Question about MM > PM
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s0uchi02 ▴ 30
@s0uchi02-473
Last seen 10.3 years ago
Hello. I have a question about MM > PM. In the "Textual descrtion of affy," it is stated that if the following command is used: > mean(mm(Dilution) > pm(Dilution)) [1] 0.2746048 Is this 0.2746048 % or 27.46048% of the MM are larger than PM? Thank you a lot in advance, best Shizuka
affy affy • 774 views
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Laurent Gautier ★ 2.3k
@laurent-gautier-29
Last seen 10.3 years ago
On Sun, Oct 12, 2003 at 03:06:45PM -0400, s0uchi02 wrote: > Hello. I have a question about MM > PM. > In the "Textual descrtion of affy," it is stated that if the following command > is used: > > > mean(mm(Dilution) > pm(Dilution)) > > [1] 0.2746048 > > Is this 0.2746048 % or 27.46048% of the MM are larger than PM? 'mm(Dilution) > pm(Dilution)' returns a vector of booleans ('logical' in R terminology). Calling 'mean()' on the results implies a cast to integers (TRUEs become 1 and FALSEs become 0) then the mean is computed. In other words, each time MM>PM you have 1, and zero when MM<pm. you="" sum="" all="" the="" ones="" and="" zeros="" and="" divide="" by="" length="" of="" the="" vector...="" which="" should="" bring="" you="" to="" the="" conclusion="" that="" roughly="" 30%="" of="" the="" mm="" are="" larger="" the="" pm.="" hopin'="" it="" helps,="" l.<="" div="">
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@wolfgang-huber-3550
Last seen 4 months ago
EMBL European Molecular Biology Laborat…
On Sun, 12 Oct 2003, s0uchi02 wrote: > > mean(mm(Dilution) > pm(Dilution)) > [1] 0.2746048 > Is this 0.2746048 % or 27.46048% of the MM are larger than PM? It is 27.46%. Best regards Wolfgang
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