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我manova()在R中运行。当我使用时，summary.aov()我收到以下错误：Error in 1L:object$rank

这是数据

Subj Buty    dN1 nN1        S1    dR1 nR1    dN2
1     1    0 215.45  22 109.11051  80.00   5 205.45
2     1    1 311.20  25  98.22383  82.22   9 245.33
3     2    0 146.11  18 111.93897  30.00   1 233.59
4     2    1 279.03  31 116.19284  93.75   8 245.14
5     3    0 230.00  25 100.73992  78.18  11 282.90
6     3    1 338.89  27  93.31388 112.00   5 247.44
7     4    0 181.71  35 107.41365  83.33   3 207.88
8     4    1 296.06  33  97.01108  93.33   9 290.34
9     5    0 178.93  28 112.66732  95.00   8 209.60
10    5    1 237.03  37  99.81596  91.58  19 212.19
11    6    0 259.57  23 117.59802 127.50  12 296.00
12    6    1 387.50  20 112.60321 140.91  11 214.19
13    7    0 149.12  34 110.28006  64.00   5 181.28
14    7    1 315.17  29 122.76490  72.73  11 168.28
15    8    0 127.50  40 108.69826  74.00   5 190.70
16    8    1 245.76  33 145.16311  73.00  10 169.74

（Subj并且Buty是因素）

manova(cbind(dN1, nN1, S1, dR1, nR1) ~ Buty + Error(Subj/Buty), data = d)

summary(), coefficients, 和residuals所有工作，但不是summary.aov()。这是一个问题，因为我需要单个数据列的读数（例如，系数的 p 值）。

知道如何解决吗？非常感激！

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1 回答 1

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根据 中的示例?summary.aov，您可以使用summary(aov())获得多层拟合。

summary(aov(cbind(dN1, nN1, S1, dR1, nR1) ~ Buty + Error(Subj/Buty), data = d))

#Error: Subj
# Response dN1 :
#          Df Sum Sq Mean Sq F value Pr(>F)
#Residuals  7  28155  4022.2               

# Response nN1 :
#          Df Sum Sq Mean Sq F value Pr(>F)
#Residuals  7    430  61.429               

# Response S1 :
#          Df Sum Sq Mean Sq F value Pr(>F)
#Residuals  7   1309  187.01               

# Response dR1 :
#          Df Sum Sq Mean Sq F value Pr(>F)
#Residuals  7 7061.1  1008.7               

# Response nR1 :
#          Df Sum Sq Mean Sq F value Pr(>F)
#Residuals  7    119      17               


#Error: Subj:Buty
# Response dN1 :
#          Df Sum Sq Mean Sq F value   Pr(>F)    
#Buty       1  53159   53159  110.25 1.55e-05 ***
#Residuals  7   3375     482                     
#---
#Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

# Response nN1 :
#          Df Sum Sq Mean Sq F value Pr(>F)
#Buty       1   6.25   6.250  0.2593 0.6263
#Residuals  7 168.75  24.107               

# Response S1 :
#          Df Sum Sq Mean Sq F value Pr(>F)
#Buty       1   2.76   2.757  0.0196 0.8926
#Residuals  7 985.06 140.723               

# Response dR1 :
#          Df Sum Sq Mean Sq F value  Pr(>F)  
#Buty       1 1016.2 1016.18  4.0084 0.08536 .
#Residuals  7 1774.6  253.51                  
#---
#Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

# Response nR1 :
#          Df Sum Sq Mean Sq F value  Pr(>F)  
#Buty       1     64  64.000  4.6667 0.06758 .
#Residuals  7     96  13.714                  
#---
#Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
于 2020-12-26T08:44:46.137 回答