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  • EVELE
    Junior Member
    • Jun 2015
    • 9

    #1

    Use of test="Wald" in the results of a time-course experiment

    Hi all,

    Concerning the example of the manual :
    The Bioconductor project aims to develop and share open source software for precise and repeatable analysis of biological data. We foster an inclusive and collaborative community of developers and data scientists.

    in the "Time Course Experiments" paragraph,

    library("fission")
    data("fission")
    ddsTC <- DESeqDataSet(fission, ~ strain + minute + strain:minute)
    ddsTC <- DESeq(ddsTC, test="LRT", reduced = ~ strain + minute)

    resultsNames(ddsTC)

    ## [1] "Intercept" "strain_mut_vs_wt" "minute_15_vs_0" "minute_30_vs_0"
    ## [5] "minute_60_vs_0" "minute_120_vs_0" "minute_180_vs_0" "strainmut.minute15"
    ## [9] "strainmut.minute30" "strainmut.minute60" "strainmut.minute120" "strainmut.minute180"

    res30 <- results(ddsTC, name="strainmut.minute30", test="Wald")
    res30[which.min(resTC$padj),]

    ## log2 fold change (MLE): strainmut.minute30
    ## Wald test p-value: strainmut.minute30
    ## DataFrame with 1 row and 6 columns
    ## baseMean log2FoldChange lfcSE stat pvalue padj
    ## <numeric> <numeric> <numeric> <numeric> <numeric> <numeric>
    ## SPBC2F12.09c 174.6712 -2.601034 0.6314737 -4.11899 3.805364e-05 0.2572426

    When should we use the test="Wald" in the results function?

    What are the results I get when I write:
    res30 <- results(ddsTC, name="strainmut.minute30")
    and what is the difference in the results by adding test="Wald"?
    res30 <- results(ddsTC, name="strainmut.minute30", test="Wald")?

    Thanks in advance for any ideas
  • dpryan
    Devon Ryan
    • Jul 2011
    • 3478

    #2
    Normally you get a Wald test unless you specify otherwise.

    Comment

    • EVELE
      Junior Member
      • Jun 2015
      • 9

      #3
      It is not the same.
      I can see differences when I apply both ways to my data.

      Common part of code:
      > times=factor(c(1,rep(2,2),rep(3,2),0,1,rep(2,2),rep(3,2),0), levels=c(0,1,2,3))
      > condition = factor(c(rep("factors", 6), rep("controls",6)), levels=c("controls","factors"))

      > colData = data.frame( condition=condition, times=times)
      > rownames(colData)=colnames(duo)

      > dds<-DESeqDataSetFromMatrix(countData= as.matrix(duo), colData= colData ,design=formula(~condition+times+condition:times))
      > dds <- DESeq(dds, test="LRT", reduced = ~ condition + times)
      estimating size factors
      estimating dispersions
      gene-wise dispersion estimates
      mean-dispersion relationship
      final dispersion estimates
      fitting model and testing

      > resultsNames(dds)
      [1] "Intercept" "condition_factors_vs_controls"
      [3] "times_1_vs_0" "times_2_vs_0"
      [5] "times_3_vs_0" "conditionfactors.times1"
      [7] "conditionfactors.times2" "conditionfactors.times3"


      The way without test="Wald"

      > res1 <- results(dds, name="conditionfactors.times1")
      > res2 <- results(dds, name="conditionfactors.times2")
      > res3 <- results(dds, name="conditionfactors.times3")


      # Here I filter padj <= 0.05
      > res1 <- res1[!is.na(res1$padj) & res1$padj <=0.05,]
      > res2 <- res2[!is.na(res2$padj) & res2$padj <=0.05,]
      > res3 <- res3[!is.na(res3$padj) & res3$padj <=0.05,]

      For the gene PTET.51.1.G0050374 I obtain:

      > res1["PTET.51.1.G0050374",]

      log2 fold change (MLE): conditionfactors.times1
      LRT p-value: '~ condition + times + condition:times' vs '~ condition + times'
      DataFrame with 1 row and 6 columns
      baseMean log2FoldChange lfcSE stat pvalue
      <numeric> <numeric> <numeric> <numeric> <numeric>
      PTET.51.1.G0050374 481.4894 5.916279 0.5890732 140.9655 2.340315e-30
      padj
      <numeric>
      PTET.51.1.G0050374 6.717171e-26



      The way with test="Wald"

      > res1W <- results(dds, name="conditionfactors.times1",test="Wald")
      > res2W <- results(dds, name="conditionfactors.times2",test="Wald")
      > res3W <- results(dds, name="conditionfactors.times3",test="Wald")

      > res1W <- res1W[!is.na(res1W$padj) & res1W$padj <=0.05,]
      > res2W <- res2W[!is.na(res2W$padj) & res2W$padj <=0.05,]
      > res3W <- res3W[!is.na(res3W$padj) & res3W$padj <=0.05,]

      For the same gene I obtain different values in stat pvalue and padj columns

      > res1W["PTET.51.1.G0050374",]

      log2 fold change (MLE): conditionfactors.times1
      Wald test p-value: conditionfactors.times1
      DataFrame with 1 row and 6 columns
      baseMean log2FoldChange lfcSE stat pvalue
      <numeric> <numeric> <numeric> <numeric> <numeric>
      PTET.51.1.G0050374 481.4894 5.916279 0.5890732 10.04337 9.826083e-24
      padj
      <numeric>
      PTET.51.1.G0050374 1.210279e-19

      Furthermore, the number of genes (after applying the same filter) vary between res1W, res2W and res3W
      > dim(res1W)[1]
      [1] 46
      > dim(res2W)[1]
      [1] 2
      > dim(res3W)[1]
      [1] 345


      while the numbers are the same for res1, res2, res3 (without test="Wald")

      > dim(res1)[1]
      [1] 780
      > dim(res2)[1]
      [1] 780
      > dim(res3)[1]
      [1] 780

      I cannot see what's happening here, but I'd like to know which of the 2 methods I should use in order to obtain all genes having a time-specific differential expression and a padj value <=0.05 in my time-course experiment.

      Thank you

      Comment

      • EVELE
        Junior Member
        • Jun 2015
        • 9

        #4
        Actually when you don't put test="Wald" I can see that it is a LRT applied and we get LRT pvalues while when putting test="Wald" it is the Wald test pvalues we obtain.

        I don't know which approach would give me what I want but I tend to believe in the LRT method for my case (genes with time-specific differential expression).

        In other words, I think I should not put test="Wald".

        Comment

        • dpryan
          Devon Ryan
          • Jul 2011
          • 3478

          #5
          Ah, I had missed that you explicitly requested an LRT with DESeq(). The default is Wald there too. The models that you're testing are completely different between the two methods. With a Wald test, you can test individual coefficient, with the LRT you're testing a more generic interaction.

          Edit: Yes, an LRT is likely what you want, since it asks the question, "what genes generally interact in a time:condition dependent manner, regardless of the exact time point".
          Last edited by dpryan; 09-16-2015, 08:39 AM.

          Comment

          • EVELE
            Junior Member
            • Jun 2015
            • 9

            #6
            In the manual link I sent, it goes like that:

            Wald tests for the log2 fold changes at individual time points can be investigated using the test argument to results:

            res30 <- results(ddsTC, name="strainmut.minute30", test="Wald")

            which seems to be consistent with what you say "With a Wald test, you can test individual coefficient".
            However, I remain with my questions because I cannot really see what a Wald Test and LRT test do.

            For the moment, I will just apply an LRT test and I'll try to read some things about Wald and LRT tests.

            Thank you Devon

            Comment

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