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我有一个小标题如下:

uuu <- structure(list(IsCharacter = c("a", "b"),
                      ShouldBeCharacter = list("One", "Another"),
                      IsList = list("Element1", c("Element2", "Element3"))
               ),
           .Names = c("IsCharacter", "ShouldBeCharacter", "IsList"),
            row.names = c(NA, -2L), class = c("tbl_df", "tbl", "data.frame"))
uuu
## A tibble: 2 × 3
#  IsCharacter ShouldBeCharacter    IsList
#        <chr>            <list>    <list>
#1           a         <chr [1]> <chr [1]>
#2           b         <chr [1]> <chr [2]>

我想将“ShouldBeCharacter”之类的列转换为所有元素都具有相同长度和类型的列,类似于“IsCharacter”,其余列保持不变。

到目前为止,我有以下功能可以解决这个问题,但对我来说它看起来很hacky。我想知道是否有更好的解决方案我没有考虑:

lists_to_atomic <- function(data) {
  # Elements of length larger than one should be kept as lists.
  # So we compute the maximum length for each column
  length_column_elements <- apply(data, 2,
                                  function(x) max(sapply(x, function(y) length(y))))
  # to_simplify will contain column names of class list and with all elements of length 1
  to_simplify <- colnames(data)[length_column_elements == 1 & sapply(data, class) == "list"]
  # Do the conversion
  data[,to_simplify] <- tibble::as_tibble(lapply(as.list(data[,to_simplify]), function(x) {do.call(c, x)}))
  return(data)  
}

这是我得到的结果,注意 ShouldBeCharacter 的类型是如何改变的:

lists_to_atomic(uuu)
## A tibble: 2 × 3
#  IsCharacter ShouldBeCharacter    IsList
#        <chr>             <chr>    <list>
#1           a               One <chr [1]>
#2           b           Another <chr [2]>

as_tibble(lapply(as.list(... do.call(c,...)))条线对我来说看起来太复杂了,但我找不到更简单的替代方案。

是否有任何简化使我的lists_to_atomic功能更可靠?

更新

我没有考虑tidyr::unnest在列表类型的列和长度为 1 的元素上使用,但是按照@taavi-p 的回答,我已经能够将函数简化为:

lists_to_atomic <- function(data) {
  # Elements of length larger than one should be kept as lists.
  # So we compute the maximum length for each column
  length_column_elements <- apply(data, 2,
                                  function(x) max(sapply(x, function(y) length(y))))
  # to_simplify will contain column names of class list and with all elements of length 1
  to_simplify <- colnames(data)[length_column_elements == 1 & 
                                vapply(data,
                                       FUN = function(x) "list" %in% class(x),
                                       FUN.VALUE = logical(1))]

  # Do the conversion
  data2 <- tidyr::unnest_(data, unnest_cols = to_simplify)
  data2 <- data2[, colnames(data)] # Preserve original column order
  return(data2)
}
4

1 回答 1

3

你可以试试:

     library(tidyr)
     uuu %>% unnest(ShouldBeCharacter) 

有关如何处理列表列的更多示例,请参见“R for Data Science”:http ://r4ds.had.co.nz/many-models.html#list-columns-1

于 2016-10-21T09:06:12.173 回答