R Anova_test()错误:变量只有一个级别

R Anova_test()错误:变量只有一个级别,r,anova,R,Anova,我正在尝试使用ANOVA_test()函数运行三向重复测量ANOVA。不幸的是,我提示内部因素只有一个级别(参见下面的错误代码),即使在评估数据帧的结构时,我发现它有多个级别。有人能帮忙吗 agn.rmanova <- anova_test( data = df_final, dv = rt, wid = subject, within = c(sleep, drug, target.type) ) 我甚至在第一次尝试进行故障排除时特别指定了级别(见下文),但都没有成功 df_f

我正在尝试使用ANOVA_test()函数运行三向重复测量ANOVA。不幸的是,我提示内部因素只有一个级别(参见下面的错误代码),即使在评估数据帧的结构时,我发现它有多个级别。有人能帮忙吗

agn.rmanova <- anova_test(
  data = df_final, dv = rt, wid = subject,
  within = c(sleep, drug, target.type)
)
我甚至在第一次尝试进行故障排除时特别指定了级别(见下文),但都没有成功

df_final <- summary %>%
  filter(id != 2)%>%
  filter(id != 6) %>%
  mutate(sleep = factor(condition, levels = c("50", "normal")),
         drug = factor(supplement, levels = c("A", "B", "C")),
         target.type = factor(target.type, levels = c("negative","neutral","positive")),
         subject = factor(subject, levels = c("1","4","7","8","14","16","19","20",
                                              "21","23","24","25","27","29","32",
                                              "33","34","35")))
df_最终百分比
过滤器(id!=2)%>%
过滤器(id!=6)%>%
突变(睡眠=因子(条件,水平=c(“50”,“正常”)),
药物=因子(补充,水平=c(“A”、“B”、“c”)),
target.type=因子(target.type,级别=c(“负”、“中性”、“正”)),
科目=系数(科目,级别=c(“1”、“4”、“7”、“8”、“14”、“16”、“19”、“20”),
"21","23","24","25","27","29","32",
"33","34","35")))
任何帮助都将不胜感激。谢谢


我也有这个问题。我的同事帮助解决了这个问题,将数据框子集为只包含相关数据,我还将非时间变量放在了错误的类别中(应该在内部,而不是之间)。不确定是什么解决了这个问题,但我想我会分享,因为这个问题还没有解决

这是我的原始代码:

    rowtotaltr.aov <- anova_test(
      data = weedcts_row_cm, dv = cm_RowTotalWeeds.tr, wid = id,
      within = c(RowMulch, Date)
      )

rowtotaltr.aov如果您有一个分组的df,请尝试df\u final%>%ungroup%>%anova\u测试(…)。我怀疑它仍然会抛出错误。如果数据仍然不起作用,你能以某种方式共享数据吗?请不要通过图像共享代码。
    rowtotaltr.aov <- anova_test(
      data = weedcts_row_cm, dv = cm_RowTotalWeeds.tr, wid = id,
      within = c(RowMulch, Date)
      )
    test<-as.data.frame(weedcts_row_cm[,c(1,3:5,12, 13)])
    test<-test[,c(1,3:5,12, 13)]
    test$Plot<-as.factor(test$Plot) 
    t<-anova_test(data = test, dv=cm_RowTotalWeeds.tr, wid=id, between=c(RowMulch), 
    within=c(Date))