Question

你有一个数据框。如何最好地检查特定列的值的所有组合是否经常发生？

（当处理来自具有因子设计的实验的数据文件时，有时需要这样做。每列是一个独立变量，我们想要检查所有独立变量的组合是否经常出现。）

Answer 1

replications()怎么样？

tmp <- transform(ToothGrowth, dose = factor(dose))

replications( ~ supp + dose, data = tmp)
replications( ~ supp * dose, data = tmp)

> replications( ~ supp + dose, data = tmp)
supp dose 
  30   20 
> replications( ~ supp * dose, data = tmp)
     supp      dose supp:dose 
       30        20        10

从?replications我们有一个平衡测试：

!is.list(replications(~ supp + dose, data = tmp))

> !is.list(replications(~ supp + dose, data = tmp))
[1] TRUE

来自replications()的输出并不是您所期望的，但使用它的测试会给出您想要的答案。

Answer 2

checkAllCombosOccurEquallyOften<- function(df,colNames,dropZeros=FALSE) {
    #in data.frame df, check whether the factors in the list colNames reflect full factorial design (all combinations of levels occur equally often)
    #
    #dropZeros is useful if one of the factors nested in the others. E.g. testing different speeds for each level of    
    # something else, then a lot of the combos will occur 0 times because that speed not exist for that level.
    #but it's dangerous to dropZeros because it won't pick up on 0's that occur for the wrong reason- not fully crossed
    #
    #Returns:
    # true/false, and prints informational message
    #
    listOfCols <- as.list( df[colNames] )
    t<- table(listOfCols)

    if (dropZeros) {  
        t<- t[t!=0]   
    }           
    colNamesStr <- paste(colNames,collapse=",")
    if ( length(unique(t)) == 1 ) { #if fully crossed, all entries in table should be identical (all combinations occur equally often)
          print(paste(colNamesStr,"fully crossed- each combination occurred",unique(t)[1],'times'))
          ans <- TRUE
      } else {
          print(paste(colNamesStr,"NOT fully crossed,",length(unique(t)),'distinct repetition numbers.'  ))
          ans <- FALSE
      } 
    return(ans)
}

加载数据集并调用上述函数

library(datasets)
checkAllCombosOccurEquallyOften(ToothGrowth,c("supp","dose")) #specify dataframe and columns

输出提供了答案，它完全交叉：

[1] "supp,dose fully crossed- each combination occurred 10 times"
[1] TRUE

Answer 3

使用相同的ToothGrowth数据：

library(datasets)
library(data.table)

dt = data.table(ToothGrowth)

setkey(dt, supp, dose)
dt[CJ(unique(supp), unique(dose)), .N] # note: using hidden by-without-by
#   supp dose  N
#1:   OJ  0.5 10
#2:   OJ  1.0 10
#3:   OJ  2.0 10
#4:   VC  0.5 10
#5:   VC  1.0 10
#6:   VC  2.0 10

然后，您可以检查所有N是否相等或者您喜欢的其他内容。

检查所有组合是否经常出现在数据帧的指定列中

3 个答案: