2

我正在这样做：

myfun <- function(inputvar_vec){
# inputvar_vec is input vector
# do something
# result = output vector
return(result)
}

DT[, result := lapply(.SD, myfun), by = byvar, .SDcols = inputvar]

我收到以下警告：

Warning message:
`In `[.data.table`(df1, , `:=`(prop, lapply(.SD, propEventInLastK)),  :
Invalid .internal.selfref detected and fixed by taking a copy of the whole table, 
so that     := can add this new column by reference. At an earlier point, this 
data.table has been copied by R (or been created manually using structure() 
or similar). (and then some more stuff) .... `

我的猜测是因为我正在堆叠result向量（在 by 操作之后），正在制作副本？

任何人都可以建议一种删除此警告的方法吗？我已经使用 apply 函数完成了这项工作，并认为它也应该在这里可以扩展。

我的另一个问题是：您可以从数据框中传递一大块行（通过使用 by 语句进行子集），然后调用函数 myfun 对其进行操作吗？

根据要求添加示例

# generate data
N = 10000
default=NA
value = 1
df = data.table(id = sample(1:5000, N, replace=TRUE),
                trial = sample(c(0,1,2), N, replace=TRUE),
                ts = sample(1:200, N, replace=TRUE))

#set keys
setkeyv(df, c("id", "ts"))

df[["trial"]] = as.numeric(df[["trial"]]==value)

testfun <- function(x){
  L=length(x)
  x = x[L:1]
  x = fts(data=x)
  y = rep(default, L)
  if(L>=K){
    y1 = as.numeric(moving.sum(x,K))
    y = c(y1, rep(default,L-length(y1)))
  } 
  return(y[L:1]/K)
}

df[, prop:= lapply(.SD, testfun), by = id, .SDcols = "trial"]

仍然收到相同的警告消息：

Warning message:
In `[.data.table`(df, , `:=`(prop, lapply(.SD, testfun)), by = id,  :
  Invalid .internal.selfref detected and fixed by taking a copy of the whole table, so that := can add this new column by reference. At an earlier point, this data.table has been copied by R (or been created manually using structure() or similar). Avoid key<-, names<- and attr<- which in R currently (and oddly) may copy the whole data.table. Use set* syntax instead to avoid copying: setkey(), setnames() and setattr(). Also, list(DT1,DT2) will copy the entire DT1 and DT2 (R's list() copies named objects), use reflist() instead if needed (to be implemented). If this message doesn't help, please report to datatable-help so the root cause can be fixed.
4

1 回答 1

3

问题出现在

df[["trial"]] = as.numeric(df[["trial"]]==value)

这不是一种data.table方法

data.table 方法是使用:=

 df[, trial := as.numeric(trial == value)]

应该避免这个问题。

了解为什么要制作副本（因此内部自我引用可能无效）请参阅准确了解 data.table 何时是对另一个 data.table 的引用（相对于副本）

重要的是要意识到 s 没有[[<-方法data.table，因此[[<-.data.frame被调用，它将复制整个对象，而且不会做任何data.table方法（例如[<-.data.table）所做的小心的事情（返回一个有效的data.table.

于 2013-08-02T03:27:33.503 回答