names = c(NA, -5L),class =&#34; data.frame&#34;)
>dput(roomdata2)
结构(列表(房间=结构(1:3,.Label = c(&#34; 0M02A&#34;,&#34; 0M03A&#34;, &#34; 0M03B&#34;),class =&#34; factor&#34;),Capacity = c(16L,8L,12L),Video.Conference = c(1L, 0L,1L)),。Name = c(&#34; Room&#34;,&#34; Capacity&#34;,&#34; Video.Conference&#34;),row.names = c(NA, -3L),class =&#34; data.frame&#34;)
答案 0 :(得分:3)
使用您的数据:
meetingtime2 <- read.csv(text = "ST,ET
2014-12-22 07:00:00,2014-12-22 07:30:00
2014-12-22 07:30:00,2014-12-22 08:00:00
2014-12-22 08:00:00,2014-12-22 08:30:00
2014-12-22 08:30:00,2014-12-22 09:00:00
2014-12-22 09:00:00,2014-12-22 09:30:00")
roomdata2 <- read.csv(text = "Room,Capacity,Video_Conference
0M02A,16,1
0M03A,8,0
0M03B,12,1")
然后merge
方便地返回笛卡尔积,因为没有列匹配。
merge(meetingtime2, roomdata2)[, c(3:5, 1:2)]
## Room Capacity Video_Conference ST ET
## 1 0M02A 16 1 2014-12-22 07:00:00 2014-12-22 07:30:00
## 2 0M02A 16 1 2014-12-22 07:30:00 2014-12-22 08:00:00
## 3 0M02A 16 1 2014-12-22 08:00:00 2014-12-22 08:30:00
## 4 0M02A 16 1 2014-12-22 08:30:00 2014-12-22 09:00:00
## 5 0M02A 16 1 2014-12-22 09:00:00 2014-12-22 09:30:00
答案 1 :(得分:0)
这很难看,但应该完成工作。鉴于以下数据:
ST <- c('2014-12-22 07:00:00', '2014-12-22 07:30:00', '2014-12-22 08:00:00', '2014-12-22 08:30:00', '2014-12-22 09:00:00')
ET <- c('2014-12-22 07:30:00', '2014-12-22 08:00:00', '2014-12-22 08:30:00', '2014-12-22 09:00:00', '2014-12-22 09:30:00')
RoomName <- c('0M02A', '0M03A', '0M03B')
Capacity <- c(16, 8, 12)
VideoCap <- c(1, 0, 1)
Times <- data.frame(ST, ET, stringsAsFactors = FALSE)
Rooms <- data.frame(RoomName, Capacity, VideoCap,stringsAsFactors = FALSE)
下面的功能应该做你想要的:
Smash <- function(DF1, DF2){
nm <- dim(DF1)
pq <- dim(DF2)
maxrow <- nm[[1]] * pq[[1]]
maxcol <- nm[[2]] + pq[[2]]
MAT <- matrix('A', nrow = maxrow, ncol = maxcol)
currow <- 1
for (i1 in seq_len(nm[[1]])) {
for (i2 in seq_len(pq[[1]])) {
curcol <- 1
for (j in seq_len(nm[[2]])) {
MAT[currow, curcol] <- DF1[i1, j]
curcol <- curcol + 1
}
for (j in seq_len(pq[[2]])) {
MAT[currow, curcol] <- DF2[i2, j]
curcol <- curcol + 1
}
currow <- currow + 1
}
}
DF <- data.frame(MAT)
names(DF) <- c(names(DF1), names(DF2))
return(DF)
}
Smash(Rooms,Times)返回:
> Smash(Rooms, Times)
RoomName Capacity VideoCap ST ET
1 0M02A 16 1 2014-12-22 07:00:00 2014-12-22 07:30:00
2 0M02A 16 1 2014-12-22 07:30:00 2014-12-22 08:00:00
3 0M02A 16 1 2014-12-22 08:00:00 2014-12-22 08:30:00
4 0M02A 16 1 2014-12-22 08:30:00 2014-12-22 09:00:00
5 0M02A 16 1 2014-12-22 09:00:00 2014-12-22 09:30:00
6 0M03A 8 0 2014-12-22 07:00:00 2014-12-22 07:30:00
7 0M03A 8 0 2014-12-22 07:30:00 2014-12-22 08:00:00
8 0M03A 8 0 2014-12-22 08:00:00 2014-12-22 08:30:00
9 0M03A 8 0 2014-12-22 08:30:00 2014-12-22 09:00:00
10 0M03A 8 0 2014-12-22 09:00:00 2014-12-22 09:30:00
11 0M03B 12 1 2014-12-22 07:00:00 2014-12-22 07:30:00
12 0M03B 12 1 2014-12-22 07:30:00 2014-12-22 08:00:00
13 0M03B 12 1 2014-12-22 08:00:00 2014-12-22 08:30:00
14 0M03B 12 1 2014-12-22 08:30:00 2014-12-22 09:00:00
15 0M03B 12 1 2014-12-22 09:00:00 2014-12-22 09:30:00