如何使这个 Spearman-Brown 系数代码起作用？它给了我一个输出 NA
spearman_brown(df[,1], df[,2], short_icc, type = "ICC1", lmer = FALSE)
数据
structure(list(var1 = c("5", "2", "5", "2", "3", "1", "6", "1",
"4", "5", "5", "2", "2", "3", "1", "3", NA, "5", "7", "5", "2",
"2", "2", NA, "2", "3", "2", "2", "5", "2", "4", "2", "3", "5",
"5", "5", "5", "2", "4", NA, "6", "7", "7", "3", "2", "3", "3",
NA), var2 = c("2", "1", "2", "2", "2", "1", "1", "1", "2", "2",
"3", "1", "2", "2", "1", "2", NA, "3", "2", "1", "2", "2", "1",
NA, "1", "2", "1", "1", "2", "1", "1", "2", "2", "2", "4", "3",
"2", "2", "1", NA, "2", "2", "4", "1", "2", "3", "2", NA)), row.names = c(NA,
-48L), class = c("tbl_df", "tbl", "data.frame"))
我尝试按照本页 https://search.r-project.org/CRAN/refmans/splithalfr/html/spearman_brown.html 上的示例进行操作
最佳答案
这是一个tibble，因此我们需要使用[[(或$)而不是[来提取>，因为 [ 不会删除 tibble 或 data.table 的尺寸，即 df[,1]或 df[,2] 仍将是具有单列的 tibble，其中 spearman_brown x 和 y 应该是向量s
spearman_brown(df[[1]], df[[2]], short_icc, type = "ICC1", lmer = FALSE)
关于r - Spearman-Brown 系数代码生成 N/A，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/74481050/