在 R 中对序数数据运行 Friedman 测试时遇到问题

Question

我正在尝试对 R 中的序数数据运行 Friedman 测试，但遇到了错误。数据可以在 Dropbox https://www.dropbox.com/s/gh8crh18y1ueriy/seltoutput.xlsx?dl=0 上找到。

作为数据的描述：

group1：小组作业，2 个级别
time1：时间点，2级
loameasure：序数数据，5 个级别
distmeasure：连续数据
vectemp：参与者 ID

导入数据后，我运行以下命令以正确格式化：

  selt$loameasure<-factor(selt$loameasure)  
  selt$distmeasure<-as.numeric(selt$distmeasure)  
  selt$time1<-factor(selt$time1)  

然后我运行：

 friedman_test(formula = loameasure ~ time1 | vectemp, data = selt)

然后我得到错误：
Friedman.test.default(c(3L, 2L, 3L, 2L, 2L, 5L, 2L, 1L, 3L, 4L, : 不是一个不可复制的完整块设计

我认为 loameasure 和 time1 必须是因素，但我确实尝试将它们作为数字，但我得到了类似的错误：
Friedman.test.default(c(3, 2, 3, 2, 2, 5, 2, 1, 3, 4, 2, 2, 4, : 不是一个不可复制的完整块设计

我已经玩了好几天了，但无法弄清楚我的问题是什么。我很想得到一些帮助！提前谢谢！

Answer 1

据我所知，弗里德曼测试不适合您的情况。我建议使用 Type-III 平方和法对不平衡设计执行 双向 ANOVA 检验。 给出了残差正态性和同质性的假设。 我试图指导您如何执行测试以及一些步骤的含义。它不完整（缺乏解释等。）但这应该是你的开始和方向。

1. 我们想知道 loameasure 是否依赖于 group1 和 time1
2. 我们将执行具有两个因素的双向方差分析
3. 因变量：loameasure
4. 自变量：group1 和 time1

library(readxl)
# load your data
df <- read_excel("C:/Users/coding/Downloads/seltoutput.xlsx", 
                 col_types = c("numeric", "numeric", "numeric")) 

# Prepare data
# group1 to factor 
df$group1 <- factor(df$group1,
                    levels = c(0, 1),
                    labels = c("Group_0", "Group_1"))
# time1 to factor
df$time1 <- factor(df$time1,
                      levels = c(1, 2),
                      labels = c("Time_1", "Time_2"))
----------------------------------------------------------------------------
# Visualize
library("ggpubr")
ggboxplot(df, x = "time1", y = "loameasure", color = "group1",
          palette = c("#00AFBB", "#E7B800"))

ggline(df, x = "time1", y = "loameasure", color = "group1",
       add = c("mean_se", "dotplot"),
       palette = c("#00AFBB", "#E7B800"))
-----------------------------------------------------------------------------
# first decide if balanced or unbalnced design
table(df$group1, df$time1)
# Output

# Time_1 Time_2
# Group_0     20     20
# Group_1     29     29

# Here it is a unbalance design 
# An unbalanced design has unequal numbers of subjects in each group!

## We will perform two-way ANOVA test in R for unbalanced designs !!!!!!!!!!!
# The recommended method are the Type-III sums of squares.
# you need `car` package
library(car)
# Our 2 way anova of unbalanced design (SS Type-III)

df_anova <- aov(loameasure ~ group1 * time1, data = df)
Anova(df_anova, type = "III")

## Output
# Anova Table (Type III tests)
# Response: loameasure
# Sum Sq Df F value    Pr(>F)    
# (Intercept)  120.050  1 83.9312 1.116e-14 ***
#   group1         0.700  1  0.4891   0.48607    
# time1         62.500  1 43.6960 2.301e-09 ***
#   group1:time1   5.716  1  3.9963   0.04849 *  
#   Residuals    134.452 94                      
# ---
#   Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


# Nomrality check ---------------------------------------------------------


# call residuals (difference between each indivdual and their group1/time1 combination mean)
res <- df_anova$residuals

# Histogram of residuals: Residuals should be normally distributed
hist(res,main="Histogram of residuals", xlab = "Residuals")
# # Extract the residuals

# Run Shapiro-Wilk test
shapiro.test(x = res )
# Output
# data:  res
# W = 0.97708, p-value = 0.08434
# P is > 0.05 therefore normality can be assumed.


# Homogenity test ---------------------------------------------------------

# Levene's test for equality of variances (in `car` package)
library(car)
leveneTest(loameasure~ time1 * group1,data=df)

# Output:
# Levene's Test for Homogeneity of Variance (center = median)
#       Df F value Pr(>F)
# group  3  0.3196 0.8112
#       94  
# P is > 0.05 therefore equal variances can be assumed.