Wilcoxon秩和檢驗 - 在比較測量前後，您能否使R子類型的治療方法不同？

Question

我很新的R編程，但我一直在學習:)

我想進行測量前後的配對Wilcoxon秩和檢驗。我的受試者已經獲得4種不同類型的治療（即作爲因子w/4水平的治療）。我的數據是序數。

我有一個包含3列的數據幀：

• 之前
• 後
• 治療的類型

能否運行wilcox.test而子集劃分的不同類型的處理？

我跑了wilcox.test如下：

wilcox.test(Before,After,paired=TRUE) 

如何修改這個代碼，以子集我的類型的治療？

希望有一個人能幫助我... :)

之前|在|之後治療
------ | ------ | ---------
2 | 2 | SH
2 | 1 | SH
2 | 2 | SH
1 | 1 | SH
2 | 2 | OH
2 | 2 | OH
2 | 2 | OH
2 | 1 | OH
2 | 1 | SA
2 | 1 | SA
1 | 1 | SA
2 | 2 | SA
2 | 2 | OA
2 | 3 | OA
1 | 2 | OA
2 | 2 | OA

Answer 1

使用split，使您的組的列表，然後lapply在列表...

x <- read.csv(text = "Before|After|Treatment 
2|2|SH 
2|1|SH 
2|2|SH 
1|1|SH 
2|2|OH 
2|2|OH 
2|2|OH 
2|1|OH 
2|1|SA 
2|1|SA 
1|1|SA 
2|2|SA 
2|2|OA 
2|3|OA 
1|2|OA 
2|2|OA", sep = "|") 

x <- split(x, x$Treatment) 
lapply(x, function(g) wilcox.test(g$Before, g$After, paired = T)) 

$OA

Wilcoxon signed rank test with continuity correction

data: g$Before and g$After V = 0, p-value = 0.3458 alternative hypothesis: true location shift is not equal to 0

$OH

Wilcoxon signed rank test with continuity correction

data: g$Before and g$After V = 1, p-value = 1 alternative hypothesis: true location shift is not equal to 0

$SA

Wilcoxon signed rank test with continuity correction

data: g$Before and g$After V = 3, p-value = 0.3458 alternative hypothesis: true location shift is not equal to 0

$SH

Wilcoxon signed rank test with continuity correction

data: g$Before and g$After V = 1, p-value = 1 alternative hypothesis: true location shift is not equal to 0