Question

我有数据框需要转换为列表，列表名称基于列

df <- data.frame("Name" = c("John", "Dora", "Dora", "Dora"), "V3" = c("Some text", "Some text 1","Some text 2","Some text 3"),  stringsAsFactors = FALSE)

到目前为止，我已经尝试过

df.list <- split(df, seq(nrow(df)))

但这会根据行分别转换为列表

需要什么类似的东西

df.list <- structure(list(`John` = structure(list(V3 = structure(1L, .Label = "Some text", 
                                                               class = "factor")), .Names = "V3", row.names = c(NA, -1L), class = "data.frame"), 
                        `Dora` = structure(list(V3 = structure(c(1L, 2L, 3L), .Label = c("Some text 1", 
                                                                                         "Some text 2",
                                                                                         "Some text 3"), 
                                                               class = "factor")), .Names = "V3", row.names = c(NA, -3L), class = "data.frame")), .Names = c("John", "Dora"))

Answer 1

OP的列在初始数据集中为character类。如果我们需要具有list列的输出factor，请使用as.factor/factor在split之后或之前进行转换

lst <- lapply(split(df['V3'], df$Name), transform, V3 = factor(V3)) 
all.equal(lst, df.list[names(lst)], check.attributes = FALSE)
#[1] TRUE

df基于R中的列值进入列表名称

1 个答案: