将两行合并为R中的一列

Question

我如何将两行合并为一行

...
type1,-2.427,-32.962,-61.097
type2,0.004276057,0.0015271631,-0.005192355
type1,-2.427,-32.962,-60.783
type2,0.0018325958,0.0033597588,-0.0021380284
...

至

type1,-2.427,-32.962,-61.097,type2,0.004276057,0.0015271631,-0.005192355
type1,-2.427,-32.962,-60.783,type2,0.0018325958,0.0033597588,-0.0021380284

要么

-2.427,-32.962,-61.097,0.004276057,0.0015271631,-0.005192355
-2.427,-32.962,-60.783,0.0018325958,0.0033597588,-0.0021380284

在GNU R中？

Answer 1

他们是否总是交替1/2？ 如果是这样，您可以只使用基本索引。 使用此样本data.frame

dd<-structure(list(V1 = structure(c(1L, 2L, 1L, 2L), .Label = c("type1", 
"type2"), class = "factor"), V2 = c(-2.427, 0.004276057, -2.427, 
0.0018325958), V3 = c(-32.962, 0.0015271631, -32.962, 0.0033597588
), V4 = c(-61.097, -0.005192355, -60.783, -0.0021380284)), .Names = c("V1", 
"V2", "V3", "V4"), row.names = c(NA, 4L), class = "data.frame")

你可以做

cbind(dd[seq(1,nrow(dd), by=2),], dd[seq(2,nrow(dd), by=2),])

#      V1     V2      V3      V4    V1          V2          V3           V4
# 1 type1 -2.427 -32.962 -61.097 type2 0.004276057 0.001527163 -0.005192355
# 3 type1 -2.427 -32.962 -60.783 type2 0.001832596 0.003359759 -0.002138028

包括“类型”列，也可以

cbind(dd[seq(1,nrow(dd), by=2),-1], dd[seq(2,nrow(dd), by=2),-1])

#       V2      V3      V4          V2          V3           V4
# 1 -2.427 -32.962 -61.097 0.004276057 0.001527163 -0.005192355
# 3 -2.427 -32.962 -60.783 0.001832596 0.003359759 -0.002138028

离开它

Answer 2

这是使用@MrFlick的示例数据的替代方法：

## Use `ave` to create an indicator variable
dd$ind <- with(dd, ave(as.numeric(V1), V1, FUN = seq_along))

## use `reshape` to "merge" your rows by indicator
reshape(dd, direction = "wide", idvar = "ind", timevar = "V1")
#   ind V2.type1 V3.type1 V4.type1    V2.type2    V3.type2     V4.type2
# 1   1   -2.427  -32.962  -61.097 0.004276057 0.001527163 -0.005192355
# 3   2   -2.427  -32.962  -60.783 0.001832596 0.003359759 -0.002138028

Answer 3

您可以使用split按类型拆分数据，然后使用cbind将它们组合在一起。 以下方法从结果中删除第一列（ [-1] ），并且还使用MrFlick的数据dd ，

> do.call(cbind, split(dd[-1], dd[[1]]))
#   type1.V2 type1.V3 type1.V4    type2.V2    type2.V3     type2.V4
# 1   -2.427  -32.962  -61.097 0.004276057 0.001527163 -0.005192355
# 3   -2.427  -32.962  -60.783 0.001832596 0.003359759 -0.002138028

在我的机器上，我认为这是当前三个答案中最快的。