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我将不胜感激帮助解决以下问题。我正在使用组内 (4) 数据进行 Kruskal-Wallis 测试，并按照此处的说明进行操作：http ://rcompanion.org/handbook/F_08.html 。第一步是按中位数对组进行排序：

$senseofpresence$Condition = factor(senseofpresence$condition, levels=c("A", "B", $"C", "D", "E"))

但是，当按中位数对组进行排序时，我收到以下错误：

$<-.data.frame( *tmp*, "Condition", value = integer(0))中的错误：替换有 0 行，数据有 128

我不确定导致错误的原因以及 KW 测试和 Dunn 事后测试是否有效，无论错误消息如何。代码和结果如下。感谢您帮助解决问题。这是我在 StackOverflow 上的第一篇文章，我对 R 也很陌生。

> senseofpresence$Condition = factor(senseofpresence$Condition,
>+                       levels=unique(senseofpresence$Condition))
> senseofpresence$PresentPlace1.f = factor(senseofpresence$PresentPlace1,
>+                        ordered = TRUE)
> xtabs( ~ Condition + PresentPlace1.f,
>+        data = senseofpresence)
         PresentPlace1.f
>Condition  1  2  3  4  5  6  7
        A  0  0  0  2  5 11 10
        B  1  1  1  1 11  9  1
        C  2  3  5  3  5  8  1
        D  0  2  0  1 10  7  6
        E  1  4  1  1  7  4  4
> 
> XT = xtabs( ~ Condition + PresentPlace1.f,
>+             data = senseofpresence)
> 
> prop.table(XT,
>+            margin = 1)
         PresentPlace1.f
>Condition          1          2          3          4          5          6          >7
>        A 0.00000000 0.00000000 0.00000000 0.07142857 0.17857143 0.39285714 >0.35714286
>        B 0.04000000 0.04000000 0.04000000 0.04000000 0.44000000 0.36000000 >0.04000000
        C 0.07407407 0.11111111 0.18518519 0.11111111 0.18518519 0.29629630 >0.03703704
>        D 0.00000000 0.07692308 0.00000000 0.03846154 0.38461538 0.26923077 >0.23076923
        E 0.04545455 0.18181818 0.04545455 0.04545455 0.31818182 0.18181818 >0.18181818
> library (lattice)
> 
> histogram( ~ PresentPlace1.f | Condition,
>+           data=senseofpresence,
>+           layout=c(1,3)  
>+           )
> 
> library(FSA)
> Summarize(PresentPlace1 ~ Condition,
>+           data=senseofpresence, 
>+           digits=3)
>  Condition  n  mean    sd min   Q1 median Q3 max
1         A 28 6.036 0.922   4 5.75    6.0  7   7
2         B 25 5.040 1.338   1 5.00    5.0  6   7
3         C 27 4.259 1.745   1 3.00    5.0  6   7
4         D 26 5.462 1.334   2 5.00    5.5  6   7
5         E 22 4.682 1.887   1 3.25    5.0  6   7
> kruskal.test(PresentPlace1 ~ Condition,
>+              data = senseofpresence)

>   Kruskal-Wallis rank sum test

>data:  PresentPlace1 by Condition
>Kruskal-Wallis chi-squared = 19.754, df = 4, p-value = 0.0005585

> 
> senseofpresence$Condition = factor(senseofpresence$condition, 
>+                       levels=c("A", "B", "C", "D", "E"))
>Error in `$<-.data.frame`(`*tmp*`, "Condition", value = integer(0)) : 
>  replacement has 0 rows, data has 128
> 
> levels(senseofpresence$Condition)
>[1] "A" "B" "C" "D" "E"
> 
> library(FSA)
> 
> DT = dunnTest(PresentPlace1 ~ Condition,
>+               data=senseofpresence,
>+               method="bh")      # Adjusts p-values for multiple comparisons;
> DT
>Dunn (1964) Kruskal-Wallis multiple comparison
>  p-values adjusted with the Benjamini-Hochberg method.

>   Comparison          Z      P.unadj        P.adj
>1       A - B  2.7334681 6.267119e-03 0.0208903954
>2       A - C  4.1353289 3.544465e-05 0.0003544465
>3       B - C  1.3087557 1.906171e-01 0.2723101491
>4       A - D  1.6053479 1.084173e-01 0.2168345127
>5       B - D -1.1242981 2.608866e-01 0.3261082038
>6       C - D -2.4681705 1.358056e-02 0.0339514018
>7       A - E  2.9473396 3.205210e-03 0.0160260486
>8       B - E  0.2995078 7.645526e-01 0.7645526055
>9       C - E -0.9599074 3.371018e-01 0.3745575517
>10      D - E  1.3893854 1.647156e-01 0.2745259737
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