用心理计算Cohen's kappa时的警告

Question

在我的db数据帧中，我通过scoreA和scoreB两种方法计算分数。然后我得到它们的五分位数，我想得到它们之间的Cohen's kappa系数。

我在CRAN中发现了几个库，psych似乎是最常用的一个，但我对其他可能性持开放态度。

下面是一些要测试的代码：

set.seed(1)
df=data.frame(
  scoreA = rnorm(n = 200, mean = 1, sd = .75)*10,
  scoreB = rnorm(n = 200, mean = 5, sd = 3)*10
)
df$quintA = cut(df$scoreA, breaks=quantile(df$scoreA, probs=seq(0,1, by=0.2), na.rm = TRUE), labels=c("Q1", "Q2", "Q3", "Q4", "Q5"), include.lowest = T)
df$quintB = cut(df$scoreB, breaks=quantile(df$scoreB, probs=seq(0,1, by=0.2), na.rm = TRUE), labels=c("Q1", "Q2", "Q3", "Q4", "Q5"), include.lowest = T)
plot(df$quintA, df$quintB, xlab="A", ylab="B")
psych::cohen.kappa(table(df$quintA, df$quintB))

Call: cohen.kappa1(x = x, w = w, n.obs = n.obs, alpha = alpha, levels = levels)

Cohen Kappa and Weighted Kappa correlation coefficients and confidence boundaries 
                  lower estimate upper
unweighted kappa -0.058    0.012 0.083 #My real world estimates are about 0.5
weighted kappa   -0.215   -0.070 0.075

 Number of subjects = 200 
Warning message:
In any(abs(bounds)) : coercing argument of type 'double' to logical

我应该担心这个警告吗？这是什么意思？

编辑:我刚刚尝试了fmsb，得到了与未加权的kappa完全相同的值(即使是小数)。这个问题仍然在继续，以防带轮子的kappa出现问题。

library(fmsb)
Kappa.test(db$scoreA, db$scoreB)
$Result

    Estimate Cohen kappa statistics and test the null hypothesis that the
    extent of agreement is same as random (kappa=0)

data:  db$scoreA and db$scoreB
Z = 383.67, p-value < 2.2e-16
95 percent confidence interval:
 0.9077865 0.9131626
sample estimates:
[1] 0.9104745

$Judgement
[1] "Almost perfect agreement"