Rbind R 中两个或多个列表中的对应元素 [英] Rbind corresponding elements in two or more lists in R
问题描述
我有 3 个列表,每个列表有 500 个元素.这里出于演示目的,我有 2 个列表,每个列表有 1 个元素:
I have 3 lists, each with 500 elements. Here for demonstrative purposes, I have 2 lists with 1 element each:
structure(list(timeseries = c(1, 7, 59), t = c(1, 3, 7)), .Names = c("timeseries", "t"), row.names = c(NA, 3L), class = "data.frame")
structure(list(timeseries = c(5, 6, 7), t = c(8, 9, 10)), .Names = c("timeseries", "t"), row.names = c(NA, 3L), class = "data.frame")
我的目标是将列表 1 中的第一个元素与列表 2 和 3 中的第一个元素进行绑定.然后,将列表 1 中的第二个元素与列表 2 和 3 中的第二个元素进行绑定.以此类推.
My aim is to rbind the first element in list 1 with the first element in list 2 and 3. Then, the second element in list 1 with the second element in list 2 and 3. And so on.
在我的例子中,我最终会得到这个表单的列表
In my example, I would end up with a list of this form
structure(list(timeseries = c(1,7,59,5, 6, 7), t = c(1,3,7,8, 9, 10)), .Names = c("timeseries", "t"), row.names = c(NA, 6L), class = "data.frame")
我该怎么做?
谢谢!
****EDIT*** 预期结果的改进示例.我有a和b.我想获得C.
****EDIT*** Improved example of the intended outcome. I have a and b. I want to obtain C.
a<-list(structure(list(timeseries = c(1, 7, 59), t = c(1, 3, 7)), .Names = c("timeseries", "t"), row.names = c(NA, 3L), class = "data.frame"),
structure(list(timeseries = c(1, 7, 59), t = c(1, 3, 7)), .Names = c("timeseries", "t"), row.names = c(NA, 3L), class = "data.frame"))
b<-list(structure(list(timeseries = c(2, 3, 5), t = c(2, 4, 6)), .Names = c("timeseries", "t"), row.names = c(NA, 3L), class = "data.frame"),
structure(list(timeseries = c(60, 70, 80), t = c(20, 30, 40)), .Names = c("timeseries", "t"), row.names = c(NA, 3L), class = "data.frame"))
c<-list(structure(list(timeseries = c(1, 7, 59, 2,3, 5), t = c(1, 3, 7, 2, 4, 6)), .Names = c("timeseries", "t"), row.names = c(NA, 6L), class = "data.frame"),
structure(list(timeseries = c(1, 7, 59, 60, 70, 80), t = c(1, 3, 7, 20, 30, 40)), .Names = c("timeseries", "t"), row.names = c(NA, 6L), class = "data.frame"))
推荐答案
假设 a
和 b
的长度是一样的
Assuming length of a
and b
is the same we can do
lapply(seq_along(a), function(x) rbind(a[[x]], b[[x]]))
#[[1]]
# timeseries t
#1 1 1
#2 7 3
#3 59 7
#4 2 2
#5 3 4
#6 5 6
#[[2]]
# timeseries t
#1 1 1
#2 7 3
#3 59 7
#4 60 20
#5 70 30
#6 80 40
seq_along
生成从 1 到对象长度的序列.如果你这样做
seq_along
generates sequence from 1 to length of the object. If you do
seq_along(a) #you would get output as
#[1] 1 2
因为 length(a)
是 2.所以我们rbind
一个一个的数据帧 rbind(a[[1]], b[[1]])
首先是 rbind(a[[2]], b[[2]])
等等.lapply
确保最终输出是一个列表.
as length(a)
is 2. So we rbind
the dataframe one by one rbind(a[[1]], b[[1]])
first, then rbind(a[[2]], b[[2]])
and so on. lapply
ensures the final output is a list.
这篇关于Rbind R 中两个或多个列表中的对应元素的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!