我试图从R中的aov摘要中提取给定效果的名称和p值。为了使我的问题更清楚，aov摘要如下所示：
Error: subj
Df Sum Sq Mean Sq F value Pr(>F)
Group 1 9.6 9.585 1.403 0.241
Residuals 58 396.3 6.832
Error: subj:StimProb
Df Sum Sq Mean Sq F value Pr(>F)
StimProb 1 739.0 739.0 2939.367 <2e-16 ***
StimProb:Group 1 0.2 0.2 0.688 0.41
Residuals 58 14.6 0.3
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Error: subj:StimVal
Df Sum Sq Mean Sq F value Pr(>F)
StimVal 1 0.126 0.12585 0.558 0.458
StimVal:Group 1 0.026 0.02609 0.116 0.735
Residuals 58 13.074 0.22541
Error: subj:StimProb:StimVal
Df Sum Sq Mean Sq F value Pr(>F)
StimProb:StimVal 1 0.255 0.25512 0.820 0.369
StimProb:StimVal:Group 1 0.036 0.03586 0.115 0.735
Residuals 58 18.044 0.31110
Error: Within
Df Sum Sq Mean Sq F value Pr(>F)
Residuals 480 3283 6.839
我正在尝试获取名称/ p值列表，例如：
Group = 2.41
Residuals = NA
StimProb = <2e-16
StimProb:Group = 0.41
...and so forth
我能够编写提取p值的代码（如果您搜索以前的论坛，这似乎是一个普遍存在的问题）：
# p values
p <- NA
for(i in 1:length(amp_aov_3)){
tmp_p <- lapply(amp_aov_3[[i]], function(aov_sum){aov_sum$'Pr(>F)'})
tmp_p <- unlist(tmp_p)
p <- c(p, tmp_p)
}
但我无法弄清楚给定效果的名称存储在何处。我只能使用names函数访问主标题（例如，'Error：subj'）。有什么建议？
以下是amp_aov_3（即aov摘要）变量的转储，以便人们可以使用代码。
structure(list(`Error: subj` = structure(list(structure(list(
Df = c(1, 58), `Sum Sq` = c(9.58542761189546, 396.251513143065
), `Mean Sq` = c(9.58542761189546, 6.83192264039767), `F value` = c(1.40303515078115,
NA), `Pr(>F)` = c(0.24104729974717, NA)), .Names = c("Df",
"Sum Sq", "Mean Sq", "F value", "Pr(>F)"), class = c("anova",
"data.frame"), row.names = c("Group ", "Residuals"))), class = c("summary.aov",
"listof")), `Error: subj:StimProb` = structure(list(structure(list(
Df = c(1, 1, 58), `Sum Sq` = c(738.998618354635, 0.173077631291876,
14.5820237916664), `Mean Sq` = c(738.998618354635, 0.173077631291876,
0.251414203304593), `F value` = c(2939.36702318812, 0.688416282838996,
NA), `Pr(>F)` = c(2.16974416659602e-51, 0.410105585145585,
NA)), .Names = c("Df", "Sum Sq", "Mean Sq", "F value", "Pr(>F)"
), class = c("anova", "data.frame"), row.names = c("StimProb ",
"StimProb:Group", "Residuals "))), class = c("summary.aov",
"listof")), `Error: subj:StimVal` = structure(list(structure(list(
Df = c(1, 1, 58), `Sum Sq` = c(0.12584744523128, 0.0260871639459221,
13.0738086981129), `Mean Sq` = c(0.12584744523128, 0.0260871639459221,
0.225410494795049), `F value` = c(0.558303398187847, 0.115731807295137,
NA), `Pr(>F)` = c(0.45796319404567, 0.734939632080671, NA
)), .Names = c("Df", "Sum Sq", "Mean Sq", "F value", "Pr(>F)"
), class = c("anova", "data.frame"), row.names = c("StimVal ",
"StimVal:Group", "Residuals "))), class = c("summary.aov",
"listof")), `Error: subj:StimProb:StimVal` = structure(list(structure(list(
Df = c(1, 1, 58), `Sum Sq` = c(0.255118030657232, 0.035859237207785,
18.0436753630829), `Mean Sq` = c(0.255118030657232, 0.035859237207785,
0.311097851087636), `F value` = c(0.820057193469219, 0.115266746724276,
NA), `Pr(>F)` = c(0.368910019571438, 0.735452223019943, NA
)), .Names = c("Df", "Sum Sq", "Mean Sq", "F value", "Pr(>F)"
), class = c("anova", "data.frame"), row.names = c("StimProb:StimVal ",
"StimProb:StimVal:Group", "Residuals "))), class = c("summary.aov",
"listof")), `Error: Within` = structure(list(structure(list(Df = 480,
`Sum Sq` = 3282.85398452856, `Mean Sq` = 6.83927913443451,
`F value` = NA_real_, `Pr(>F)` = NA_real_), .Names = c("Df",
"Sum Sq", "Mean Sq", "F value", "Pr(>F)"), class = c("anova",
"data.frame"), row.names = "Residuals")), class = c("summary.aov",
"listof"))), .Names = c("Error: subj", "Error: subj:StimProb",
"Error: subj:StimVal", "Error: subj:StimProb:StimVal", "Error: Within"
), class = "summary.aovlist")
答案 0 :(得分：1)
这是我的解决方案。此函数从R。
中的aov函数中提取p值df.filter($"Email" match {case ".*@.*".r => true case _ => false})
答案 1 :(得分：0)
这个问题已经蛰伏了很长一段时间，但是从@Jaap所说的，我见过的最好的整洁解决方案是broom。来自here
a <- anova(lm(mpg ~ wt + qsec + disp, mtcars))
a如下所示：
a
Analysis of Variance Table
Response: mpg
Df Sum Sq Mean Sq F value Pr(>F)
wt 1 847.73 847.73 121.4366 1.082e-11 ***
qsec 1 82.86 82.86 11.8694 0.001817 **
disp 1 0.00 0.00 0.0001 0.990423
Residuals 28 195.46 6.98
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
很高兴只是看一下，但不是一个非常整洁的方法，而且信息不易被提取。 broom始终将内容导出到数据框中，其中的名称位于各自的列中。
#from broom package
tidy(a)
term df sumsq meansq statistic p.value
1 wt 1 8.477252e+02 8.477252e+02 1.214366e+02 1.081618e-11
2 qsec 1 8.285831e+01 8.285831e+01 1.186944e+01 1.817334e-03
3 disp 1 1.023935e-03 1.023935e-03 1.466785e-04 9.904229e-01
4 Residuals 28 1.954626e+02 6.980807e+00 NA NA