Question

我有一个像

这样的列表的例子

lst<-list(setNames(c(1,10,50,60,70,80),c("id","id1","math","phy","che","bio")),setNames(c(2,20,45),c("id","id1","phy")),setNames(c(3,30,75),c("id","id1","bio")))

我的预期结果：

df<-rbind(c(1,10,50,60,70,80),c(2,20,NA,45,NA,NA),c(3,30,NA,NA,NA,75))
colnames(df)<-c("id","id1","math","phy","che","bio")
df

任何建议将不胜感激。提前谢谢。

Answer 1

您可以在＆＃39; data.table＆＃39;的开发版中试用rbindlist。即v1.9.5。它可以从here

安装

library(data.table) #v1.9.5+
rbindlist(lapply(lst, as.data.frame.list), fill=TRUE)
#    id id1 math phy che bio
#1:  1  10   50  60  70  80
#2:  2  20   NA  45  NA  NA
#3:  3  30   NA  NA  NA  75

Answer 2

另一种方式：

nms = unique(unlist(lapply(lst, names)))
do.call(rbind, lapply(lst, "[", nms))
#     id id1 math phy che bio
#[1,]  1  10   50  60  70  80
#[2,]  2  20   NA  45  NA  NA
#[3,]  3  30   NA  NA  NA  75

Answer 3

另一种方法：

上述问题的基础R解决方案

dataFrame <- do.call(rbind,List)

OR dplyr解决方案

library(dplyr)
dataFrame <- rbind_all(List)

按列名组合列表元素以生成数据帧

3 个答案: