很多时候，我想转换一个列表，其中每个索引具有相同的元素类型到数据帧。例如，我可能有一个列表：
Very often I want to convert a list wherein each index has identical element types to a data frame. For example, I may have a list:
> my.list [[1]] [[1]]$global_stdev_ppb [1] 24267673 [[1]]$range [1] 0.03114799 [[1]]$tok [1] "hello" [[1]]$global_freq_ppb [1] 211592.6 [[2]] [[2]]$global_stdev_ppb [1] 11561448 [[2]]$range [1] 0.08870838 [[2]]$tok [1] "world" [[2]]$global_freq_ppb [1] 1002043我想将此列表转换为数据帧， index元素是一列。自然（对我）的事情是使用 do.call ：
I want to convert this list to a data frame where each index element is a column. The natural (to me) thing to go is to is use do.call:
> my.matrix<-do.call("rbind", my.list) > my.matrix global_stdev_ppb range tok global_freq_ppb [1,] 24267673 0.03114799 "hello" 211592.6 [2,] 11561448 0.08870838 "world" 1002043直截了当，但是当我尝试将此矩阵转换为数据帧时，列仍然是列表元素，而不是向量：
Straightforward enough, but when I attempt to cast this matrix as a data frame, the columns remain list elements, rather than vectors:
> my.df<-as.data.frame(my.matrix, stringsAsFactors=FALSE) > my.df[,1] [[1]] [1] 24267673 [[2]] [1] 11561448目前，为了正确地获取数据框架，我使用 unlist 和 as.vector ，然后重新创建数据框：
Currently, to get the data frame cast properly I am iterating over each column using unlist and as.vector, then recasting the data frame as such:
new.list<-lapply(1:ncol(my.matrix), function(x) as.vector(unlist(my.matrix[,x]))) my.df<-as.data.frame(do.call(cbind, new.list), stringsAsFactors=FALSE)但是，似乎效率很低。有没有更好的方法呢？
This, however, seem very inefficient. Is there are better way to do this?
推荐答案我想你想要的：
> do.call(rbind, lapply(my.list, data.frame, stringsAsFactors=FALSE)) global_stdev_ppb range tok global_freq_ppb 1 24267673 0.03114799 hello 211592.6 2 11561448 0.08870838 world 1002043.0 > str(do.call(rbind, lapply(my.list, data.frame, stringsAsFactors=FALSE))) 'data.frame': 2 obs. of 4 variables: $ global_stdev_ppb: num 24267673 11561448 $ range : num 0.0311 0.0887 $ tok : chr "hello" "world" $ global_freq_ppb : num 211593 1002043更多推荐
将列表作为数据框架最有效的方法是什么？
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