将两个数据帧列表组合在一起，dataframe为dataframe

I have what is no doubt a simple problem. I have spent the last hour or so looking around for a solution but am clearly missing something. If this is indeed a duplicate please link me to the right way to do this:

我有一个简单的问题。我花了大约一个小时左右寻找解决方案，但显然遗漏了一些东西。如果这确实是重复，请将我链接到正确的方法：

Example data:

示例数据：

names <- c("Cycling1.opr", "Cycling2.opr", "Cycling3.opr")
mydf1 <- data.frame(V1=c(1:5), V2=c(21:25)) 
mydf2 <- data.frame(V1=c(1:10), V2=c(21:30))
mydf3 <- data.frame(V1=c(1:30), V2=c(21:50))
opr <- list(mydf1,mydf2,mydf3)
mydf4 <- data.frame(timestamp=c(1:5))
mydf5 <- data.frame(timestamp=c(1:10))
mydf6 <- data.frame(timestamp=c(1:30))
timestamp <- list(mydf4,mydf5,mydf6)
names(opr) <- names
names(timestamp) <- names

Each list (opr and timestamp) always has the same number of data.frames and when having the same name, each of these data.frames is always the same length. What I would like to do is merge each similarly named dataframe into a single dataframe as part of a final list (perhaps named finalopr) such that its structure is as follows.

每个列表（opr和timestamp）始终具有相同数量的data.frames，并且当具有相同的名称时，这些data.frames中的每一个始终具有相同的长度。我想做的是将每个类似命名的数据帧合并为一个数据帧作为最终列表（可能名为finalopr）的一部分，使其结构如下所示。

dput(finalopr)
list(structure(list(V1 = 1:5, V2 = 21:25, timestamp = 1:5), .Names = c("V1", 
"V2", "timestamp"), row.names = c(NA, -5L), class = "data.frame"), 
structure(list(V1 = 1:10, V2 = 21:30, timestamp = 1:10), .Names = c("V1", 
"V2", "timestamp"), row.names = c(NA, -10L), class = "data.frame"), 
structure(list(V1 = 1:30, V2 = 21:50, timestamp = 1:30), .Names = c("V1", 
"V2", "timestamp"), row.names = c(NA, -30L), class = "data.frame"))

1 个解决方案

#1

> mapply(cbind, opr, timestamp, SIMPLIFY=FALSE)
$Cycling1.opr
  V1 V2 timestamp
1  1 21         1
2  2 22         2
3  3 23         3
4  4 24         4
5  5 25         5

$Cycling2.opr
   V1 V2 timestamp
1   1 21         1
2   2 22         2
3   3 23         3
4   4 24         4
5   5 25         5
6   6 26         6
7   7 27         7
8   8 28         8
9   9 29         9
10 10 30        10

$Cycling3.opr
   V1 V2 timestamp
1   1 21         1
2   2 22         2
3   3 23         3
4   4 24         4
5   5 25         5
6   6 26         6
7   7 27         7
8   8 28         8
9   9 29         9
10 10 30        10
11 11 31        11
12 12 32        12
13 13 33        13
14 14 34        14
15 15 35        15
16 16 36        16
17 17 37        17
18 18 38        18
19 19 39        19
20 20 40        20
21 21 41        21
22 22 42        22
23 23 43        23
24 24 44        24
25 25 45        25
26 26 46        26
27 27 47        27
28 28 48        28
29 29 49        29
30 30 50        30

Here's the structure:

这是结构：

> str(mapply(cbind, opr, timestamp, SIMPLIFY=FALSE))
List of 3
 $ Cycling1.opr:'data.frame':   5 obs. of  3 variables:
  ..$ V1       : int [1:5] 1 2 3 4 5
  ..$ V2       : int [1:5] 21 22 23 24 25
  ..$ timestamp: int [1:5] 1 2 3 4 5
 $ Cycling2.opr:'data.frame':   10 obs. of  3 variables:
  ..$ V1       : int [1:10] 1 2 3 4 5 6 7 8 9 10
  ..$ V2       : int [1:10] 21 22 23 24 25 26 27 28 29 30
  ..$ timestamp: int [1:10] 1 2 3 4 5 6 7 8 9 10
 $ Cycling3.opr:'data.frame':   30 obs. of  3 variables:
  ..$ V1       : int [1:30] 1 2 3 4 5 6 7 8 9 10 ...
  ..$ V2       : int [1:30] 21 22 23 24 25 26 27 28 29 30 ...
  ..$ timestamp: int [1:30] 1 2 3 4 5 6 7 8 9 10 ...