[sage-support] Re: substitution in list

Besides the pure python options pointed above, if your list is filled by symbolic expressions, you can also turn it into a vector and do the substitutions straightforwardly. Given x = list(var('x', n=10)) ep = list(var('epsilon', n=10)) a random sublist my_list = [x[randint(0,8)] for _ in

[sage-support] Re: substitution in list

That's a python question. See for instance, https://stackoverflow.com/questions/2582138/finding-and-replacing-elements-in-a-list If you scroll down, there are some suggestions there that deal with multiple replacements as well On Tuesday, 28 December 2021 at 06:09:35 UTC-8 cyrille piatecki

[sage-support] Re: substitution of lambda variable

On Thursday, May 28, 2009 at 9:31:46 PM UTC+5:30, Jason Grout wrote: Mike Hansen wrote: > Hello, > > On Thu, May 28, 2009 at 8:38 AM, Paul Sargent wrote: >> # Subs for lambda (have to use "lambda", but that's a keyword) lambda is a reserved keyword in Python, which

[sage-support] Re: substitution of lambda variable

On Wednesday, September 16, 2015 at 1:54:40 AM UTC-7, Mandeep Singh wrote: > > Can we declare lambda as a variable? I want to use it in eigenvalues > (linear algebra). > like: > l = var("lambda"); > It'll give error. So how can I declare it as a variable. > This works. I don't know if it will

[sage-support] Re: Substitution in infinite polynomial rings

Hi Bruno, I am sorry that I (as author of the InfinitePolynomialRing stuff) did not answer before. On 2014-04-16, BJ brunojo...@gmail.com wrote: The output looks something like this: [-e_8 + e_4^2, -e_10 + e_6*e_4, -1382*e_12 + 2205*e_8*e_4 + 500*e_6^2 - 1323*e_4^3, -10*e_14 + 21*e_10*e_4 +

[sage-support] Re: Substitution in infinite polynomial rings

On Wednesday, April 16, 2014 4:16:30 PM UTC-7, BJ wrote: I have the following code, which produces a list of polynomials in the infinite number of variables e_0, e_1, ... M.e = InfinitePolynomialRing(QQ, implementation=sparse) However, I've been having a lot of trouble figuring out how

[sage-support] Re: Substitution in infinite polynomial rings

On Thursday, April 17, 2014 9:39:09 AM UTC-7, Nils Bruin wrote: but it's flawed: sage: f(e_4=2) KeyError: 'e_4' And also flawed in a different way: sage: f(e_2=e[4]) TypeError: unsupported operand parent(s) for '+': 'Multivariate Polynomial Ring in e_4, e_2, e_1 over Rational Field' and

[sage-support] Re: substitution

On 01/09/2013 09:27 AM, Michael Beeson wrote: sage: K.d,e,p,g,m,f,u,v,j,N = FractionField(PolynomialRing(QQ,10,'depgmfuvjN')) sage: R.s = K[] sage: w=u sage: u=0 sage: w u Why doesn't Sage answer 0 for the value of w here? More generally, if I have some complicated expression and I assign a

Re: [sage-support] Re: substitution

Your first line defines a function field of several variables and assigns values to the identifiers d,p,e, etc so that in particular, u is assigned a value which is the 6th generator of that field. When you set w=u you have a new identifier called w whose value is a copy of u's value, so is also

[sage-support] Re: substitution in sums and multiplications

Hi Burcin, On Jan 25, 4:41 pm, Burcin Erocal bur...@erocal.org wrote: ... In your example, (k1 + k2) is not a subexpression of f, so there is nothing to substitute. In other words, (k1+k2)*2 is automatically turned into 2*k1+2*k2, but (k1+k2)^2 is not turned into k1^2+2*k1*k2+k2^2 ? Why?

[sage-support] Re: Substitution

Thanks. Unfortunately, for my particular example, it didn't work as is -- | Sage Version 4.2, Release Date: 2009-10-24 | | Type notebook() for the GUI, and license() for information.|

[sage-support] Re: Substitution

This simple minded approach seems to work for me: sage: a(x)=2*x sage: b(x)=a(x)^2 sage: g(x)=diff(b(x),x) sage: g(2) 16 -Alasdair On Jan 21, 2:14 am, Stochastix laurent.decreusef...@gmail.com wrote: I have the following problem : I have two nested functions a=lambda x: 2*x b=lambda x:

