R中第二列的`str`报告是什么？这列中的`atomic`是什么意思？

Question

使用str(survey_OM)我的数据帧上表明我所有的数值数据为atomic.如果我使用class(survey_OM$perc.OM)它返回numeric.

我一直认为第二列str显示了数据类,但它看起来并不那么简单...所以我的问题是:

1. 第二列str报告是什么？
2. atomic它与什么有什么不同numeric？
3. 在这种情况下,为什么会数据显示atomic,不num还是int？

谢谢.
dput(head(survey_OM, 20))规定:

> dput(head(survey_OM, 20))
  structure(list(lake = structure(c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 
  3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("E-2", 
  "E-4", "E pond", "EX 1", "GTH 110", "GTH 112", "GTH 114", "GTH 156", 
  "GTH 91", "GTH 98", "N-1", "NE-10", "NE-11", "NE-3", "NE-8", 
  "NE-9", "NE-9b", "S-10", "S-11", "S-3", "S-6", "S-7"), class = "factor"), 
  date = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
  2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("2007/06/15", 
  "2007/06/18", "2007/06/19", "2007/06/20", "2007/06/21", "2007/06/27", 
  "2007/06/29", "2007/07/07", "2007/07/19", "2007/07/20", "2008/07/26", 
  "2008/07/30", "2008/08/04", "2008/08/06"), class = "factor"), 
  depth = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
  2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("E", 
  "epi", "H", "hypo"), class = "factor"), 
  depth.m = structure(c(6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), .Label = c("", "10.9", "12.9", "1.5", "2", 
  "2.1", "2.2", "2.3", "2.4", "2.5", "2.6", "2.7", "3", "3.1", 
  "3.5", "4", "4.2", "4.8", "4.9", "5", "5.1", "5.5", "6", 
  "6.5", "7", "7.2", "9.9", "not recorded"), class = "factor"), 
  rep = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
  2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A", 
  "B", "C"), class = "factor"), 
  sed = c(0L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 0L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L), 
  notes = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
  1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", 
  "col on SE side", "lg snail shell", "not collected", "very hard sediments"
  ), class = "factor"), 
  dry.mass = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), 
  perc.OM = c(47.1300248455119, 47.4260808104607, 47.7349307375515, 46.4501104675465, 44.1513415737111, 43.5608499678045, 42.9921259842519, 42.2674677347574, 39.6643311064039, 
  39.0968130690949, 46.7768514928267, 46.9211608642763, 46.7877013177158, 
  47.0709930313588, 44.3241152581706, 43.7905468025952, 41.706074101281, 
  36.5061097383474, 37.4329041152142, 37.7757939038389)), .Names = c("lake", 
  "date", "depth", "depth.m", "rep", "sed", "notes", "dry.mass", 
  "perc.OM"), comment = c("working data frame of the sediment char from the 2007 sed    survey       created:", "Wed Apr 27 14:23:33 2011"), row.names = c(NA, 20L), class = "data.frame")

并且完整的输出str(survey_OM)是:

> str(survey_OM)
'data.frame':   780 obs. of  9 variables:
 $ lake    : Factor w/ 22 levels "E-2","E-4","E pond",..: 3 3 3 3 3 3 3 3 3 3 ...
  ..- attr(*, "comment")= chr "names of the lakes"
 $ date    : Factor w/ 14 levels "2007/06/15","2007/06/18",..: 2 2 2 2 2 2 2 2 2 2 ...
  ..- attr(*, "comment")= chr "date that the cores were collected"
 $ depth   : Factor w/ 4 levels "E","epi","H",..: 2 2 2 2 2 2 2 2 2 2 ...
  ..- attr(*, "comment")= chr "relative depth ID; epi = shallowest corable Z, hypo =   deepest Z, S, M, D = shallow, med, deep"
 $ depth.m : Factor w/ 28 levels "","10.9","12.9",..: 6 6 6 6 6 6 6 6 6 6 ...
  ..- attr(*, "comment")= chr "depth that core was collected in m"
 $ rep     : Factor w/ 3 levels "A","B","C": 1 1 1 1 1 1 1 1 1 1 ...
  ..- attr(*, "comment")= chr "replicate ID for core"
 $ sed     : atomic  0 1 2 3 4 5 6 7 8 9 ...
  ..- attr(*, "comment")= chr "depth of sample from sed/water interface in cm"
 $ notes   : Factor w/ 5 levels "","col on SE side",..: 1 1 1 1 1 1 1 1 1 1 ...
  ..- attr(*, "comment")= chr "comments on sample"
 $ dry.mass: atomic  0 0 0 0 0 0 0 0 0 0 ...
  ..- attr(*, "comment")= chr "dry mass of the sediment at that sed Z in g/m^2"
 $ perc.OM : atomic  47.1 47.4 47.7 46.5 44.2 ...
  ..- attr(*, "comment")= chr "percent OM of the samp. based on LOI at 550d C"
 - attr(*, "comment")= chr  "working data frame of the sediment char from the 2007 sed survey created:" "Wed Apr 27 14:23:33 2011"

Answer 1

看着utils:::str.default,我们看到,我们得到的通常的输出int,num等等,如果下面的if说法是正确的:

if (     is.vector(object) 
     || (is.array(object) && is.atomic(object))
     ||  is.vector(object, mode = "language") 
     || is.vector(object, mode = "symbol")
   )

我们得到atomic,如果这种说法是错误的(这否则将是int,num等).

查看帮助页面is.vector,我们看到只有当它是一个没有名称以外的属性的向量时它才返回true.这是一个数据框,其中b有一个额外的属性:

d <- data.frame(a=1:4, b=5:8)
attr(d$b, "tmp") <- "surprise!"

并呼吁str它给人atomic的b替代int.

> str(d)
'data.frame':   4 obs. of  2 variables:
 $ a: int  1 2 3 4
 $ b: atomic  5 6 7 8
  ..- attr(*, "tmp")= chr "surprise!"

我在你的编辑中看到你在数据框的元素上有额外的属性,并且你正在获得关于你的属性的额外线条,所以这似乎解释了它.