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Kernel: Julia 1.0

Julia 1.0 Kernel in CoCalc

Here is the other common building block for programming: Evaluating a block of code more than once, e.g. by either counting the iterations or until a condition is met (or no longer met). This is one of the most confusing parts of programming. So don't be shy to take your time inspecting this. Try to evaluate this little program in your head by reading the lines over and over again...

using Printf

s = 0
for i = [1 2 5 100 -1 5]
    s = s + i
    @printf("i = %4d  →  s = %4d\n", i, s)
end
i = 1 → s = 1 i = 2 → s = 3 i = 5 → s = 8 i = 100 → s = 108 i = -1 → s = 107 i = 5 → s = 112 
@printf("%.2f", pi)
3.14
1+1+1+23
26
VERSION
v"1.0.5"
ENV["JULIA_DEPOT_PATH"]
"/home/user/.julia:/ext/julia/depot/"
Pkg.installed()

using Pkg
for (k, v) in Pkg.installed()
    println(k, ":::", (if nothing == v "N/A" else v end))
end

using Printf

s = 0
for i = [1 2 5 100 -1 5]
    s = s + i
    @printf("i = %4d  →  s = %4d\n", i, s)
end

[sin(3.14), sin(3.141), sin(3.142)]

println("Hello", 99)
x = 10
println("Interpolation $(5 + x)")
@printf("pi = %.7f\n", float(pi))

Printf.@printf("%f %F %f %F\n", Inf, Inf, NaN, NaN)

using CSV

using DataFrames

#using Gadfly

using Nemo

using Statistics


Statistics.median([8 9 8 6 87 6 7 6 5.1 4 5 4 3 4 3 3 3 3 ])
5.05
using LinearAlgebra

m1 = [  1 2 -3
        3 -1 1
        1.0 1 1]

q1, r1 = LinearAlgebra.qr(m1)
LinearAlgebra.QRCompactWY{Float64,Array{Float64,2}} Q factor: 3×3 LinearAlgebra.QRCompactWYQ{Float64,Array{Float64,2}}: -0.301511 0.816497 -0.492366 -0.904534 -0.408248 -0.123091 -0.301511 0.408248 0.86164 R factor: 3×3 Array{Float64,2}: -3.31662 2.22045e-16 -0.301511 0.0 2.44949 -2.44949 0.0 0.0 2.21565 
q1 * r1
3×3 Array{Float64,2}: 1.0 2.0 -3.0 3.0 -1.0 1.0 1.0 1.0 1.0
using DifferentialEquations
α=1
β=1
u₀=1/2
f(t,u) = α*u
g(t,u) = β*u
dt = 1//2^(4)
tspan = (0.0,1.0)
prob = SDEProblem(f,g,u₀,(0.0,1.0))
sol = solve(prob,EM(),dt=dt)
using Plots
plot(sol)

using DifferentialEquations

f(x) = sin(2π.*x[:,1]).*cos(2π.*x[:,2])
gD(x) = sin(2π.*x[:,1]).*cos(2π.*x[:,2])/(8π*π)

dx = 1//2^(5)
mesh = notime_squaremesh([0 1 0 1],dx,:dirichlet)
prob = PoissonProblem(f,mesh,gD=gD)

sol = solve(prob)

using Plots
plot(sol)

using GLM

using PyPlot
x = range(0, stop = 4*pi, length=1000)
y = sin.(3*x + 1.5*cos.(2*x))
plot(x, y, color="red", linewidth=2.0, linestyle="--")
┌ Info: Precompiling PyPlot [d330b81b-6aea-500a-939a-2ce795aea3ee] └ @ Base loading.jl:1192
Image in a Jupyter notebook
1-element Array{PyCall.PyObject,1}: PyObject <matplotlib.lines.Line2D object at 0x7f857bae0c50>
using PyPlot
x = range(0; stop=2*pi, length=1000); y = sin.(3 * x + 4 * cos.(2 * x));
plot(x, y, color="red", linewidth=2.0, linestyle="--")
title("A sinusoidally modulated sinusoid")
Image in a Jupyter notebook
PyObject Text(0.5, 1, 'A sinusoidally modulated sinusoid')
using D4M

row = "a,a,a,a,a,a,a,aa,aaa,b,bb,bbb,a,aa,aaa,b,bb,bbb,"
column = "a,aa,aaa,b,bb,bbb,a,a,a,a,a,a,a,aa,aaa,b,bb,bbb,"
values = "a-a,a-aa,a-aaa,a-b,a-bb,a-bbb,a-a,aa-a,aaa-a,b-a,bb-a,bbb-a,a-a,aa-aa,aaa-aaa,b-b,bb-bb,bbb-bbb,"

A = Assoc(row,column,values)

Ar = A["a,b,",:]

Ac = A[:,"a,b,"]

Av = A > "b,"

SymPy test

using SymPy
x = symbols("x")      # or   @vars x, Sym("x"), or  Sym(:x)
y = sin(pi*x)
y(1), y(2.2), y(123456)

Unicode Plots

using Plots
unicodeplots()

# /Users/tom/.julia/v0.4/Plots/docs/example_generation.jl, line 50:
plot(sin,x-> sin(1.5x), 0, 4π, line=1, leg=false, fill=(0,:orange))