Question

我对data.table的行为感到有些惊讶。我想从data.table所有非NA值中选择一行。

有NA个值，它正在运作：

t = data.table(a=1,b=NA)
t[, !is.na(t), with=F]

没有NA值，它无法正常工作：

t = data.table(a=1, b=2)
t[, !is.na(t), with=F]

基本区别在于t[, !c(F, F), with=F]不起作用。有趣的是t[, c(T, T), with=F]做得很好。

我知道有很多方法可以实现所需的输出，但我只对此感兴趣 - 对我来说很奇怪 - data.table的行为。

Answer 1

我调查了data.table:::`[.data.table` source code

这对我来说确实看起来像个错误。基本上发生的是，!is.na()来电分为!和is.na()来电。然后，它将此向量相加，如果长度为零则返回null.data.table()。问题是，对于dt <- data.table(a = 1, b = 2)，sum(is.na(dt))始终为零。

下面是一段缩短的代码，用以说明引擎盖下的内容

sim_dt <- function(...) {

  ## data.table catches the call
  jsub <- substitute(...)
  cat("This is your call:", paste0(jsub, collapse = ""))

  ## data.table separates the `!` from the call and sets notj = TRUE instead
  ## and saves `is.na(t)` into `jsub`
  if (is.call(jsub) && deparse(jsub[[1L]], 500L, backtick=FALSE) %in% c("!", "-")) {  # TODO is deparse avoidable here?
    notj = TRUE
    jsub = jsub[[2L]]
  } else notj = FALSE

  cat("\nnotj:", notj)
  cat("\nThis is the new jsub: ", paste0(jsub, collapse = "("), ")", sep = "")

  ## data.table evaluates just the `jsub` part which obviously return a vector of `FALSE`s (because `!` was removed)
  cat("\nevaluted j:", j <- eval(jsub, setattr(as.list(seq_along(dt)), 'names', names(dt)), parent.frame()))# else j will be evaluated for the first time on next line

  ## data.table checks if `j` is a logical vector and looks if there are any TRUEs and gets an empty vector
  if (is.logical(j)) cat("\nj after `which`:", j <- which(j))

  cat("\njs length:", length(j), "\n\n")

  ## data.table checks if `j` is empty (and it's obviously is) and returns a null.data.table
  if (!length(j)) return(data.table:::null.data.table()) else return(dt[, j, with = FALSE])

}


## Your data.table
dt <- data.table(a = 1, b = 2)
sim_dt(!is.na(dt))
# This is your call: !is.na(dt)
# notj: TRUE
# This is the new jsub: is.na(dt)
# evaluted j: FALSE FALSE
# j after `which`: 
# js length: 0 
# 
# Null data.table (0 rows and 0 cols)


dt <- data.table(a = 1, b = NA)
sim_dt(!is.na(dt))

# This is your call: !is.na(dt)
# notj: TRUE
# This is the new jsub: is.na(dt)
# evaluted j: FALSE TRUE
# j after `which`: 2
# js length: 1 
# 
#     b
# 1: NA

Answer 2

@Roland已经提到is.na(t)输出是一个矩阵，你需要一个向量来选择列。

但是列选择应该在OP给出的示例中起作用，因为它在data.table中只有一行。我们需要做的就是将它包装在()中以进行评估。例如：

library(data.table)
t = data.table(a=1, b=2)

t[,(!c(FALSE,FALSE)),with=FALSE]
#    a b
# 1: 1 2

t[,(!is.na(t)),with=FALSE]
#    a b
# 1: 1 2

使用布尔值作为列选择器的data.table行为

2 个答案: