Two-way ANOVA with multtest's MTP/EBMTP

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Neil Gray ▴ 20

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Last seen 11.3 years ago

Hello - hopefully a quick question here. I've been using the MTP and EBMTP functions for multiple testing correction of microarray expression data. I've had no problems using the t-test option, but my overall design is a two-factor ANOVA (2x2 groups using two different treatments). My eSet therefore has two phenoData slots, and while it's easy enough to specify one or the other, e.g., test='t.twosamp.equalvar' and Y='treatment1', but I don't understand how to specify the 2x2 design for test='f.twoway'. It seems Y can't be a matrix, and it can only be of length equal to the number of samples. So, given that 12 samples with groups defined as treatment1=c(0,0,0,0,0,0,1,1,1,1,1,1) and treatment2=c(0,0,0,1,1,1,0,0,0,1,1,1), how do I pass this design to MTP?

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ADD COMMENT • link updated 14.9 years ago by James W. MacDonald 68k • written 14.9 years ago by Neil Gray ▴ 20

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James W. MacDonald 68k

@james-w-macdonald-5106

Last seen 16 hours ago

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Hi Neil, On 1/19/2011 4:12 PM, Neil Gray wrote: > Hello - hopefully a quick question here. I've been using the MTP and > EBMTP functions for multiple testing correction of microarray expression > data. I've had no problems using the t-test option, but my overall > design is a two-factor ANOVA (2x2 groups using two different > treatments). My eSet therefore has two phenoData slots, and while it's > easy enough to specify one or the other, e.g., test='t.twosamp.equalvar' > and Y='treatment1', but I don't understand how to specify the 2x2 design > for test='f.twoway'. It seems Y can't be a matrix, and it can only be of > length equal to the number of samples. Yes. From ?MTP: Y: A vector, factor, or 'Surv' object containing the outcome of interest. This may be class labels (F-tests and two sample t-tests) or a continuous or polycotomous dependent variable (linear regression based t-tests), or survival data (Cox proportional hazards based t-tests). For 'block.f' and 'f.twoway' tests, class labels must be ordered by block and within each block ordered by group. If 'X' is an ExpressionSet, 'Y' can be a character string referring to the column of 'pData(X)' to use as outcome. Default is 'NULL'. So for f.twoway, you need a factor, with the class labels ordered correctly. Let's say you have something like this: > trt <- rep(c("trt","untrt"), each=6) > typ <- rep(c("wt","ko"), each=3, times=2) > cbind(trt,typ) trt typ [1,] "trt" "wt" [2,] "trt" "wt" [3,] "trt" "wt" [4,] "trt" "ko" [5,] "trt" "ko" [6,] "trt" "ko" [7,] "untrt" "wt" [8,] "untrt" "wt" [9,] "untrt" "wt" [10,] "untrt" "ko" [11,] "untrt" "ko" [12,] "untrt" "ko" To get the factor, I believe you need something like this: > my.y <- factor(paste(trt,typ, sep = "")) > my.y [1] trtwt trtwt trtwt trtko trtko trtko untrtwt untrtwt untrtwt [10] untrtko untrtko untrtko Levels: trtko trtwt untrtko untrtwt Does that make sense? Best, Jim > > So, given that 12 samples with groups defined as > treatment1=c(0,0,0,0,0,0,1,1,1,1,1,1) and > treatment2=c(0,0,0,1,1,1,0,0,0,1,1,1), how do I pass this design to MTP? > > _______________________________________________ > Bioconductor mailing list > Bioconductor at r-project.org > https://stat.ethz.ch/mailman/listinfo/bioconductor > Search the archives: > http://news.gmane.org/gmane.science.biology.informatics.conductor -- James W. MacDonald, M.S. Biostatistician Douglas Lab University of Michigan Department of Human Genetics 5912 Buhl 1241 E. Catherine St. Ann Arbor MI 48109-5618 734-615-7826 ********************************************************** Electronic Mail is not secure, may not be read every day, and should not be used for urgent or sensitive issues

ADD COMMENT • link 14.9 years ago James W. MacDonald 68k

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Hi Jim, Thanks for your help, but I'm still having some trouble. The format seems to be correct, but I keep getting an error: > y = paste(rep(c("pF","pS"), each=6),rep(c("aF","aS"), each=3, times=2),sep="") > yf = factor(y) > yf [1] pFaF pFaF pFaF pFaS pFaS pFaS pSaF pSaF pSaF pSaS pSaS pSaS Levels: pFaF pFaS pSaF pSaS > m = MTP(exset,Y=yf,test='f.twoway',B=100,typeone='fdr',nulldist='boot.ctr' ,method='sd.minP',alpha=0.05) Error in rep(x, BlockNum[x]) : invalid 'times' argument > traceback() 9: FUN(1L[[1L]], ...) 8: lapply(X, FUN, ...) 7: sapply(1:l, function(x) rep(x, BlockNum[x])) 6: unlist(sapply(1:l, function(x) rep(x, BlockNum[x]))) 5: FUN(newX[, i], ...) 4: apply(X, MARGIN, FUN, ...) 3: wapply(X, 1, stat.closure, W) 2: get.Tn(X, stat.closure, W) 1: MTP(exset, Y = yf, test = "f.twoway", B = 100, typeone = "fdr", nulldist = "boot.ctr", method = "sd.minP", alpha = 0.05) I feel like I must be missing something obvious here, but I've tried organizing the group data in a number of ways... Neil On 1/19/11 5:37 PM, James W. MacDonald wrote: > Hi Neil, > > On 1/19/2011 4:12 PM, Neil Gray wrote: >> Hello - hopefully a quick question here. I've been using the MTP and >> EBMTP functions for multiple testing correction of microarray expression >> data. I've had no problems using the t-test option, but my overall >> design is a two-factor ANOVA (2x2 groups using two different >> treatments). My eSet therefore has two phenoData slots, and while it's >> easy enough to specify one or the other, e.g., test='t.twosamp.equalvar' >> and Y='treatment1', but I don't understand how to specify the 2x2 design >> for test='f.twoway'. It seems Y can't be a matrix, and it can only be of >> length equal to the number of samples. > > Yes. From ?MTP: > > Y: A vector, factor, or 'Surv' object containing the outcome of > interest. This may be class labels (F-tests and two sample > t-tests) or a continuous or polycotomous dependent variable > (linear regression based t-tests), or survival data (Cox > proportional hazards based t-tests). For 'block.f' and > 'f.twoway' tests, class labels must be ordered by block and > within each block ordered by group. If 'X' is an > ExpressionSet, 'Y' can be a character string referring to the > column of 'pData(X)' to use as outcome. Default is 'NULL'. > > So for f.twoway, you need a factor, with the class labels ordered > correctly. > > Let's say you have something like this: > > > trt <- rep(c("trt","untrt"), each=6) > > typ <- rep(c("wt","ko"), each=3, times=2) > > cbind(trt,typ) > trt typ > [1,] "trt" "wt" > [2,] "trt" "wt" > [3,] "trt" "wt" > [4,] "trt" "ko" > [5,] "trt" "ko" > [6,] "trt" "ko" > [7,] "untrt" "wt" > [8,] "untrt" "wt" > [9,] "untrt" "wt" > [10,] "untrt" "ko" > [11,] "untrt" "ko" > [12,] "untrt" "ko" > > To get the factor, I believe you need something like this: > > > my.y <- factor(paste(trt,typ, sep = "")) > > my.y > [1] trtwt trtwt trtwt trtko trtko trtko untrtwt untrtwt > untrtwt > [10] untrtko untrtko untrtko > Levels: trtko trtwt untrtko untrtwt > > Does that make sense? > > Best, > > Jim > >

