data.frame行到列表 [英] data.frame rows to a list
本文介绍了data.frame行到列表的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
到目前为止,我已经以下列方式解决了这个问题:但是我想知道是否有更好的方法来处理这个问题。
xy.df< - data.frame(x = runif (10),y = runif(10))
#预先分配列表并填充一个循环
xy.list < - vector(list,nrow(xy 。$)
for(i in 1:nrow(xy.df)){
xy.list [[i]]< - xy.df [i,]
}
解决方案
像这样:
xy.list< - split(xy.df,seq(nrow(xy.df)))
pre>
如果您希望将
xy.df
的rownames作为输出列表的名称,您可以do:xy.list< - setNames(split(xy.df,seq(nrow(xy.df))) ,rownames(xy.df))
I have a data.frame which I would like to convert to a list by rows, meaning each row would correspond to its own list elements. In other words, I would like a list that is as long as the data.frame has rows.
So far, I've tackled this problem in the following manner, but I was wondering if there's a better way to approach this.
xy.df <- data.frame(x = runif(10), y = runif(10)) # pre-allocate a list and fill it with a loop xy.list <- vector("list", nrow(xy.df)) for (i in 1:nrow(xy.df)) { xy.list[[i]] <- xy.df[i,] }
解决方案Like this:
xy.list <- split(xy.df, seq(nrow(xy.df)))
And if you want the rownames of
xy.df
to be the names of the output list, you can do:xy.list <- setNames(split(xy.df, seq(nrow(xy.df))), rownames(xy.df))
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