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Substituting multiple values

asked 2011-08-25 03:09:33 -0600

Sagenoob gravatar image

updated 2011-08-25 03:46:29 -0600

What is the best way of substituting a list (or vector or whatever) of values into an expression? For example, suppose I have

z = var('x y')
zvals = (1, 2)
w = x^2 + y^2

and want to substitute zvals for z in the expression w. I have tried the following commands:

w.subs(z=zvals)                        #doesn't work
w.subs({z:zvals})                      #doesn't work
w.subs(x=1,y=2)                        #fine, but cumbersome if z has many elements
w.subs(z[0]=zvals[0],z[1]=zvals[1])    #doesn't work
w.subs({z[0]:zvals[0],z[1]:zvals[1]})  #fine, and could turn this into a loop 
                                       #if there are many variables, but ugly
w.subs(dict(zip(z,zvals)))             #best I can come up with

As far as I can see, none of this behaviour changes if z and zvals are vectors or lists instead of tuples.

Is there a simpler way of doing this? Also, why doesn't the fourth attempt work when the fifth one does - is this down to a limitation of Python?

EDIT: I also realised that you can do w.subs(z[0]==zvals[0]) but w.subs(z[0]==zvals[0],z[1]==zvals[1]) won't work - why is this?

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answered 2011-08-25 04:00:10 -0600

Jason Grout gravatar image

I would do it this way:

sage: z = var('x y')
sage: zvals = (1, 2)
sage: w(x,y)=x^2+y^2

Note that w explicitly lists the order of arguments...

sage: w(*zvals)

Python documentation of the *mylist syntax is here.

Here is another way:

sage: w(1,2)
sage: w(x=zvals[0], y=zvals[1])
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Thanks - I had forgotten about the * syntax.

Sagenoob gravatar imageSagenoob ( 2011-08-25 09:16:14 -0600 )edit

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Asked: 2011-08-25 03:09:33 -0600

Seen: 1,908 times

Last updated: Aug 25 '11