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yap-6.3/packages/prism/exs/bloodAaBb.psm

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2011-11-10 12:24:47 +00:00
%%%%
%%%% Another hypothesis on ABO blood type inheritance --- bloodAaBb.psm
%%%%
%%%% Copyright (C) 2007,2008
%%%% Sato Laboratory, Dept. of Computer Science,
%%%% Tokyo Institute of Technology
%% ABO blood type consists of A, B, O and AB. They are observable
%% (phenotypes) and determined by a pair of blood type genes (geneotypes).
%% At present, it is known that there are three ABO genes, namely a, b and
%% o located on the 9th chromosome of a human being, but in early 20th
%% century, there was another hypothesis that we have two loci for ABO
%% blood type with dominant alleles A/a and B/b. That is, genotypes aabb,
%% A*bb, aaB* and A*B* correspond to the blood types (phenotypes) O, A, B
%% and AB, respectively, where * stands for a don't care symbol. We call
%% this hypothesis the AaBb gene model, and assume random mating.
%%-------------------------------------
%% Quick start : sample session -- the same as that of bloodABO.psm
%%
%% ?- prism(bloodAaBb),go,print_blood.
%% % Learn parameters from randomly generated
%% % 100 samples with A:B:O:AB = 38:22:31:9
%%
%% ?- probf(bloodtype(ab),E),print_graph(E).
%% ?- prob(bloodtype(ab),P).
%%
%% ?- viterbif(bloodtype(ab),P,E),print_graph(E).
%% ?- viterbi(bloodtype(ab),P).
%% % P is the probability of a most likely
%% % explanation E for bloodtype(ab).
go:- learn_bloodtype(100).
%%-------------------------------------
%% Session for model selection:
%%
%% -- we try to evaluate the plausibilities of the correct model (ABO
%% gene model) and this AaBb gene model according to the data in
%% `bloodtype.dat'. The data file `bloodtype.dat' contains 38
%% persons of blood type A, 22 persons of blood type B, 31 persons
%% of blood type O, and 9 persons of blood type AB (the ratio is
%% almost the same as that in Japanese people).
%%
%% 1. Modify bloodABO.psm and bloodAaBb.psm:
%% - Use learn/0 instead of learn/1.
%%
%% 2. Get the BIC value for the ABO gene model (bloodABO.psm)
%% ?- prism(bloodABO).
%% ?- learn.
%% ?- learn_statistics(bic,BIC).
%%
%% 3. Get the BIC value for the AaBb gene model (this file)
%% ?- prism(bloodAaBb).
%% ?- learn.
%% ?- learn_statistics(bic,BIC).
%%
:- set_prism_flag(data_source,file('bloodtype.dat')).
% When we run learn/0, the data are supplied
% by `bloodtype.dat'.
values(locus1,['A',a]).
values(locus2,['B',b]).
%%------------------------------------
%% Modeling part:
bloodtype(P) :-
genotype(locus1,X1,Y1),
genotype(locus2,X2,Y2),
( X1=a, Y1=a, X2=b, Y2=b -> P=o
; ( X1='A' ; Y1='A' ), X2=b, Y2=b -> P=a
; X1=a, Y1=a, ( X2='B' ; Y2='B') -> P=b
; P=ab
).
genotype(L,X,Y) :- msw(L,X),msw(L,Y).
%%------------------------------------
%% Utility part:
%% (the same as that in bloodABO.psm)
learn_bloodtype(N) :- % Learn parameters from N observations
random_set_seed(214857), % Set seed of the random number generator
gen_bloodtype(N,Gs),!, % Sample bloodtype/1 of size N
learn(Gs). % Perform search and graphical EM learning
% learn. % <= when using the file `bloodtype.dat'
gen_bloodtype(N,Gs) :-
N > 0,
random_select([a,b,o,ab],[0.38,0.22,0.31,0.09],X),
Gs = [bloodtype(X)|Gs1], % Sample a blood type with an empirical
N1 is N-1,!, % ratio for Japanese people.
gen_bloodtype(N1,Gs1).
gen_bloodtype(0,[]).
print_blood :-
prob(bloodtype(a),PA),prob(bloodtype(b),PB),
prob(bloodtype(o),PO),prob(bloodtype(ab),PAB),
nl,
format("P(A) = ~6f~n",[PA]),
format("P(B) = ~6f~n",[PB]),
format("P(O) = ~6f~n",[PO]),
format("P(AB) = ~6f~n",[PAB]).
print_gene :-
get_sw(locus1,[_,['A',a],[GA,Ga]]),
get_sw(locus2,[_,['B',b],[GB,Gb]]),
nl,
format("P(A) = ~6f~n",[GA]),
format("P(a) = ~6f~n",[Ga]),
format("P(B) = ~6f~n",[GB]),
format("P(b) = ~6f~n",[Gb]).