R在将列表转换为矩阵时处理NULL值

Question

假设我有如下结构：

data = structure(list(person = structure(list(name = "A, B", 
        gender = "F", dead = NULL), .Names = c("name", 
        "gender", "dead")), person = structure(list(name = "C", 
        gender = "M", dead = "RIP"), .Names = c("name", 
        "gender", "dead"))), .Names = c("person", "person"))

我想把它转换成矩阵

data = matrix(unlist(data), nrow = length(data), ncol=length(data[[1]]), byrow = TRUE)

如何避免在使用matrix 时或什至在此之前仅使用base 函数而不使用plyr 的rbind.fill 时回收元素？ 结果是：

> data
     [,1]   [,2]  [,3]  
[1,] "A, B" "F"   "C"   
[2,] "M"    "RIP" "A, B"

我想得到NA 或""，其中值为NULL。例如：

> data
     [,1]   [,2]  [,3]  
[1,] "A, B" "F"   ""   
[2,] "C"    "M"   "RIP" 

任何帮助将不胜感激。

Answer 1

您可以尝试stringi 包中的新stri_list2matrix 函数。

library(stringi)
stri_list2matrix(lapply(data, unlist), byrow=TRUE, fill="")
#      [,1]   [,2] [,3] 
# [1,] "A, B" "F"  ""   
# [2,] "C"    "M"  "RIP"

或者对于NA 而不是""，省略fill 参数

stri_list2matrix(lapply(data, unlist), byrow=TRUE)
#      [,1]   [,2] [,3] 
# [1,] "A, B" "F"  NA   
# [2,] "C"    "M"  "RIP"

或者，如果您更喜欢基本的 R 答案，为避免出现问题，您可以首先使用 length<- 使所有向量的长度相同。这会将NA 附加到所有较短的向量，并使它们与最长向量的长度相同。

len <- max(sapply(data, length)) ## get length of longest vector
t(sapply(unname(data), function(x) `length<-`(unname(unlist(x)), len)))
#      [,1]   [,2] [,3] 
# [1,] "A, B" "F"  NA   
# [2,] "C"    "M"  "RIP"