[sage-support] Re: Substitution

Hi, I think what Minh was trying to say is that these lines: Stochastix wrote: sage: a=lambda x: alpha*x-mu1+mu2 sage: f=lambda x: (a(x)*b-c+sqrt((a(x)*b-c)^2+4.0*a(x)*b*r*mu1))/(2*a (x)*mu1) sage: g=lambda x: (r+l-mu1*f(x))/mu2 sage: prev=lambda x: f(x)/(f(x)+g(x)) sage: k=lambda x:

[sage-support] Re: Substitution

Stochastix wrote: I have the following problem : I have two nested functions a=lambda x: 2*x b=lambda x: a(x)^2 Everything goes well as long as I want to compute the derivative of b but how can I evaluate this derivative at x=2. I tried g=lambda x: diff(b(x),x) g(x) returns 8*x as expected

[sage-support] Re: substitution

On Wed, 7 Oct 2009 13:20:30 -0700 (PDT) ma...@mendelu.cz ma...@mendelu.cz wrote: BTW: When looked at this problem, the notation D[0](alpha)(r)^2 seems to be unusual to me. Is it possible to fix sage so that it prints derivatives of functions in one variable as usual: alpha'(r) ? There is

[sage-support] Re: substitution

Sorry I should make my question clearer. I'm not trying to substitute for sigma^2. What I want to do is replace the various constants multiplying the terms in sigma with a single constant. For example I get the following term in sigma^2: -1/2*(1/R^2 - beta_0/(R^2*beta))*sigma(r)^2 And I want

[sage-support] Re: substitution

perhaps, no subst is necessary: var('r beta beta_0 R a h') psi=function('psi',r) sigma=function('sigma',r) H_0=function('H_0',r) H_grad = lambda psi : (1/2)*(beta/beta_0)*R^2*(psi.diff(r))^2 H_0(psi) = -(1/2)*(beta/beta_0-1)*psi^2 + beta*a*psi^4 + beta*h*psi

[sage-support] Re: substitution

Ranjit a écrit : I'm trying to figure out how to do a simple substitution. I've the following code: var('r beta beta_0 R a h') psi=function('psi',r) sigma=function('sigma',r) H_0=function('H_0',r) H_grad = lambda psi : (1/2)*(beta/beta_0)*R^2*(psi.diff(r))^2 H_0(psi) =

[sage-support] Re: substitution

This works, though it would have been nicer if epsilon actually equalled what it was replacing. I just tried (-1/2*(1/R^2 - beta_0/(R^2*beta))*sigma(r)^2).subs((1/R^2 - beta_0/ (R^2*beta))==epsilon) which works, where (-1/2*(1/R^2 - beta_0/(R^2*beta))*sigma(r)^2).subs((1/2)*(1/R^2 -

[sage-support] Re: substitution

On 8 říj, 19:41, Ranjit rjcha...@gmail.com wrote: does not. So I guess there's an upper limit to the number of operations that the expression being replaced can have to work in subs (). Perhaps true, but IMHO very risky guess which is not true, if pattern matching in Maxima is behind the

[sage-support] Re: substitution

On 7 říj, 16:21, Ranjit rjcha...@gmail.com wrote: I'm trying to figure out how to do a simple substitution. I've the following code: var('r beta beta_0 R a h') psi=function('psi',r) sigma=function('sigma',r) H_0=function('H_0',r) H_grad = lambda psi :

[sage-support] Re: substitution of lambda variable

Mike Hansen wrote: Hello, On Thu, May 28, 2009 at 8:38 AM, Paul Sargent psa...@gmail.com wrote: # Subs for lambda (have to use lambda, but that's a keyword) sage: e1.subs(lambda = 3) File ipython console, line 1

[sage-support] Re: Substitution with symbolic equations

Not sure if this is what you are after, but the following would give you the solution: sage: solve([e1,e2],c,a) [[c == (d - b*e)/e, a == d/e]] You can give n equations to solve and solve for n variables. Solve will insert one into another automatically. An equation has a different syntax to a

[sage-support] Re: Substitution with symbolic equations

Hello, On Fri, May 15, 2009 at 5:06 AM, Paul Sargent psa...@gmail.com wrote: Lets give ourselves two symbolic equations: sage: var(a b c d e) sage: e1 = a == b + c sage: e2 = d == e * a Now, lets say I want to know what c is in terms of b, d e. By hand I'd substitute e1 in e2, and then

[sage-support] Re: Substitution with symbolic equations

On 15 May 2009, at 16:33, Mike Hansen wrote: In Sage 4.0 which will be released within the week, you'll be able to do the following: sage: var(a b c d e) (a, b, c, d, e) sage: e1 = a == b + c sage: e2 = d == e * a sage: e3 = e2.subs(e1); e3 d == (b + c)*e Well, that's what I call

