*上面的程序可以大幅减化（首数是样本量，末数是可能的秩和，求给定首数时末数对应的概率）
clear
cap pr drop rankext
pr rankext
args n e
if wordcount("`*'")!=2|int(`e')!=`e'|int(`n')!=`n'|`e'<0|`n'<0|`e'>`n'*(`n'+1)/2 {
n di as error "Two and only two nonnegative integers, of which"
n di as error "the last is smaller than the accumlated sum by the first, are allowed"
exit
}
qui{
clear
set ob `=2^`n''
forv i=1/`n'{
g v`i'=`i'*mod(int((_n-1)/2^(`n'-`i')),2)
}
egen exp=rowtotal(v*)
collapse (count)t=v1,by(exp)
egen p=pc(t),prop
su p if exp<=`e'
n di
n di in yellow "exact P-value for the ranksum of `e' given the sample size of `n' = " r(sum)
clear
}
end
clear
cap pr drop rankext
pr rankext
args e n
if wordcount("`*'")!=2|int(`e')!=`e'|int(`n')!=`n'|`e'<0|`n'<0|`e'>`n' {
n di in red "Two and only two nonnegative integers, of which the 2nd is larger, are allowed"
exit
}
clear
qui{
set ob `=2^`n''
forv i=1/`n'{
g v`i'=0
replace v`i'=`i' if mod(int((_n-1)/2^(`n'-`i')),2)
}
egen exp=rowtotal(v*)
collapse (count)t=v1,by(exp)
egen p=pc(t),prop
keep if exp==`e'
n di
n di in yellow "exact P Value for `e'/`n' = " p
}
clear
end
谢谢！
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