Hello,

I'm doing some deconvolution analysis with CIBERSORT in R with the LM22 file but when I look at the results the p-value's column is all 9999. What I'm doing wrong?

```
results <- cibersort(LM22, at41)
```

p-value in CIBERSORT

0

Entering edit mode

7 weeks ago

michelafrancesconi9
▴
20

Hello,

I'm doing some deconvolution analysis with CIBERSORT in R with the LM22 file but when I look at the results the p-value's column is all 9999. What I'm doing wrong?

```
results <- cibersort(LM22, at41)
```

2

Entering edit mode

6 weeks ago

Aspire
▴
340

You have not explicitly specified the number of permutations (which are used to calculate the p-value). The default value of the permutations is 0. It is not possible to calculate a p-value by permutations when no permutations are used. Hence I believe that the authors of the package deliberately used such a number which is clearly not a probability, to show that clearly.

```
cibersort <- function(sig_matrix, mixture_file, perm = 0, QN = TRUE){
..
P <- perm #number of permutations
..
pval <- 9999
..
if (P > 0) {
pval <- 1 - (which.min(abs(nulldist - mix_r)) / length(nulldist))
}
..
}
```

So, you can see from the code that if perm==0 (default), the pval is left with the value of 9999

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