Comments on: Post hoc analysis for Friedman’s Test (R code)

By: Jan

Jan — Mon, 16 Jan 2012 13:04:00 +0000

Nice coding & example!

By: RRookie

RRookie — Tue, 10 Jan 2012 18:45:00 +0000

Hi~
Your code is so charming!but I got some trouble with it…I have an data frame like this S T TD
1 0.00 1 D1
2 0.00 1 D2
3 0.00 1 D3
4 0.70 1 D4
5 0.74 1 D5
6 0.63 1 D6
7 0.61 1 D7
8 0.56 1 D8
9 0.54 1 D9
10 0.60 1 D10
11 0.00 2 D1
12 0.68 2 D2
13 0.57 2 D3
14 0.56 2 D4
15 0.56 2 D5
16 0.48 2 D6
17 0.57 2 D7
18 0.58 2 D8
19 0.56 2 D9
20 0.56 2 D10
21 0.00 3 D1
22 0.76 3 D2
23 0.68 3 D3
24 0.70 3 D4
25 0.59 3 D5
26 0.57 3 D6
27 0.62 3 D7
28 0.50 3 D8
29 0.59 3 D9
30 0.00 3 D10
31 0.00 4 D1
32 0.76 4 D2
33 0.70 4 D3
34 0.73 4 D4
35 0.68 4 D5
36 0.66 4 D6
37 0.80 4 D7
38 0.55 4 D8
39 0.53 4 D9
40 0.52 4 D10
41 0.00 5 D1
42 0.82 5 D2
43 0.67 5 D3
44 0.50 5 D4
45 0.67 5 D5
46 0.58 5 D6
47 0.00 5 D7
48 0.00 5 D8
49 0.00 5 D9
50 0.00 5 D10
51 0.00 6 D1
52 0.00 6 D2
53 0.59 6 D3
54 0.48 6 D4
55 0.68 6 D5
56 0.62 6 D6
57 0.48 6 D7
58 0.49 6 D8
59 0.48 6 D9
60 0.42 6 D10
61 0.00 7 D1
62 0.00 7 D2
63 0.69 7 D3
64 0.00 7 D4
65 0.00 7 D5
66 0.00 7 D6
67 0.00 7 D7
68 0.57 7 D8
69 0.49 7 D9
70 0.61 7 D10
71 0.00 8 D1
72 0.72 8 D2
73 0.75 8 D3
74 0.00 8 D4
75 0.00 8 D5
76 0.00 8 D6
77 0.00 8 D7
78 0.66 8 D8
79 0.74 8 D9
80 0.56 8 D10and I trend to compute ” friedman.test(S ~ TD | T) “but I can’t run your procedure successfully…it just gave me a wrong message :”In if (to.post.hoc.if.signif) { :
the condition has length > 1 and only the first element will be used”how can I do next ?

By: Tal Galili

Tal Galili — Sun, 04 Dec 2011 19:48:00 +0000

My pleasure

By: guest

guest — Sun, 04 Dec 2011 19:40:00 +0000

This was really helpful! thanks

By: Gaël

Gaël — Mon, 14 Nov 2011 15:26:41 +0000

Hi Ben,

Friedman’s test takes into account the dependance between the three columns in your “randomdata” object. Applying it to pairs of columns from this data set thus seems wrong.

However, I’d advise (as in the comment left on October 6 2010) using:
pairwise.wilcox.test(x=data$measure,g=data$group,p.adjust.method=”holm”,paired=T)

By: Ben M

Ben M — Sun, 13 Nov 2011 17:22:48 +0000

This may sound stupid but why not follow up the overall Friedman test with new Friedman tests for each pair of contrasts? Would this be wrong? Example:

randomdata <- cbind(rnorm(30,0),rnorm(30,1),rnorm(30,1))
friedman.test(randomdata)

friedman.test(randomdata[,c(1,2)])
friedman.test(randomdata[,c(1,3)])
friedman.test(randomdata[,c(2,3)])

With an appropriate correction for multiple testing, obviously.

By: Henric Winell

Henric Winell — Sun, 10 Jul 2011 17:08:48 +0000

No.

‘Tukey’ simply corresponds to the particular set of contrasts that represents all pairwise comparisons of the groups. So, for three groups (A, B and C) you’d want the contrasts representing B vs A, C vs A and C vs B.

By: Meech

Meech — Thu, 07 Jul 2011 07:50:01 +0000

You use Tukey in your code-isn’t this usually reserved for parametric analysis?

Thanks

By: rbest

rbest — Thu, 02 Jun 2011 21:55:38 +0000

Okay, so cancel the bit about SAS vs. R. I’m getting the same results now. But I would still love to know why your p-values are slightly different from the friedman.test function (what is the maxT method?), and especially why I’m seeing two very slightly different p-values associated with the same maxT statistic (one before and one after the multiple comparisons). Anyway. Sorry for the previous post, which was mostly wrong!

By: rbest

rbest — Thu, 02 Jun 2011 21:29:50 +0000

Hi there

Question about your test statistic. I’m getting really different p-values running a Friedman’s Test in SAS vs. the friedman.test function in R. I know that SAS is using the somewhat updated (as of Iman and Davenport 1980) version of the test statistic, as presented in 2nd and 3rd editions of Conover. The R help does not cite a formula, but does cite an older text (1973). So I’m wondering if the function in R is using the old test statistic? You appear to use a newer version of the same Hollander and Wolfe text, but when I use your code I get a p-value more similar to that provided by the friedman.test function. But not exactly the same. Any insight on (a) what test statistic is being used in R, and why your test and the friedman.test function are not identical for the global test? Thanks very much!