[julia-users] Re: How to write data to new line in outfile.txt

Hello Use data frame that is much better than txt file. Em quinta-feira, 13 de outubro de 2016 08:08:53 UTC-3, program...@gmail.com escreveu: > > How to write data to new line in outfile.txt file, in this script data i > still writing in the same line > > io=open("outfile.txt", "w" > o=open("d

[julia-users] Problem using FastAnonymous.jl in Julia 0.5

using FastAnonymous INFO: Precompiling module FastAnonymous. WARNING: symbol is deprecated, use Symbol instead. in depwarn(::String, ::Symbol) at ./deprecated.jl:64 in symbol(::String, ::Vararg{String,N}) at ./deprecated.jl:30 in include_from_node1(::String) at ./loading.jl:488 (repeats 2 tim

Re: [julia-users] Problem using FastAnonymous.jl in Julia 0.5

Hello Tom In fact, AF already exist in julia 0.5. Thanks for this ;) Em domingo, 13 de novembro de 2016 16:59:46 UTC-2, Tom Breloff escreveu: > > Don't use it? Anonymous functions are fast in 0.5 now. > > On Sunday, November 13, 2016, > > wrote: > >> using FastAnonymous >> >> INFO: Precompiling

[julia-users] (Lagrange Poly )simple doubt, but...

Lagrange Poly 1 function lg(X,Y,Xint) 2 3 n = length(X) 4 L = ones(1:n) 5 6 for k = collect(1:n) 7 8 9 for c = collect(1:n) 10 if c!=k 11 * L[k] = (L[k]*( Xint - X[c] ))/( X[k] - X[c] ) # problem* 12 end 13 end 14

Re: [julia-users] (Lagrange Poly )simple doubt, but...

now it's working! thanks Yu! It was the L vector Em sexta-feira, 27 de maio de 2016 22:06:46 UTC-3, Yichao Yu escreveu: > > On Fri, May 27, 2016 at 8:56 PM, > > wrote: > > Lagrange Poly > > > > 1 function lg(X,Y,Xint) > > 2 > > 3 n = length(X) > > 4 L = ones(1:n) > > You probably wa

[julia-users] why the numerical result is different (RK45 julia and matlab)?

this is the test for equation differential using runge-kutta45: f(x,y)= (-5*x - y/5)^1/8 + 10 why the numerical result is different? I used : function Rk_JL() f(x,y)=

[julia-users] Re: why the numerical result is different (RK45 julia and matlab)?

Hi daniel in fact, there is a relationship with the tolerance between the methods. Em sábado, 11 de junho de 2016 11:51:53 UTC-3, Daniel Carrera escreveu: > > Probably because the Julia module has different default values for the > absolute and relative tolerance. For Julia, the defaults are rel

Re: [julia-users] why the numerical result is different (RK45 julia and matlab)?

Hi erick in fact erick, there is an error!hehe. I did not realize before. Em sexta-feira, 10 de junho de 2016 23:38:11 UTC-3, Erik Schnetter escreveu: > > When you write `(...)^1/8`, you probably mean `(...)^(1/8)` instead. > > -erik > > On Fri, Jun 10, 2016 at 10:29 PM, > > wrote: > >> this is t

Re: [julia-users] Re: why the numerical result is different (RK45 julia and matlab)?

Hi Daniel I did the test and the difference decreased. interesting, I had not noticed. Em domingo, 12 de junho de 2016 09:53:40 UTC-3, Daniel Carrera escreveu: > > I don't know what you are trying to say, but did you try running the two > programs with the same relative and absolute tolerances?

