我试图对我的数据执行图基的HSD测试或LSD测试。我有两个因素,收集(2个处理)和灌溉(5个处理),并想做测试蔗糖的反应,从每个组合,所以10个总处理。
数据:
structure(list(Collection = structure(c(1L, 1L, 1L, 1L, 1L, 2L
), .Label = c("1", "2"), class = "factor"), Irrigation = structure(c(1L,
2L, 3L, 4L, 5L, 1L), .Label = c("Rate1", "Rate2", "Rate3", "Rate4",
"Rate5"), class = "factor"), meanSuc = c(0.585416666666667, 0.5032,
0.61375, 0.602775, 0.688466666666667, 0.545133333333333)), row.names =
c(NA,
-6L), groups = structure(list(Collection = structure(1:2, .Label = c("1",
"2"), class = "factor"), .rows = list(1:5, 6L)), row.names = c(NA,
-2L), class = c("tbl_df", "tbl", "data.frame"), .drop = TRUE), class =
c("grouped_df",
"tbl_df", "tbl", "data.frame"))尝试将处理组合成一个列,并使用农业进行测试:
Tukey_data <- dataAvgSucCI %>%
mutate(Tukey_ID = paste(Collection, Irrigation, sep="_"))
TukeyAov <- aov(meanSuc ~ Tukey_ID,Tukey_data)
HSD.test(TukeyAov, "Tukey_ID", group=TRUE)错误消息:
如果(pvaluek <= 0.001) sigk <-“*”中出错,否则错误(pvaluek <= ) 除了:警告消息: qtukey(1 - alpha,ntr,DFerror):生成NaNs
如何编辑代码以使其工作?
还是我最好写点完全不同的东西?
发布于 2019-06-19 21:53:57
数据必须是这样的(单向方差分析):
Collection = rep(1:2, times = 1, each = 5)
Irrigation = rep(1:5, times = 2, each = 1)
meanSuc = rnorm(10, mean = 0, sd = 1)
d = data.frame(Collection, Irrigation, meanSuc)
fit = aov(meanSuc ~ as.factor(Collection), data=d)
TukeyHSD(fit)或者双向方差分析:
fit2 = aov(meanSuc ~ as.factor(Collection) + as.factor(Irrigation), data = d)
TukeyHSD(fit2)我想你喜欢做双向变异。就像AkselA说的,如果你像你那样做了一个单向的方差分析,你的目标变量(meanSuc)就不会有任何变化。
https://stackoverflow.com/questions/56675135
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