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  • DEseq2 gives constant adjusted p-values

    I am doing some exploratory analyses on two datasets without replicates. i found it strange that irrespective of the calculated p-value, the adjusted p-value has constant value across all features. this value was 0.9999295.

    is there a known cause to this? does it relate to the fact that no replicates are being used?

    Thanks

  • #2
    No replicates? Yeah, you shouldn't expect to get anything significant in that case. Anyway, the actual answer is that the resulting adjusted values isn't strictly related to the lack of replicates, but is instead dependent on the distribution of the unadjusted p-values. DESeq2 uses the BH method from the p.adjust() function, as is the case for pretty much every other package. If you have a more or less uniform distribution of raw p-values, you should expect the adjusted p-values to be ~1.

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    • #3
      I am well aware of the consequences of having no replicates, as it has rightly been stated in multiple threads already

      i just find it odd that every single adjusted p-value has the same exact value. i just checked the distribution of p-values and it is rather uniform (with a downward trend when approaching zero), but i still don't understand how having exactly the same value at all genes is possible.

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      • #4
        Well, given what you describe, they should all be about 1. Any deviation from that is just due to how things are implemented (plus the joys of floating point arithmetic).

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        • #5
          ooooh, so the value would be 1 everywhere but because of floating point approximation shenanigans it turns out to be 0.99 bla bla.

          it makes sense, i tried other correction methods and they all give either 1 or the same 0.999 value i had with the benjamini. so i would say it makes sense.

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          • #6
            Yup. If you're curious, this is how R does the BH adjustment method:
            Code:
            i <- lp:1L
            o <- order(p, decreasing = TRUE)
            ro <- order(o)
            pmin(1, cummin(n/i * p[o]))[ro]
            You input "p" and the adjusted values are the output of cummin() with a max value of 1. "lp" is just the length of the input vector.

            Comment

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