[julia-users] Tutorial Julia language brazilian portuguese

Hello everyone I am writing a tutorial Julia language Portuguese (over 200 pages): https://github.com/jmarcellopereira/juliatutorialbr

[julia-users] Re: Tutorial Julia language brazilian portuguese

participantes aqui no forum. obrigado pela boa recepção da ideia. ;) Em sexta-feira, 24 de junho de 2016 20:57:25 UTC-3, jmarcell...@ufpi.edu.br escreveu: > > Hello everyone > > I am writing a tutorial Julia language Portuguese (over 200 pages): > > https://github.com/jmarcellopereira/juliatutorialbr >

[julia-users] Error clone juliabox

When I tried to clone a project , this error occurs : Server error: Traceback (most recent call last): File "/usr/local/lib/python2.7/dist-packages/IPython/html/base/handlers.py", line 441, in wrapper result = yield gen.maybe_future(method(self, *args, **kwargs)) File "/usr/local/lib/python2.7/

[julia-users] Re: Error clone juliabox

I solved thus: no rename your notebooks with ~,`, ç , ^, ' ;) Em sábado, 16 de julho de 2016 12:24:35 UTC-3, jmarcell...@ufpi.edu.br escreveu: > > When I tried to clone a project , this error occurs : > > Server error: Traceback (most recent call last): File > "/usr/local/lib/python2.7/dist-pa

[julia-users] mapreduce two variable (x,y)

how to write mapreduce for two variable? Ex: mapreduce((x,y)-> 2*y.^2 - 1 + x, + ,1:3,1:3) LoadError: wrong number of arguments while loading In[182], in expression starting on line 1 in anonymous at ./In[182]:1

Re: [julia-users] mapreduce two variable (x,y)

;) ok, thanks Schnetter Em segunda-feira, 22 de agosto de 2016 20:39:11 UTC-3, Erik Schnetter escreveu: > > This function could be added... > > Currently, you'll have to `zip` the two collections, and unzip them in > your function: > > ```Julia > mapreduce(a -> begin x,y=a; 2*y^2 - 1 + x end, +,

[julia-users] Re: Tutorial Julia language brazilian portuguese

olá felipe. É uma boa ideia. Se você criar eu compartilho com o pessoal da unb. Hello Felipe. It's a great idea. If you marry I share with the staff of unb. Em segunda-feira, 12 de setembro de 2016 17:54:50 UTC-3, Phelipe Wesley escreveu: > > O que acham de criarmos um grupo Julia-Brasil no sla

[julia-users] CurveFit Parckage. Why the code below does not work?

Code x = [0.0 0.2 0.4 1.0 1.6 1.8 2.0 2.6 2.8 3.0 3.8 4.8 5.0 5.2 6.0 6.2 7.4 7.6 7.8 8.6 8.8 9.0 9.2 9.4 10.0 10.6 10.8 11.2 11.6 11.8 12.2 12.4]; y = [-0.183 -0.131 0.027 0.3 0.579 0.853 0.935 1.133 1.269 1.102 1.092 1.143 0.811 0.91 0.417 0.46 -0.516 -0.334 -0.504 -0.946 -0.916 -0.975 -1.

[julia-users] CurveFit Package. why the code below does not work?

Code: x = [0.0 0.2 0.4 1.0 1.6 1.8 2.0 2.6 2.8 3.0 3.8 4.8 5.0 5.2 6.0 6.2 7.4 7.6 7.8 8.6 8.8 9.0 9.2 9.4 10.0 10.6 10.8 11.2 11.6 11.8 12.2 12.4]; y = [-0.183 -0.131 0.027 0.3 0.579 0.853 0.935 1.133 1.269 1.102 1.092 1.143 0.811 0.91 0.417 0.46 -0.516 -0.334 -0.504 -0.946 -0.916 -0.975 -1

[julia-users] How to data Geom.smooth Gadfly

Geom.smooth create an fitting line on the points. How to get the data from this line? ex x = [0.0,0.2,0.4,0.8,1.0,1.2,1.4,1.6,2.0,2.2,2.6,2.8,3.0,3.4] y = [-0.18344,-0.1311,0.02689,0.11053,0.25394,0.25719,0.53189,0.57905,0.93518,0.9166,1.13329,1.26893,1.10203,1.13392] using Gadfly Gadfly.plot(

[julia-users] I wish someone would publish a non-linear fitting code

I wish someone would publish a non-linear fitting code using this data, how we can use CurveFit? x = [0.0 0.2 0.4 1.0 1.6 1.8 2.0 2.6 2.8 3.0 3.8 4.8 5.0 5.2 6.0 6.2 7.4 7.6 7.8 8.6 8.8 9.0 9.2 9.4 10.0 10.6 10.8 11.2 11.6 11.8 12.2 12.4]; y = [-0.183 -0.131 0.027 0.3 0.579 0.853 0.935 1.133 1.

