在R中对序数数据运行Friedman测试时遇到问题

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问题描述

我正在尝试对R中的序数数据进行Friedman测试，但出现错误.可以在保管箱 https://www.dropbox.com/s/gh8crh18y1ueriy/seltoutput.xlsx?dl = 0 .

I am attempting to run a friedman test on ordinal data in R and am getting errors. The data can be found here on dropbox https://www.dropbox.com/s/gh8crh18y1ueriy/seltoutput.xlsx?dl=0.

作为数据描述:

group1:小组分配，2级
time1:时间点，2级
方法:顺序数据，共5个级别
distmeasure:连续数据
vectemp:参与者ID

group1: group assignments, 2 levels
time1: time points, 2 levels
loameasure: ordinal data, 5 levels
distmeasure: continuous data
vectemp: participant IDs

导入数据后，我运行以下命令正确格式化:

After importing the data I run the following to correctly format:

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

然后我跑:

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

然后我得到了错误:
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模块设计

Then I get the error:
Error in friedman.test.default(c(3L, 2L, 3L, 2L, 2L, 5L, 2L, 1L, 3L, 4L, :not an unreplicated complete block design

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

I thought that loameasure and time1 had to be factors but I did try them as numeric and I get a similar error:
Error in friedman.test.default(c(3, 2, 3, 2, 2, 5, 2, 1, 3, 4, 2, 2, 4, :not an unreplicated complete block design

我已经玩了好几天了，还没弄清我的问题是什么.我希望得到一些帮助！预先谢谢你！

I've been playing around with this for days and haven't been able to figure out what my problem is. I would love some assistance! Thank you in advance!

推荐答案

据我所知，弗里德曼测试不适用于您的情况.我建议使用III型平方和求和方法对不平衡设计进行双向ANOVA测试.给出了残差和均一性的正态性假设.我试图指导您如何执行测试以及某些步骤的含义.它不完整(缺乏解释等.)但这应该是您的起点和方向.

As far as I can anticipate a Friedman test is not appropriate in your situation. I would suggest to perform a two-way ANOVA test for unbalanced designs with Type-III sums of square method.The assumptions of Normality of residuals and homogenity are given.I have tried to guide you how to perform the test and the meaning of some steps. It is not complete (lacking of interpretation etc..) But this should be a begin and direction for you.

我们想知道Loameasure是否取决于group1和time1
我们将基于两个因素进行双向方差分析
因变量: loameasure
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.

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