忽略卡方检验的函数变量

Question

我正在r中构建一个用户定义的函数。它看起来像这样（数据框只是一个例子）：

library(dplyr)
library(lazyeval)

df<-data.frame(type1=c("a","b","c","a","b","c",NA),
               type2=c("d","e","f","d","e","f","f"),
               type3=c("d","f","g","e","f","g","e"),
               type4=c("e","f","g","e","f","g","e"))

f<-function(x){
  y<-df%>%
    dplyr::filter_(lazyeval::interp(~!is.na(x),x=as.name(x)))%>%
    dplyr::group_by_(x)%>%
    dplyr::summarize("Sum2"=sum(type2=="d"),
                     "Sum3"=sum(type3=="d"),
                     "Sum4"=sum(type4=="d"))
  y<-dplyr::bind_rows(y,setNames(data.frame("Total",
                                            sum(y$Sum2),
                                            sum(y$Sum3),
                                            sum(y$Sum4),
                                            paste(
                                              round(chisq.test(y$Sum2,y$Sum3)$p.value,3),
                                              "(Chi-sq.)")), 
                                 c(x, "Sum2","Sum3","Sum4","pvalue")))
  return(y)
}

result_f<-f("type1")

问题在于我想在卡方中不仅包括两个变量而是三个变量（Sum2，Sum3和Sum4），因为在这种情况下只有{{ 1}}和Sum2是相关的，但在其他情况下，函数可能需要分析三个变量。

是否有任何方法可以包含Sum3，即Sum 4而不是chisq.test(y$Sum2,y$Sum3,y$Sum4)，但是使用某种条件选择排除那些归零的变量（例如）？