ADD REPLY • link 14.9 years ago Neil Gray ▴ 20

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Hi Neil, Well, I've played around with this a bit, googled around for an example, and looked at the code to see if I could divine exactly what the expectation for the Y argument is, and I must say that I am stumped. I was thinking that you actually wanted > y [1] 11 11 11 12 12 12 21 21 21 22 22 22 Levels: 11 12 21 22 as the code is grepping for "12", but it doesn't work out the way it should. Anyway, I think you would be better served by contacting the maintainer directly. She is the expert, after all. See packageDescription("multtest")$Main for her email address. Best, Jim On 1/20/2011 4:42 PM, Neil Gray wrote: > Hi Jim, > Thanks for your help, but I'm still having some trouble. The format > seems to be correct, but I keep getting an error: > > > y = paste(rep(c("pF","pS"), each=6),rep(c("aF","aS"), each=3, > times=2),sep="") > > yf = factor(y) > > yf > [1] pFaF pFaF pFaF pFaS pFaS pFaS pSaF pSaF pSaF pSaS pSaS pSaS > Levels: pFaF pFaS pSaF pSaS > > m = > MTP(exset,Y=yf,test='f.twoway',B=100,typeone='fdr',nulldist='boot.ct r',method='sd.minP',alpha=0.05) > > Error in rep(x, BlockNum[x]) : invalid 'times' argument > > > traceback() > 9: FUN(1L[[1L]], ...) > 8: lapply(X, FUN, ...) > 7: sapply(1:l, function(x) rep(x, BlockNum[x])) > 6: unlist(sapply(1:l, function(x) rep(x, BlockNum[x]))) > 5: FUN(newX[, i], ...) > 4: apply(X, MARGIN, FUN, ...) > 3: wapply(X, 1, stat.closure, W) > 2: get.Tn(X, stat.closure, W) > 1: MTP(exset, Y = yf, test = "f.twoway", B = 100, typeone = "fdr", > nulldist = "boot.ctr", method = "sd.minP", alpha = 0.05) > > I feel like I must be missing something obvious here, but I've tried > organizing the group data in a number of ways... > > Neil > > On 1/19/11 5:37 PM, James W. MacDonald wrote: >> Hi Neil, >> >> On 1/19/2011 4:12 PM, Neil Gray wrote: >>> Hello - hopefully a quick question here. I've been using the MTP and >>> EBMTP functions for multiple testing correction of microarray expression >>> data. I've had no problems using the t-test option, but my overall >>> design is a two-factor ANOVA (2x2 groups using two different >>> treatments). My eSet therefore has two phenoData slots, and while it's >>> easy enough to specify one or the other, e.g., test='t.twosamp.equalvar' >>> and Y='treatment1', but I don't understand how to specify the 2x2 design >>> for test='f.twoway'. It seems Y can't be a matrix, and it can only be of >>> length equal to the number of samples. >> >> Yes. From ?MTP: >> >> Y: A vector, factor, or 'Surv' object containing the outcome of >> interest. This may be class labels (F-tests and two sample >> t-tests) or a continuous or polycotomous dependent variable >> (linear regression based t-tests), or survival data (Cox >> proportional hazards based t-tests). For 'block.f' and >> 'f.twoway' tests, class labels must be ordered by block and >> within each block ordered by group. If 'X' is an >> ExpressionSet, 'Y' can be a character string referring to the >> column of 'pData(X)' to use as outcome. Default is 'NULL'. >> >> So for f.twoway, you need a factor, with the class labels ordered >> correctly. >> >> Let's say you have something like this: >> >> > trt <- rep(c("trt","untrt"), each=6) >> > typ <- rep(c("wt","ko"), each=3, times=2) >> > cbind(trt,typ) >> trt typ >> [1,] "trt" "wt" >> [2,] "trt" "wt" >> [3,] "trt" "wt" >> [4,] "trt" "ko" >> [5,] "trt" "ko" >> [6,] "trt" "ko" >> [7,] "untrt" "wt" >> [8,] "untrt" "wt" >> [9,] "untrt" "wt" >> [10,] "untrt" "ko" >> [11,] "untrt" "ko" >> [12,] "untrt" "ko" >> >> To get the factor, I believe you need something like this: >> >> > my.y <- factor(paste(trt,typ, sep = "")) >> > my.y >> [1] trtwt trtwt trtwt trtko trtko trtko untrtwt untrtwt untrtwt >> [10] untrtko untrtko untrtko >> Levels: trtko trtwt untrtko untrtwt >> >> Does that make sense? >> >> Best, >> >> Jim >> >> > > _______________________________________________ > Bioconductor mailing list > Bioconductor at r-project.org > https://stat.ethz.ch/mailman/listinfo/bioconductor > Search the archives: > http://news.gmane.org/gmane.science.biology.informatics.conductor -- James W. MacDonald, M.S. Biostatistician Douglas Lab University of Michigan Department of Human Genetics 5912 Buhl 1241 E. Catherine St. Ann Arbor MI 48109-5618 734-615-7826 ********************************************************** Electronic Mail is not secure, may not be read every day, and should not be used for urgent or sensitive issues

ADD REPLY • link 14.9 years ago James W. MacDonald 68k

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