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Dear R user community,

I have many data.frames in a list, as follows (only one data.frame in the list of 21 shown for convenience):

> str(datal)
List of 21
 $ BallitoRaw.DAT                :'data.frame': 1083 obs. of  3 variables:
  ..$ Filename: Factor w/ 21 levels "BallitoRaw.DAT",..: 1 1 1 1 1 1 1 1 1 1 ...
  ..$ date    :Class 'Date'  num [1:1083] 7318 7319 7320 7321 7322 ...
  ..$ temp    : num [1:1083] NA 25.8 NA NA NA NA NA NA NA 24.4 ...

If I work on each data.frame in the list individually I can create a zoo object from temp and date, as such:

> BallitoRaw.zoo <- zoo(datal$BallitoRaw.DAT$temp, datal$BallitoRaw.DAT$date)

The zoo object looks like this:

> head(BallitoRaw.zoo)
1990-01-14 1990-01-15 1990-01-16 1990-01-17 1990-01-18 1990-01-19 
        NA       25.8         NA         NA         NA         NA

How do I use llply or apply (or similar) to work on the whole list at once?

The output needs to go into a new list of data.frames, or a series of independent data.frames (each one named as in the zoo example above). Note that the date column, although a regular time series (days), contains missing dates (in addition to NAs for temps of existing dates); the missing dates will be filled by the zoo function. The output data.frame with the zoo object will thus be longer than the original one.

Help kindly appreciated.

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2 回答 2

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makeNamedZoo <- function(dfrm){ dfrmname <- deparse(substitute(dfrm))
  zooname <-dfrmname
   assign(zooname,   zoo(dfrm$temp, dfrm$date))
   return(get(zooname)) }
ListOfZoos <- lapply(dflist, makeNamedZoo)
names(ListOfZoos) <- paste( sub("DAT$", "", names(dflist) ), "zoo", sep="")

这是一个简单的测试用例：

df1 <- data.frame(a= letters[1:10], date=as.Date("2011-01-01")+0:9, temp=rnorm(10) )
df2 <- data.frame(a= letters[1:10], date=as.Date("2011-01-01")+0:9, temp=rnorm(10) )
dflist <- list(dfone.DAT=df1,dftwo.DAT=df2)
ListOfZoos <- lapply(dflist, makeNamedZoo) 
names(ListOfZoos) <- paste( sub("DAT$", "", names(dflist) ), "zoo", sep="")

$dfone.zoo
2011-01-01 2011-01-02 2011-01-03 2011-01-04 2011-01-05 2011-01-06 2011-01-07 
 0.7869056  1.6523928 -1.1131432  1.2261783  1.1843587  0.2673762 -0.4159968 
2011-01-08 2011-01-09 2011-01-10 
-1.2686391 -0.4135859 -1.4916291 

$dftwo.zoo
2011-01-01 2011-01-02 2011-01-03 2011-01-04 2011-01-05 2011-01-06 2011-01-07 
 0.7356612 -0.1263861 -1.6901240 -0.6441732 -1.4675871  2.3006544  1.0263354 
2011-01-08 2011-01-09 2011-01-10 
-0.8577544  0.6079986  0.6625564
于 2011-09-25T14:50:26.063 回答
1

这是实现我需要的更简单的方法：

tozoo <- function(x) zoo(x$temp, x$date) 
data1.zoo <- do.call(merge, lapply(split(data1, data1$Filename), tozoo))

结果是一个不错的动物园对象。

于 2012-02-27T13:45:11.733 回答