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从汽车包装中提取Anova或Manova函数输出的多变量测试

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我想知道如何从以下MWE中的fm1输出中提取 Multivariate Tests: Site 部分 . 任何帮助将受到高度赞赏 . 谢谢

library(car)
fm1 <- summary(Anova(lm(cbind(Al, Fe, Mg, Ca, Na) ~ Site, data=Pottery)))
fm1

Type II MANOVA Tests:

Sum of squares and products for error:
           Al          Fe          Mg          Ca         Na
Al 48.2881429  7.08007143  0.60801429  0.10647143 0.58895714
Fe  7.0800714 10.95084571  0.52705714 -0.15519429 0.06675857
Mg  0.6080143  0.52705714 15.42961143  0.43537714 0.02761571
Ca  0.1064714 -0.15519429  0.43537714  0.05148571 0.01007857
Na  0.5889571  0.06675857  0.02761571  0.01007857 0.19929286

------------------------------------------

Term: Site 

Sum of squares and products for the hypothesis:
            Al          Fe          Mg         Ca         Na
Al  175.610319 -149.295533 -130.809707 -5.8891637 -5.3722648
Fe -149.295533  134.221616  117.745035  4.8217866  5.3259491
Mg -130.809707  117.745035  103.350527  4.2091613  4.7105458
Ca   -5.889164    4.821787    4.209161  0.2047027  0.1547830
Na   -5.372265    5.325949    4.710546  0.1547830  0.2582456

Multivariate Tests: Site
                 Df test stat  approx F num Df   den Df     Pr(>F)    
Pillai            3   1.55394   4.29839     15 60.00000 2.4129e-05 ***
Wilks             3   0.01230  13.08854     15 50.09147 1.8404e-12 ***
Hotelling-Lawley  3  35.43875  39.37639     15 50.00000 < 2.22e-16 ***
Roy               3  34.16111 136.64446      5 20.00000 9.4435e-15 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

2 回答

  • 2

    我也找不到如何提取测试表但作为一种解决方法,您可以通过在所有测试类型上运行 Anova 命令来计算结果 .

    但是打印方法 print.Anova.mlm 没有返回结果,所以这需要稍微调整一下 .

    library(car)
    
    # create new print function
    outtests <- car:::print.Anova.mlm
    
    # allow the function to return the results and disable print
    body(outtests)[[16]] <- quote(invisible(tests))
    body(outtests)[[15]] <- NULL
    
    # Now run the regression
    mod <- lm(cbind(Al, Fe, Mg, Ca, Na) ~ Site, data=Pottery)
    
    # Run the Anova over all tests  
    tab <- lapply(c("Pillai", "Wilks", "Hotelling-Lawley", "Roy"), 
                      function(i)  outtests(Anova(mod, test.statistic=i)))
    
    tab <- do.call(rbind, tab)
    row.names(tab) <- c("Pillai", "Wilks", "Hotelling-Lawley", "Roy")
    tab  
    
       # Type II MANOVA Tests: Pillai test statistic
       #               Df test stat approx F num Df den Df    Pr(>F)    
     #Pillai            3     1.554    4.298     15 60.000 2.413e-05 ***
     #Wilks             3     0.012   13.089     15 50.091 1.840e-12 ***
     #Hotelling-Lawley  3    35.439   39.376     15 50.000 < 2.2e-16 ***
     #Roy               3    34.161  136.644      5 20.000 9.444e-15 ***
     #---
     #Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
    

    由于输出表是 anovadata.frame 类,因此可以使用 xtable .

    xtable:::xtable(tab)
    
  • 3

    fm1$multivariate.tests 将您带到 fm1 输出的 Site 部分 .

    然后你可以使用 catcapture.output 的组合进行漂亮的打印,或者只使用 capture.output 作为字符向量 .

    > cat(capture.output(fm1$multivariate.tests)[18:26], sep = "\n")
    #
    # Multivariate Tests: Site
    #                  Df test stat  approx F num Df   den Df     Pr(>F)    
    # Pillai            3   1.55394   4.29839     15 60.00000 2.4129e-05 ***
    # Wilks             3   0.01230  13.08854     15 50.09147 1.8404e-12 ***
    # Hotelling-Lawley  3  35.43875  39.37639     15 50.00000 < 2.22e-16 ***
    # Roy               3  34.16111 136.64446      5 20.00000 9.4435e-15 ***
    # ---
    # Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
    

    Update: 从结果来看

    unlist(fm1$multivariate.tests, recursive = FALSE)
    

    它看起来不像数字值那样容易访问结果 . 因此,正如您所要求的,这是将结果操作到矩阵中所需的内容 . 完成此操作然后看到user20650的答案后,我建议您遵循该建议并通过ANOVA表获取值 .

    co <- capture.output(fm1$multivariate.tests)[20:24]
    s <- strsplit(gsub("([*]+$)|[<]", "", co[-1]), "\\s+")
    dc <- do.call(rbind, lapply(s, function(x) as.numeric(x[-1])))
    row.names(dc) <- sapply(s, "[", 1)
    s2 <- strsplit(co[1], " ")[[1]]
    s2 <- s2[nzchar(s2)]
    s3 <- s2[-c(1, length(s2))]
    colnames(dc) <- c(s2[1], paste(s3[c(TRUE, FALSE)], s3[c(FALSE, TRUE)]), s2[10])
    dc
    #                  Df test stat  approx F num Df   den Df     Pr(>F)
    # Pillai            3   1.55394   4.29839     15 60.00000 2.4129e-05
    # Wilks             3   0.01230  13.08854     15 50.09147 1.8404e-12
    # Hotelling-Lawley  3  35.43875  39.37639     15 50.00000 2.2200e-16
    # Roy               3  34.16111 136.64446      5 20.00000 9.4435e-15
    

    如果有人想改善我的第二个代码块,请随意 .

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