我可以將TukeyHSD的結果放入表格嗎？ R中

杜克事後檢驗返回類似我可以將TukeyHSD的結果放入表格嗎？ R中

  diff   lwr  upr  p adj 
2-1 2.125000e-01 -0.13653578 0.5615358 0.4873403 
3-1 2.250000e-01 -0.12403578 0.5740358 0.4219408 
4-1 3.875000e-01 0.03846422 0.7365358 0.0206341 
5-1 6.875000e-01 0.33846422 1.0365358 0.0000020 
6-1 2.250000e-01 -0.12403578 0.5740358 0.4219408 
3-2 1.250000e-02 -0.31064434 0.3356443 0.9999974 
4-2 1.750000e-01 -0.14814434 0.4981443 0.6147144 
5-2 4.750000e-01 0.15185566 0.7981443 0.0006595 
6-2 1.250000e-02 -0.31064434 0.3356443 0.9999974 
4-3 1.625000e-01 -0.16064434 0.4856443 0.6866539 
5-3 4.625000e-01 0.13935566 0.7856443 0.0009888 
6-3 1.776357e-15 -0.32314434 0.3231443 1.0000000 
5-4 3.000000e-01 -0.02314434 0.6231443 0.0844160 
6-4 -1.625000e-01 -0.48564434 0.1606443 0.6866539 
6-5 -4.625000e-01 -0.78564434 -0.1393557 0.0009888

這是罰款，而是難以閱讀的結果。如果結果可以排列在一個較低的對角線表格中，那麼將會更好。

喜歡的東西

1 2 3 4 5 6 
1 
2 p 
3 p p 
4 p p p 
5 p p p p 
6 p p p p p

其中p是適當的P值。這可能嗎？

來源

2017-07-14 Demetri P

這是我對使用tidyverse進行手動轉換的建議。我已經把它作爲一個函數裝箱了，你可以通過傳遞除p_adj以外的東西來更改度量。請注意，我假設輸入（tbl）爲數據幀。

transformTable <- function(tbl, metric) { 
    # Takes table of TurkeyHSD output metrics 
    # and transforms them into a pairwise comparison matrix. 
    # tbl is assumed to be a data.frame or tibble, 
    # var is assumed to be a character string 
    # giving the variable name of the metric in question 
    # (here: "diff", "lwr", "upr", or "p_adj") 
    tbl <- tbl %>% 
    # Split comparison into individual variables 
    mutate(
     Var1 = as.numeric(substr(X, 1, 1)), 
     Var2 = as.numeric(substr(X, 3, 3))) %>% 
    # Only keep relevant fields 
    select(Var1, Var2, matches(metric)) %>% 
    # Filter out NA's 
    filter(!is.na(metric)) %>% 
    # Make into "wide" format using Var2 
    spread_(key = 'Var2', value = metric, fill = '') 

    # Let's change the row names to Var1 
    row.names(tbl) <- tbl$Var1 
    # And drop the Var1 column 
    tbl <- select(tbl, -Var1) 

    return(tbl) 
} 


transformTable(df, 'p_adj')

輸出：

  1   2   3   4   5 
2 0.4873403           
3 0.4219408 0.9999974        
4 0.0206341 0.6147144 0.6866539      
5  2e-06 0.0006595 0.0009888 0.084416   
6 0.4219408 0.9999974   1 0.6866539 0.0009888

再生的數據集：

df <- structure(list(X = structure(c(1L, 2L, 4L, 7L, 11L, 3L, 5L, 8L, 
12L, 6L, 9L, 13L, 10L, 14L, 15L), .Label = c("2-1", "3-1", "3-2", 
"4-1", "4-2", "4-3", "5-1", "5-2", "5-3", "5-4", "6-1", "6-2", 
"6-3", "6-4", "6-5"), class = "factor"), diff = c(0.213, 0.225, 
0.388, 0.688, 0.225, 0.0125, 0.175, 0.475, 0.0125, 0.163, 0.463, 
1.78e-15, 0.3, -0.163, -0.463), lwr = c(-0.13653578, -0.12403578, 
0.03846422, 0.33846422, -0.12403578, -0.31064434, -0.14814434, 
0.15185566, -0.31064434, -0.16064434, 0.13935566, -0.32314434, 
-0.02314434, -0.48564434, -0.78564434), upr = c(0.5615358, 0.5740358, 
0.7365358, 1.0365358, 0.5740358, 0.3356443, 0.4981443, 0.7981443, 
0.3356443, 0.4856443, 0.7856443, 0.3231443, 0.6231443, 0.1606443, 
-0.1393557), p_adj = c(0.4873403, 0.4219408, 0.0206341, 2e-06, 
0.4219408, 0.9999974, 0.6147144, 0.0006595, 0.9999974, 0.6866539, 
0.0009888, 1, 0.084416, 0.6866539, 0.0009888)), .Names = c("X", 
"diff", "lwr", "upr", "p_adj"), class = "data.frame", row.names = c(NA, 
-15L))

來源

2017-07-14 08:43:15 JanLauGe

我可以將TukeyHSD的結果放入表格嗎？ R中

回答

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