[R] Pairwise testing with pairwise.prop.test

[Jenny Liu]

Thank you Stephen and Peter for your code and your answers!
Stephen, your NaN explanation makes sense - thank you for putting it so clearly.
Cheers,
Jenny






On Tue, Apr 2, 2019 11:10 AM, peter dalgaard pdalgd using gmail.com  wrote:
>> My questions:1) Why is there a "-" instead of a numerical result for pairs
1-2,

>> 1-16, and 2-16?

> 

> When the difference between the pair is zero, the p.value is NaN (not a
number).

> 




Not quite: When both groups have 0 successes (or both 0 failures), the test stat
has a divide-by-zero condition.




>> 2) Is there an easy way to export/convert the result to a list with two
columns

>> instead of the matrix? Column 1 would be the pair being compared, and column
2

>> would be the p-value. For example, Column 1 would say "6-8" so column 2 would

>> say "0.9532".

> 

> Fairly easy:

> 

> idx <- expand.grid(2:16, 1:15)

> pair <- as.matrix(idx[idx[,1] > idx[,2], ])

> mode(pair) <- "character"

> comp <- apply(pair, 1, paste0, collapse="-")

> tbl <- data.frame(comp, p.value=EggResults$p.value[pair])

> head(tbl)

> # comp p.value

> # 1 2-1 NaN

> # 2 3-1 2.706354e-24

> # 3 4-1 1.487240e-23

> # 4 5-1 1.946384e-31

> # 5 6-1 4.888537e-25

> # 6 7-1 7.683167e-41

> 




Somewhat neater:




> out <- pairwise.prop.test(smokers, patients)

[....]

> pmat <- out$p.value

> ix <- lower.tri(pmat, diag=TRUE)

> R <- rownames(pmat)[row(pmat)[ix]]

> C <- colnames(pmat)[col(pmat)[ix]]

> data.frame(row.vs.col = paste(R,C,sep="-"), adj.p = pmat[ix])

  row.vs.col adj.p

1 2-1 1.00000000

2 3-1 1.00000000

3 4-1 0.11856482

4 3-2 1.00000000

5 4-2 0.09321728

6 4-3 0.12376805




Also, labeling seems to work if you label the original data appropriately, e.g.




> names(smokers) <- names(patients) <- as.roman(1:4)

> out <- pairwise.prop.test(smokers, patients)

[....]

> ix <- lower.tri(pmat, diag=TRUE)

> pmat <- out$p.value

> R <- rownames(pmat)[row(pmat)[ix]]

> C <- colnames(pmat)[col(pmat)[ix]]

> data.frame(row.vs.col = paste(R,C,sep="-"), adj.p = pmat[ix])

  row.vs.col adj.p

1 II-I 1.00000000

2 III-I 1.00000000

3 IV-I 0.11856482

4 III-II 1.00000000

5 IV-II 0.09321728

6 IV-III 0.12376805




-pd




-- 

Peter Dalgaard, Professor,

Center for Statistics, Copenhagen Business School

Solbjerg Plads 3, 2000 Frederiksberg, Denmark

Phone: (+45)38153501

Office: A 4.23

Email: pd.mes using cbs.dk Priv: PDalgd using gmail.com
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