Question

我有3个列表，每个列表包含500个元素。这里出于说明目的，我有2个列表，每个列表1个元素：

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中的第一个元素绑定在一起。然后，将列表1中的第二个元素与列表2和3中的第二个元素绑定。以此类推。

在我的示例中，我最终得到了这种形式的列表

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！

****编辑***预期结果的改进示例。我有a和b。我想获得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"))

Answer 1

假设a和b的长度是我们可以做的

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。如果您这样做

seq_along(a) #you would get output as
#[1] 1 2

因为length(a)是2。所以我们首先rbind一个rbind(a[[1]], b[[1]])一个rbind(a[[2]], b[[2]])一个数据帧，然后是lapply，依此类推。 def make_single(op,size,end,sign): name = '_%s%s%s%s' % (op, size, end, sign) fmt = sizes[size] end = '>' if end == 'b' else '<'确保最终输出是列表。

Answer 2

只需尝试map2函数：

purrr::map2(a,b,rbind) -> d
identical(c,d)
# [1] TRUE

将R中两个或多个列表中的对应元素绑定

2 个答案: