我正在努力研究如何进行重复测量方差分析。我的数据结构如下
means <- structure(list(col = c("c", "v1", "b1", "v2", "b2"),
`1` = c(8.55,9.73, 8.93, 9.52, 9.91),
`2` = c(8.4, 9.97, 9.08, 9.66, 9.97),
`3` = c(8.48, 10.04, 9.13, 9.73, 10.04),
`4` = c(8.42, 9.63,8.9, 9.34, 9.82),
`5` = c(8.42, 9.59, 8.87, 9.39, 9.69),
`6` = c(8.52, 9.74, 9.02, 9.58, 9.84),
`7` = c(8.37, 9.67,8.98, 9.47, 9.74),
`8` = c(8.42, 9.67, 9.02, 9.52, 9.77),
`9` = c(8.56, 9.79, 9.36, 9.6, 9.78),
`10` = c(8.44, 9.63,9.15, 9.52, 9.67),
`11` = c(8.3, 9.58, 9.05, 9.49, 9.63),
`12` = c(8.03, 9.33, 8.82, 9.23, 9.38),
`13` = c(7.95, 9.08, 8.7, 9.04, 9.19),
`14` = c(8, 8.34, 8.37, 8.43, 8.54),
`15` = c(8.04,8.26, 8.4, 8.45, 8.61),
`16` = c(8.08, 8.09, 8.18, 8.16,8.28),
`17` = c(7.99, 8.06, 8.09, 8.15, 8.26),
`18` = c(8.06, 8.06, 8.09, 8.1, 8.22),
`19` = c(7.96, 7.96, 7.99, 8.03, 8.1),
`20` = c(7.96, 7.98, 7.99, 7.99, 8.11),
`21` = c(8.16, 8.22, 8.22, 8.26, 8.33),
`22` = c(8.08, 8.16, 8.13, 8.2, 8.2),
`23` = c(7.94, 7.97, 7.94, 7.98, 8.07),
`24` = c(8.02,8.03, 8, 8.08, 8.1),
`25` = c(8.03, 8.08, 8.09, 8.12, 8.15),
`26` = c(7.92, 7.95, 7.95, 7.96, 7.98)),
.Names = c("col","1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12",
"13", "14", "15", "16", "17", "18", "19", "20", "21", "22", "23","24", "25", "26"),
class = c("data.table", "data.frame"))
其中＆＃34; col＆＃34;代表不同的基质（处理），标题中的数字是随时间的测量值。这只是数据的一部分。
为了进行重复测量ANOVA（希望是正确的统计测试），我试着按照我在网上找到的几个例子，例如： http://rtutorialseries.blogspot.de/2011/02/r-tutorial-series-one-way-repeated.html
# step 1 Define the levels:
levels <- c(1:26)
# define factor
factor <- as.factor(levels)
#define the frame
frame <- data.frame(factor)
# bind the colums
bind <- cbind (means$`1`,means$`2`,means$`3`,means$`4`,means$`5`,means$`6`,means$`7`,means$`8`,means$`9`,means$`10`,means$`11`,means$`12`,means$`13`,means$`14`,means$`15`,means$`16`,means$`17`,means$`18`,means$`19`,means$`20`,means$`21`,means$`22`,means$`23`,means$`24`,means$`25`,means$`26`)
# define the model
model <- lm(ph_bind ~ 1)
# ANOVA
analysis <- Anova(model, idata=frame, idesign= ~factor)
这导致：
> analysis <- Anova(model, idata = factor, idesign = ~factor)
Warning message:
In Anova.lm(model, idata = factor, idesign = ~factor) :
the model contains only an intercept: Type III test substituted
> summary (analysis)
Sum Sq Df F value Pr(>F)
Min. : 61.59 Min. : 1 Min. :20519 Min. :0
1st Qu.:2495.43 1st Qu.: 33 1st Qu.:20519 1st Qu.:0
Median :4929.27 Median : 65 Median :20519 Median :0
Mean :4929.27 Mean : 65 Mean :20519 Mean :0
3rd Qu.:7363.10 3rd Qu.: 97 3rd Qu.:20519 3rd Qu.:0
Max. :9796.94 Max. :129 Max. :20519 Max. :0
NA's :1 NA's :1
这不是我希望的预期输出。我做错了什么？
感谢任何帮助：）