[julia-users] CurveFit Pakage: non-linear fitting code

I wish someone would publish a nonlinear fitting code using package CurveFit using this data, how we can use CurveFit? x = [0.0 0.2 0.4 1.0 1.6 1.8 2.0 2.6 2.8 3.0 3.8 4.8 5.0 5.2 6.0 6.2 7.4 7.6 7.8 8.6 8.8 9.0 9.2 9.4 10.0 10.6 10.8 11.2 11.6 11.8 12.2 12.4]; y = [-0.183 -0.131 0.027 0.3 0.579

Re: [julia-users] CurveFit Pakage: non-linear fitting code

hello tshort I published this post on the "stack". the solution presented has errors in the coefficients. Em sábado, 23 de janeiro de 2016 14:08:07 UTC-2, tshort escreveu: > > One link: > > > http://stackoverflow.com/questions/34840875/julia-using-curvefit-package-non-linear > On Jan 23, 2016

[julia-users] Solve ode with initial conditions

I want to solve an ode with initial conditions. For example using SymPy f = SymFunction("f") x,y = Sym("x,y") edo1 = SymPy.diff(f(x),x) + x SymPy.dsolve(edo1) f(x) == C1 - x^2/2 How to solve C1?

[julia-users] Re: Solve ode with initial conditions

Hi j verzani SymPy documentation has the "ics", but does not show example: dsolve(*eq*, *func=None*, *hint='default'*, *simplify=True*, *ics=None*, *xi=None*, *eta=None*, *x0=0*, *n=6*, ***kwargs*) any tips? Em quinta-feira, 4 de fevereiro de 2016 02:01:20 UTC-2, j verzani escreveu: > > There

[julia-users] Re: Solve ode with initial conditions

Hi j in fact, the solution is not easy. I got your tip and made this way: C1 = Sym("C1") # eq1 na forma: substituir x por 0 na parte direita da sol_simb # e subtrair a expressao pelo valor de y0 eq1 = subs(rhs(sol_simb),x=>0) - 1 SymPy.solve(eq1,C1) Em quinta-feira, 4 de fevereiro de 2016 1

[julia-users] Re: Solve ode with initial conditions

Hi j very good,thanks! very practical. For a second order ode, as would be? Em sexta-feira, 5 de fevereiro de 2016 01:50:34 UTC-2, j verzani escreveu: > > It can be a bit cleaner to just get the constant if you substitute as the > following: > > eqn = diff(u(x),x) - 2u(x) # for example > out =

[julia-users] Re: Solve ode with initial conditions

Very good. I hope looking forward to the new version. Idea Code: using SymPy y = SymFunction("y") x = Sym("x") edo2 = SymPy.diff(y(x),x,x) - 4*y(x)-x sol_simb_ord2 = SymPy.dsolve(edo2) C1,C2 = Sym("C1,C2") eq1 = sol_simb_ord2(y(x) => 1, x=> 0) eq2 = SymPy.diff((sol_simb_ord2),x)(diff(y(

Re: [julia-users] Re: Non-uniform interpolation in 1D

I did not understand how to use the "A". "A" is a vector tuple ...? Em domingo, 28 de fevereiro de 2016 18:32:23 UTC-3, Tomas Lycken escreveu: > > Gridded here is the *interpolation scheme*, as opposed to (implicitly > uniform) BSpline interpolation; it’s simply the way Interpolations.jl > let

Re: [julia-users] Re: Non-uniform interpolation in 1D

Hello Tim, good morning In this case, where we have discrete points (P_NOM = [1.5, 2.2, 3.7, 5.6, 7.5, 11.2, 14.9] and ETA = [93, 94, 94, 95, 95, 95.5, 95.5 ]), as would be "A" for P_NOM(2.8) value ? Since function f (x) is more intuitive. Em quarta-feira, 9 de março de 2016 23:22:54 UTC-3, Tim