3

具有分类变量 catA、catB 和 catC 的示例数据框。Obs 是一些观察值。

catA <- rep(factor(c("a","b","c")), length.out=100)
catB <- rep(factor(1:4), length.out=100)
catC <- rep(factor(c("d","e","f")), length.out=100)
obs <- runif(100,0,100)
dat <- data.frame(catA, catB, catC, obs)

分类变量的所有可能的数据子集。

allsubs <- expand.grid(catA = c(NA,levels(catA)), catB = c(NA,levels(catB)),
    catC = c(NA,levels(catC)))
> head(allsubs, n=10)
   catA catB catC
 1  <NA> <NA> <NA>
 2     a <NA> <NA>
 3     b <NA> <NA>
 4     c <NA> <NA>
 5  <NA>    1 <NA>
 6     a    1 <NA>
 7     b    1 <NA>
 8     c    1 <NA>
 9  <NA>    2 <NA>
 10    a    2 <NA>

现在，创建一个输出数据框的最简单方法是什么，其结果列包含一个函数的结果，该函数应用于 dat 的相应子集（在每一行中由 cat 变量的组合定义）。因此，输出应类似于以下数据框“whatiwant”，其中结果列将包含应用于每个子集的函数的结果。

> whatiwant
    catA catB catC results
 1  <NA> <NA> <NA>       *
 2     a <NA> <NA>       *
 3     b <NA> <NA>       *
 4     c <NA> <NA>       *
 5  <NA>    1 <NA>       *
 6     a    1 <NA>       *
 7     b    1 <NA>       *
 8     c    1 <NA>       *
 9  <NA>    2 <NA>       *
 10    a    2 <NA>       *

因此，如果应用的函数是“平均”，则结果应该是：

dat$results[1] = mean(subset(dat,)$obs)
dat$results[2] = mean(subset(dat, catA=="a")$obs)

等等等等。

4

4 回答 4

4 
ans <- with(dat, tapply(obs, list(catA, catB, catC), mean))
ans <- data.frame(expand.grid(dimnames(ans)), results=c(ans))
names(ans)[1:3] <- names(dat)[1:3]

str(ans)
# 'data.frame':  36 obs. of  4 variables:
#  $ catA   : Factor w/ 3 levels "a","b","c": 1 2 3 1 2 3 1 2 3 1 ...
#  $ catB   : Factor w/ 4 levels "1","2","3","4": 1 1 1 2 2 2 3 3 3 4 ...
#  $ catC   : Factor w/ 3 levels "d","e","f": 1 1 1 1 1 1 1 1 1 1 ...
#  $ results: num  69.7 NA NA 55.3 NA ...
于 2013-05-29T22:11:38.920 回答
2

另一种方法是，一个函数获取变量的所有组合，另一个函数将函数应用于所有子集。组合功能是从另一个帖子中窃取的...

## return all combinations of vector up to maximum length n
multicombn <- function(dat, n) {
    unlist(lapply(1:n, function(x) combn(dat, x, simplify=F)), recursive=F)
}

对于 allsubs，vars 的形式是c("catA","catB","catC"), out.name = "mean". func 需要以 ddply 的形式编写，

func=function(x) mean(x$obs, na.rm=TRUE)

library(plyr)
allsubs <- function(indat, vars, func=NULL, out.name=NULL) {
    results <- data.frame()
    nvars <- rev(multicombn(vars,length(vars)))
    for(i in 1:length(nvars)) {
        results <-
            rbind.fill(results, ddply(indat, unlist(nvars[i]), func))
    }
    if(!missing(out.name)) names(results)[length(vars)+1] <- out.name
    results
}

这个答案和 shwaund 的一个区别是，这不会返回空子集的行，因此结果列中没有 NA。

allsubs(dat, c("catA","catB","catc"), func, out.name="mean")
> head(allsubs(dat, vars, func, out.name = "mean"),20)
   catA catB catC     mean
1     a    1    d 56.65909
2     a    2    d 54.98116
3     a    3    d 37.52655
4     a    4    d 58.29034
5     b    1    e 52.88945
6     b    2    e 50.43122
7     b    3    e 52.57115
8     b    4    e 59.45348
9     c    1    f 52.41637
10    c    2    f 34.58122
11    c    3    f 46.80256
12    c    4    f 51.58668
13 <NA>    1    d 56.65909
14 <NA>    1    e 52.88945
15 <NA>    1    f 52.41637
16 <NA>    2    d 54.98116
17 <NA>    2    e 50.43122
18 <NA>    2    f 34.58122
19 <NA>    3    d 37.52655
20 <NA>    3    e 52.57115
于 2013-05-31T18:50:51.797 回答
1

这不是最干净的解决方案，但我认为它接近你想要的。

getAllSubs <- function(df, lookup, fun) {

  out <- lapply(1:nrow(lookup), function(i) {

    df_new <- df

    if(length(na.omit(unlist(lookup[i,]))) > 0) {

      for(j in colnames(lookup)[which(!is.na(unlist(lookup[i,])))]) {
        df_new <- df_new[df_new[,j] == lookup[i,j],]
      }  
    } 
    fun(df_new)  
  })

  if(mean(sapply(out, length) ==1) == 1) {
    out <- unlist(out)
  } else {
    out <- do.call("rbind", out)
  }

  final <- cbind(lookup, out)
  final[is.na(final)] <- NA
  final
}

正如当前编写的那样，您必须事先构造查找表，但您可以轻松地将构造移动到函数本身。我在最后添加了几行以确保它可以容纳不同长度的输出，因此将 NaN 转换为 NA，只是因为这似乎可以创建更清晰的输出。正如当前编写的那样，在所有列都是 NA 的情况下，它将函数应用于整个原始数据帧。

dat_out <- getAllSubs(dat, allsubs, function(x) mean(x$obs, na.rm = TRUE))

head(dat_out,20)

   catA catB catC      out
1  <NA> <NA> <NA> 47.25446
2     a <NA> <NA> 51.54226
3     b <NA> <NA> 46.45352
4     c <NA> <NA> 43.63767
5  <NA>    1 <NA> 47.23872
6     a    1 <NA> 66.59281
7     b    1 <NA> 32.03513
8     c    1 <NA> 40.66896
9  <NA>    2 <NA> 45.16588
10    a    2 <NA> 50.59323
11    b    2 <NA> 51.02013
12    c    2 <NA> 33.15251
13 <NA>    3 <NA> 51.67809
14    a    3 <NA> 48.13645
15    b    3 <NA> 57.92084
16    c    3 <NA> 49.27710
17 <NA>    4 <NA> 44.93515
18    a    4 <NA> 40.36266
19    b    4 <NA> 44.26717
20    c    4 <NA> 50.74718
于 2013-05-31T00:42:34.153 回答
1

仅使用向量化函数和基础 R

# Find all possible subsets of your data
combVars <- c("catA", "catB", "catC")
subsets <- lapply(0:length(combVars), combn, x = combVars, simplify = FALSE)
subsets <- do.call(c, subsets)
# Calculate means by each subset
meanValues <- lapply(subsets, function(x) aggregate(dat[["obs"]], by = dat[x], FUN = mean))
# Pull them all into one dataframe
Reduce(function(x,y) merge(x,y,all=TRUE), meanValues)
于 2016-09-16T17:10:29.843 回答