Hello, I am running the following code (part of an ongoing math project):
*e = RootSystem(['E',6]).ambient_space()Roots = e.roots() a1 = vector((1/2, -1/2, -1/2, -1/2, -1/2, -1/2, -1/2, 1/2))a2 = vector((1, 1, 0, 0, 0, 0, 0, 0))a3 = vector((-1, 1, 0, 0, 0, 0, 0, 0))a4 = vector((0, -1, 1, 0, 0, 0, 0, 0))a5 = vector((0, 0, -1, 1, 0, 0, 0, 0))a6 = vector((0, 0, 0, -1, 1, 0, 0, 0))Lini = [a1, a2, a3, a4, a5]def proj(t, ai): myLone = vector(ai) result= t - myLone.dot_product(t) / (ai.dot_product(ai))*myLone return resultmyP = [proj(x, a6) for x in Lini][proja1, proja2, proja3, proja4, proja5] = myPmyR = [vector(v) for v in Roots]myL = [proj(x, a6) for x in myR]len(myL)* * # produce all combinaisons of five vectors with 2 vectors in myL and 3 in myPmyPlist1= Combinations(myP, 3).list()myLlist1= Combinations(myL, 2).list() * *#Alternatively I would have like to use 5 vectors combinaisons in another function of mine but again it seems the error of "too many values" unable me to do so :* *#myLlist= Combinations(myL, 5).list()def letsapply(listg, alist): for listel in listg: myPset1= set(tuple(v) for v in listel) print(myPset1) for listel2 in alist: myLset1= set(tuple(v) for v in listel2) print(myLset1) myLset1= myLset1.union(myPset1) if len(myLset1) >= 5: print(myLset1, "myLset1") if len(myLset1) >= 5: for myLel in myLset1: [proja1, proja2, proja3, proja4, proja5] = myLel print(myLel)print(letsapply(myPlist1,myLlist1))* Is there a way to handle this "capacity" issue? I am quite new to SAGE, I have been using it on cocalc and with Jupyter. Thanks a lot for your help. Best, Sarah Dijols (Math postdoc in Tsinghua University) -- You received this message because you are subscribed to the Google Groups "sage-support" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-support+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/sage-support/3fb8b554-3dfe-43ea-a752-c9dd7ff2aca6n%40googlegroups.com.
