在 R 中使用 NA 进行 Friedman 检验

Question

我需要对某些每月浓度进行弗里德曼测试。

不幸的是，一些月度记录丢失了，这意味着并非所有月份都有成对的措施。

structure(list(Name = c("a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m"), Month = c("March", "March", "March", "March", "March", "March", "March", "March", "March", "March", "March", "March", "March", "April", "April", "April", "April", "April", "April", "April", "April", "April", "April", "April", "April", "April", "May", "May", "May", "May", "May", "May", "May", "May", "May", "May", "May", "May", "May", "June", "June", "June", "June", "June", "June", "June", "June", "June", "June", "June", "June", "June", "July", "July", "July", "July", "July", "July", "July", "July", "July", "July", "July", "July", "July", "August", "August", "August", "August", "August", "August", "August", "August", "August", "August", "August", "August", "August", "September", "September", "September", "September", "September", "September", "September", "September", "September", "September", "September", "September", "September", "October", "October", "October", "October", "October", "October", "October", "October", "October", "October", "October", "October", "October", "November", "November", "November", "November", "November", "November", "November", "November", "November", "November", "November", "November", "November", "December", "December", "December", "December", "December", "December", "December", "December", "December", "December", "December", "December", "December", "January", "January", "January", "January", "January", "January", "January", "January", "January", "January", "January", "January", "January" ), Conc = c(0.01, 0.005, 0.005, 0.009, 0.01, NA, NA, NA, NA, NA, 0.005, 0.006, 0.007, 0.01, 0.005, 0.005, 0.005, 0.005, NA, NA, NA, NA, NA, 0.004, 0.006, 0.005, 0.004, 0.013, 0.005, 0.01, 0.012, 0.005, 0.008, 0.01, 0.009, 0.005, 0.01, 0.005, 0.018, 0.005, 0.012, 0.019, 0.014, 0.014, 0.015, 0.02, 0.013, 0.016, 0.005, 0.005, 0.005, 0.005, 0.012, 0.01, 0.01, 0.01, 0.025, NA, NA, 0.011, 0.012, 0.009, 0.01, 0.008, 0.012, 0.01, 0.009, 0.009, 0.012, 0.013, 0.01, 0.015, 0.01, 0.011, 0.009, 0.01, 0.013, 0.013, 0.012, NA, NA, 0.011, 0.015, 0.014, 0.012, 0.012, 0.012, 0.013, 0.013, 0.011, 0.01, 0.01, 0.01, 0.012, 0.017, 0.011, 0.013, 0.015, 0.014, 0.012, 0.013, 0.014, 0.013, 0.012, 0.01, 0.012, 0.01, 0.014, 0.013, 0.013, 0.013, 0.012, 0.01, 0.007, 0.008, 0.01, 0.01, 0.014, 0.014, 0.012, 0.028, 0.016, 0.016, 0.015, 0.069, 0.018, 0.007, 0.008, 0.012, 0.011, 0.008, 0.009, 0.007, 0.011, 0.011, 0.011, 0.011, 0.01, 0.006, 0.01, 0.01, 0.007, 0.008)), class = "data.frame", row.names = c(NA, -143L))

我怎样才能确保 R 只跳转这些不成对的组合，但仍然在所有月份进行测试，以便代码可以工作？我想保留措施而不必完全删除块。

抱歉，这一定是一项简单的任务，但我是 R 的新手，我很容易迷失在语法中。

任何人都可以指出事后分析的功能吗？

谢谢您的帮助

Answer 1

假设您的数据（以上）存储在变量df 中，以下 sn-p 允许您仅选择指定了 Conc 值的行...

idx <- which(!is.na(df$Conc))
df_no_missing <- df[ idx, ]

您也可以使用complete.cases()...

idx <- complete.cases(df) |> which()
df_no_missing <- df[ idx, ]