Pierre GM schrieb:
> All,
> I have a (m_1+1,m_2+1) matrix of 0 and 1, and I need to compute a second 
> matrix U by recurrence.
> A double-loop as such
> #............................
> U = empty((m_1+1, m_2+1))
> U[:,0] = C[:,0]
> U[0,:] = C[0,:]
> for i in range(1,m_1+1):
>     for j in range(1,m_2+1):
>         U[i,j] = C[i,j] * (U[i,j-1] + U[i-1,j])
> #............................
> does the trick, but I suspect there is not a simpler, more elegant solution. 
> Would anybody have any idea ?
> Thanks a lot in advance, and all my apologies for being a tad lazy...
> 

Going from two to one loop at least seems relatively easy; what about
something like (untested, especially all the parentheses...):

for i in range (1,m_1+1):
    U[i,1:] = dot(C[i,1:], vstack((U[i,:-1],U[i-1,1:])).sum(axis=0))

Actually, now that I see the structure more clearly, what about

tempU1 = U[:-1,1:] # upper right block (w/o last row and first col)
tempU2 = U[1:,:-1] # lower left (w/9 first row and last col)
U[1:,1:] = dot(C[1:,1:], tempU1+tempU2) # hopefully does it elementwise?

> (PS: FYI, it's for a reimplementation of the Kolmogorov-Smirnov test 
> statistic, where ties occur)

Glad that there are other applied statisticians (maybe even
econometrician?) around working with plain old 2d matrices... Would you
mind posting the resulting code?

cheers,
sven

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