I have the following solution:
solve(...)=[[xa1 == R*a/((a + b)*p1), xa2 == R*b/(a + b), l1 == (a + b)*(R*b/(a + b))^b*(R*a/((a + b)*p1))^a/R]]Im aware I can get the whole set of solutions back with the use of
solve(...)[0]
[xa1 == R*a/((a + b)*p1), xa2 == R*b/(a + b), l1 == (a + b)*(R*b/(a + b))^b*(R*a/((a + b)*p1))^a/R]Im only interested in getting back xa1. Is there a way to do that?
A solution I found to this problem is that you need to do as follows:
solution1=solve(...)[0]
solution1[0]
xa1 == R*a/((a + b)*p1), xa2 == R*b/(a + b)solution1(0][0] i more streamlined. And faster.
You should peruse a Python tutorial ; Python lends itself to a lot of useful idioms,which may be time consuming to discover by yourself.
I cannot recommend enough this marvelous book on Sage, even if it's a bit outdated (reporting on a Python 2-based version of Sage, whereas Sage is Python 3-based : version 9.1 and later support only Python 3).
@EmmanuelCharpentier Thanks for the tip!
Asked: 4 years ago
Seen: 272 times
Last updated: Jul 16 '20
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