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millanek avatar millanek commented on August 23, 2024 1

Dsuite outputs all combinations of populations, not all permutations. With three populations, you have one trio.

If, for some reason, you want all permutations of your three populations, you can specify the different arrangements using the --tree option. Still, there are only three arrangements meaningfully calculate:
((P1,P2),P3); ((P1,P3),P2); ((P2,P3),P1)

This is because the rest are just the mirror image: ((P1,P2),P3) is the mirror image of ((P2,P1),P3). The D statistics will be exactly the same, just the sign reversed (i.e. one positive and the other negative).

Milan

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bioshimmer avatar bioshimmer commented on August 23, 2024

In the beginning I had the same problem as you. you can seach the answer in the close issues#35.

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jiangzy26 avatar jiangzy26 commented on August 23, 2024

In the beginning I had the same problem as you. you can seach the answer in the close issues#35.

Hi,Could you share how you deal with it? Now I got P1=A, P2=B, P3=C, but I want a setting P1=B, P2=A, P3=C, and get Dstatistic and Z-score, could you share your solutions? Thanks!

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bioshimmer avatar bioshimmer commented on August 23, 2024

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jiangzy26 avatar jiangzy26 commented on August 23, 2024

I thought, for example, P1=A, P2=B, P3=C, Dstatistic=0.01, Z=10, p=0, then based on the formula in the Dsuite paper, when P1=B, P2=A, P3=C, the Dstatistic should be -0.01, Z = D/std_err(D), the Z result should be -10, and P value should be 1-P_original=1-0=1. Thanks

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