[sage-support] Re: substitution problem

On Mar 30, 6:07 pm, Jason Grout jason-s...@creativetrax.com wrote: V wrote: Hi, I'm fairly new to sage with some background in maxima. My workbook is shared at http://www.sagenb.org/home/pub/410/ the last three lines show the error I get. Basically, I derive an equilibrium

[sage-support] Re: substitution problem

V wrote: On Mar 30, 6:07 pm, Jason Grout jason-s...@creativetrax.com wrote: V wrote: Hi, I'm fairly new to sage with some background in maxima. My workbook is shared at http://www.sagenb.org/home/pub/410/ the last three lines show the error I get. Basically, I derive an equilibrium

[sage-support] Re: substitution problem

This is exactly what I wanted. Thanks for your help! :) I'll look up the maxima manuals to see if I can force it to think harder. Thanks anyway! Have a nice day! V On Mar 31, 3:35 pm, Jason Grout jason-s...@creativetrax.com wrote: V wrote: On Mar 30, 6:07 pm, Jason Grout

[sage-support] Re: substitution problem

V wrote: Hi, I'm fairly new to sage with some background in maxima. My workbook is shared at http://www.sagenb.org/home/pub/410/ the last three lines show the error I get. Basically, I derive an equilibrium condition that I would like to use in the previous stage of my game

[sage-support] Re: Substitution

Hi, On Fri, Mar 13, 2009 at 7:09 PM, hpon peter.norli...@gmail.com wrote: Hi, What is the easiest way to make a mathematical substitution in Sage? For example: We have eqn1 = F == a + b I'm not sure I understand what you are trying to do in this line. If you just want to define F as

[sage-support] Re: Substitution

Excellent. Thank you Alex! /hpon On 13 Mar, 09:16, Alex Ghitza aghi...@gmail.com wrote: Hi, On Fri, Mar 13, 2009 at 7:09 PM, hpon peter.norli...@gmail.com wrote: Hi, What is the easiest way to make a mathematical substitution in Sage? For example:  We have eqn1 = F == a + b

[sage-support] Re: substitution in and integration of piecewise functions

One of the issues right now is that sage's piecewise is a completely separate class then the rest of the calculus library. I think it should descend from symbolic expression and be on the same level as, e.g. sin and addition (and probably even have a pynac counterpart). Letting the operands

[sage-support] Re: substitution in and integration of piecewise functions

On Nov 24, 2008, at 5:04 AM, Ondrej Certik wrote: On Mon, Nov 24, 2008 at 1:25 PM, David Joyner [EMAIL PROTECTED] wrote: On Mon, Nov 24, 2008 at 6:19 AM, Ondrej Certik [EMAIL PROTECTED] wrote: Hi, when I use regular expressions, I can use .subs(): sage: e = x+y sage: e.subs(x=y)

[sage-support] Re: substitution in and integration of piecewise functions

On Tue, Nov 25, 2008 at 3:15 AM, Ondrej Certik [EMAIL PROTECTED] wrote: One of the issues right now is that sage's piecewise is a completely separate class then the rest of the calculus library. I think it Just for the record, this is because Piecewise was implemented long long ago by David

[sage-support] Re: substitution in and integration of piecewise functions

On Tue, Nov 25, 2008 at 3:13 PM, William Stein [EMAIL PROTECTED] wrote: On Tue, Nov 25, 2008 at 3:15 AM, Ondrej Certik [EMAIL PROTECTED] wrote: One of the issues right now is that sage's piecewise is a completely separate class then the rest of the calculus library. I think it Just for the

[sage-support] Re: substitution in and integration of piecewise functions

On Tue, 25 Nov 2008 12:15:41 +0100 Ondrej Certik [EMAIL PROTECTED] wrote: One of the issues right now is that sage's piecewise is a completely separate class then the rest of the calculus library. I think it should descend from symbolic expression and be on the same level as, e.g. sin

[sage-support] Re: substitution in and integration of piecewise functions

On Tue, Nov 25, 2008 at 11:00 PM, Burcin Erocal [EMAIL PROTECTED] wrote: On Tue, 25 Nov 2008 12:15:41 +0100 Ondrej Certik [EMAIL PROTECTED] wrote: One of the issues right now is that sage's piecewise is a completely separate class then the rest of the calculus library. I think it

[sage-support] Re: substitution in and integration of piecewise functions

On Mon, 24 Nov 2008 12:19:55 +0100 Ondrej Certik [EMAIL PROTECTED] wrote: Hi, when I use regular expressions, I can use .subs(): sage: e = x+y sage: e.subs(x=y) 2*y but not with Piecewise: sage: var(h H x y) (h, H, x, y) sage: u = Piecewise([((0, h), x/h), ((h, H), 1)]) sage:

