替换迭代函数的嵌套循环 - R

Question

我有一个迭代函数的嵌套循环，但我想避免运行嵌套循环。但是，我坚持执行一个将运行并保存输出的替代脚本。

下面是循环，以及运行完整脚本的其他有用信息。

# data
df = data.frame(y1 = sample(x=0L:1L, size=120, replace=TRUE),
                y2 = sample(x=0L:1L, size=120, replace=TRUE),
                x1 = sample(x=1L:3L, size=120, replace=TRUE),
                x2 = sample(x=1L:3L, size=120, replace=TRUE),
                x3 = sample(x=1L:3L, size=120, replace=TRUE),
                wgt = sample(runif(120, min=.25, max=.70), replace=TRUE))

#the function
chi2_test <- function(...){
  termlabels = sapply(substitute(...()), deparse)
  form <- reformulate(termlabels=termlabels, response="wgt")
  
  cross_table = xtabs(form, data=df)
  chi2_test = chisq.test(cross_table, correct=TRUE)

  return(list(crosstabs = cross_table,
              chi2_test = chi2_test)
  )
}

# dvs and ivs
dvars = names(df)[1:2]
ivars = names(df)[3:5]
combos = tidyr::crossing(dvars, ivars)

# the loop: where I need some help with
for (i in combos$dvars){
  for (j in combos$ivars){
    cat(i, "by", j)
    print(chi2_test(df[, i], df[, j]))
  }
}

Answer 1

不需要嵌套循环，因为完整的组合已经用

crossing

创建了。因此，只需遍历

combos

的行序列，提取与每一行对应的名称，基于此对

df

列进行子集化，并将输出分配给初始化的

list

，其长度与行数相同连击

out <- vector('list', nrow(combos))
names(out) <- do.call(paste, c(combos, sep = " by "))
for(i in seq_along(combos$dvars)) 
  out[[i]] <- chi2_test(df[, combos$dvars[i]], df[, combos$ivars[i]])

-输出

> out[1]
$`y1 by x1`
$`y1 by x1`$crosstabs
                     df[, combos$ivars[i]]
df[, combos$dvars[i]]         1         2         3
                    0 10.926195  8.558175  6.373337
                    1 13.388605  8.971302  8.481009

$`y1 by x1`$chi2_test

    Pearson's Chi-squared test

data:  cross_table
X-squared = 0.12113, df = 2, p-value = 0.9412