将data.frame转换为列表列表

How can I convert a data.frame

如何转换data.frame

df <- data.frame(id=c("af1", "af2"), start=c(100, 115), end=c(114,121))

To a list of lists

到列表列表

LoL <- list(list(id="af1", start=100, end=114), list(id="af2", start=115, end=121))

I've tried things like

我尝试过类似的东西

not.LoL <- as.list(as.data.frame(t(df)))

and I'm really not sure what I end up with after this, but it isn't quite right. My requirement is that I can access, say, the first start by the command

而且我真的不确定在此之后我最终会得到什么,但这不太对劲。我的要求是,我可以访问命令的第一次启动

> LoL[[1]]$start
[1] 100

the not.LoL that I currently have gives me the following error:

我目前拥有的not.LoL给了我以下错误:

> not.LoL[[1]]$start
Error in not.LoL[[1]]$start : $ operator is invalid for atomic vectors

Explanations and/or solutions would be greatly appreciated.

非常感谢解释和/或解决方案。

Edit: I should have made it clear that "id" here is actually non-unique - there can be multiple elements under a single ID. So I could do with a solution that doesn't depend on unique IDs to split on.

编辑:我应该明确这里的“id”实际上是非唯一的 - 在一个ID下可以有多个元素。所以我可以使用不依赖于唯一ID的解决方案来进行拆分。

3 个解决方案

#1

Using plyr , you can do this

使用plyr,您可以这样做

dlply(df,.(id),c)

To avoid grouping by id , if there are multiple ( maybe you need to change column name , id is unique for me)

要避免按ID分组,如果有多个(可能需要更改列名,id对我来说是唯一的)

dlply(df,1,c)

#2

LMAo <- lapply(split(df,df$id), function(x) as.list(x)) # is one way

# more succinctly
# LMAo <- lapply(split(df,df$id), as.list)

An edited solution as per your comment:

根据您的评论编辑的解决方案:

lapply( split(df,seq_along(df[,1])), as.list)

#3

You can use apply to turn your data frame into a list of lists like this:

您可以使用apply将数据框转换为列表列表,如下所示:

LoL <- apply(df,1,as.list)

However, this will change all your data to text, as it passes a single atomic vector to the function.

但是,这会将所有数据更改为文本,因为它将单个原子向量传递给函数。

#1