[sage-support] Re: substitution in and integration of piecewise functions

On Mon, Nov 24, 2008 at 6:19 AM, Ondrej Certik [EMAIL PROTECTED] wrote: Hi, when I use regular expressions, I can use .subs(): sage: e = x+y sage: e.subs(x=y) 2*y but not with Piecewise: sage: var(h H x y) (h, H, x, y) sage: u = Piecewise([((0, h), x/h), ((h, H), 1)]) I don't think

[sage-support] Re: substitution in and integration of piecewise functions

On Mon, Nov 24, 2008 at 7:25 AM, David Joyner [EMAIL PROTECTED] wrote: On Mon, Nov 24, 2008 at 6:19 AM, Ondrej Certik [EMAIL PROTECTED] wrote: Hi, when I use regular expressions, I can use .subs(): sage: e = x+y sage: e.subs(x=y) 2*y but not with Piecewise: sage: var(h H x y) (h, H,

[sage-support] Re: substitution in and integration of piecewise functions

On Mon, Nov 24, 2008 at 1:25 PM, David Joyner [EMAIL PROTECTED] wrote: On Mon, Nov 24, 2008 at 6:19 AM, Ondrej Certik [EMAIL PROTECTED] wrote: Hi, when I use regular expressions, I can use .subs(): sage: e = x+y sage: e.subs(x=y) 2*y but not with Piecewise: sage: var(h H x y) (h, H,

[sage-support] Re: substitution in and integration of piecewise functions

On Mon, Nov 24, 2008 at 2:15 PM, David Joyner [EMAIL PROTECTED] wrote: On Mon, Nov 24, 2008 at 7:25 AM, David Joyner [EMAIL PROTECTED] wrote: On Mon, Nov 24, 2008 at 6:19 AM, Ondrej Certik [EMAIL PROTECTED] wrote: Hi, when I use regular expressions, I can use .subs(): sage: e = x+y sage:

[sage-support] Re: substitution in and integration of piecewise functions

On Mon, 24 Nov 2008 14:04:53 +0100 Ondrej Certik [EMAIL PROTECTED] wrote: On Mon, Nov 24, 2008 at 1:25 PM, David Joyner [EMAIL PROTECTED] wrote: On Mon, Nov 24, 2008 at 6:19 AM, Ondrej Certik [EMAIL PROTECTED] wrote: Hi, when I use regular expressions, I can use .subs(): sage:

[sage-support] Re: substitution in and integration of piecewise functions

On Mon, Nov 24, 2008 at 3:08 PM, Burcin Erocal [EMAIL PROTECTED] wrote: On Mon, 24 Nov 2008 14:04:53 +0100 Ondrej Certik [EMAIL PROTECTED] wrote: On Mon, Nov 24, 2008 at 1:25 PM, David Joyner [EMAIL PROTECTED] wrote: On Mon, Nov 24, 2008 at 6:19 AM, Ondrej Certik [EMAIL PROTECTED]

[sage-support] Re: substitution in a matrix (newby)

William Stein [EMAIL PROTECTED] writes: On 7/26/07, Roger Mason [EMAIL PROTECTED] wrote: You should do m.substitute(a=1) just like you're doing -- unfortunately, nobody implemented that yet, and it's doing some dumb generic behavior. [...] Fortunately, I just implemented this for

[sage-support] Re: substitution in a matrix (newby)

David William, David Joyner [EMAIL PROTECTED] writes: I would use the lambda notation: sage: m = lambda x: matrix([[x[0],x[1]],[x[2],x[3]]]) sage: m([1,2,3,4]) [1 2] [3 4] sage: m([1,2,3,a]) [1 2] [3 a] sage: m([a,b,3,4]) [a b] [3 4] There might be better ways though.

[sage-support] Re: substitution in a matrix (newby)

On 7/26/07, Roger Mason [EMAIL PROTECTED] wrote: You should do m.substitute(a=1) just like you're doing -- unfortunately, nobody implemented that yet, and it's doing some dumb generic behavior. [...] Fortunately, I just implemented this for SAGE-2.7.1, which I'll release very very soon.

[sage-support] Re: substitution in a matrix (newby)

I would use the lambda notation: sage: m = lambda x: matrix([[x[0],x[1]],[x[2],x[3]]]) sage: m([1,2,3,4]) [1 2] [3 4] sage: m([1,2,3,a]) [1 2] [3 a] sage: m([a,b,3,4]) [a b] [3 4] There might be better ways though. On 7/24/07, Roger Mason [EMAIL