Passing two arrays in a function - julia

I am trying to work with 2 arrays in a function with a for loop but I do not know the syntax, since there are no examples in the internet. I have written the code without function but I would certainly appreciate if this was transformed to a function. The output is the array price and the array succrate. Thank you in advance.
length_of_arrays = 101
lower_limit = 0
steps_per_unit = 1
price1 = 10
succrate1 = 5
succrate2 = 7
price = Array{Float64, 1}(101)
succrate = Array{Float64, 1}(101)
for pr_A = 1:1:length_of_arrays
price[pr_A] = lower_limit + ((pr_A-1) / steps_per_unit)
if price[pr_A] == price1
succrate[pr_A] = succrate1
else
succrate[pr_A] = succrate2
end
end

There is no special syntax for passing in arrays. Just pass them as an argument like anything else. The arrays will be modified in the function (to mark this I have followed a convention of using a final ! in the name) so there's no need to return them from the function.
function calc_prices!(price, succrate, lower_limit, steps_per_unit, price1, succrate1, succrate2)
for pr_A in eachindex(price)
price[pr_A] = lower_limit + ((pr_A-1) / steps_per_unit)
if price[pr_A] == price1
succrate[pr_A] = succrate1
else
succrate[pr_A] = succrate2
end
end
end
len_of_arrays = 101
pri = Array{Float64, 1}(len_of_arrays)
succ = Array{Float64, 1}(len_of_arrays)
calc_prices!(pri, succ, 0, 1, 10, 5, 7)
println(pri[10])
println(succ[10])

I used for i in 1:length_of_array, since eachindex in nested for loops produced a bounds error
function modifyarrays!(length_of_arrays, price, lower_limit, steps_per_unit, succrate)
for pr_A in 1:101
price[pr_A] = lower_limit + ((pr_A-1) / steps_per_unit)
for d in 1:20
if price[pr_A] == price1
succrate[pr_A, d] = succrate1
else
succrate[pr_A, d] = succrate2
end
end
end
end
length_of_arrays = 101
lower_limit = 0
steps_per_unit = 1
price1 = 10
succrate1 = 5
succrate2 = 7
price = Array{Float64, 1}(101)
succrate = Array{Float64, 2}(101,20)
modifyarrays!(101, price, 0, 1, succrate)
println(succrate[11, 2])
I would be happy to get an answer on how eachindex could have worked or about cartesian index.
In addition, how do I call succrate from the modifyarrays! function into another function?

Related

Continued fractions and Pell's equation - numerical issues

Mathematical background
Continued fractions are a way to represent numbers (rational or not), with a basic recursion formula to calculate it. Given a number r, we define r[0]=r and have:
for n in range(0..N):
a[n] = floor(r[n])
if r[n] == [an]: break
r[n+1] = 1 / (r[n]-a[n])
where a is the final representation. We can also define a series of convergents by
h[-2,-1] = [0, 1]
k[-2, -1] = [1, 0]
h[n] = a[n]*h[n-1]+h[n-2]
k[n] = a[n]*k[n-1]+k[n-2]
where h[n]/k[n] converge to r.
Pell's equation is a problem of the form x^2-D*y^2=1 where all numbers are integers and D is not a perfect square in our case. A solution for a given D that minimizes x is given by continued fractions. Basically, for the above equation, it is guaranteed that this (fundamental) solution is x=h[n] and y=k[n] for the lowest n found which solves the equation in the continued fraction expansion of sqrt(D).
Problem
I am failing to get this simple algorithm work for D=61. I first noticed it did not solve Pell's equation for 100 coefficients, so I compared it against Wolfram Alpha's convergents and continued fraction representation and noticed the 20th elements fail - the representation is 3 compared to 4 that I get, yielding different convergents - h[20]=335159612 on Wolfram compared to 425680601 for me.
I tested the code below, two languages (though to be fair, Python is C under the hood I guess), on two systems and get the same result - a diff on loop 20. I'll note that the convergents are still accurate and converge! Why am I getting different results compared to Wolfram Alpha, and is it possible to fix it?
For testing, here's a Python program to solve Pell's equation for D=61, printing first 20 convergents and the continued fraction representation cf (and some extra unneeded fluff):
from math import floor, sqrt # Can use mpmath here as well.
def continued_fraction(D, count=100, thresh=1E-12, verbose=False):
cf = []
h = (0, 1)
k = (1, 0)
r = start = sqrt(D)
initial_count = count
x = (1+thresh+start)*start
y = start
while abs(x/y - start) > thresh and count:
i = int(floor(r))
cf.append(i)
f = r - i
x, y = i*h[-1] + h[-2], i*k[-1] + k[-2]
if verbose is True or verbose == initial_count-count:
print(f'{x}\u00B2-{D}x{y}\u00B2 = {x**2-D*y**2}')
if x**2 - D*y**2 == 1:
print(f'{x}\u00B2-{D}x{y}\u00B2 = {x**2-D*y**2}')
print(cf)
return
count -= 1
r = 1/f
h = (h[1], x)
k = (k[1], y)
print(cf)
raise OverflowError(f"Converged on {x} {y} with count {count} and diff {abs(start-x/y)}!")
continued_fraction(61, count=20, verbose=True, thresh=-1) # We don't want to stop on account of thresh in this example
A c program doing the same:
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int main() {
long D = 61;
double start = sqrt(D);
long h[] = {0, 1};
long k[] = {1, 0};
int count = 20;
float thresh = 1E-12;
double r = start;
long x = (1+thresh+start)*start;
long y = start;
while(abs(x/(double)y-start) > -1 && count) {
long i = floor(r);
double f = r - i;
x = i * h[1] + h[0];
y = i * k[1] + k[0];
printf("%ld\u00B2-%ldx%ld\u00B2 = %lf\n", x, D, y, x*x-D*y*y);
r = 1/f;
--count;
h[0] = h[1];
h[1] = x;
k[0] = k[1];
k[1] = y;
}
return 0;
}
mpmath, python's multi-precision library can be used. Just be careful that all the important numbers are in mp format.
In the code below, x, y and i are standard multi-precision integers. r and f are multi-precision real numbers. Note that the initial count is set higher than 20.
from mpmath import mp, mpf
mp.dps = 50 # precision in number of decimal digits
def continued_fraction(D, count=22, thresh=mpf(1E-12), verbose=False):
cf = []
h = (0, 1)
k = (1, 0)
r = start = mp.sqrt(D)
initial_count = count
x = 0 # some dummy starting values, they will be overwritten early in the while loop
y = 1
while abs(x/y - start) > thresh and count > 0:
i = int(mp.floor(r))
cf.append(i)
x, y = i*h[-1] + h[-2], i*k[-1] + k[-2]
if verbose or initial_count == count:
print(f'{x}\u00B2-{D}x{y}\u00B2 = {x**2-D*y**2}')
if x**2 - D*y**2 == 1:
print(f'{x}\u00B2-{D}x{y}\u00B2 = {x**2-D*y**2}')
print(cf)
return
count -= 1
f = r - i
r = 1/f
h = (h[1], x)
k = (k[1], y)
print(cf)
raise OverflowError(f"Converged on {x} {y} with count {count} and diff {abs(start-x/y)}!")
continued_fraction(61, count=22, verbose=True, thresh=mpf(1e-100))
Output is similar to wolfram's:
...
335159612²-61x42912791² = 3
1431159437²-61x183241189² = -12
1766319049²-61x226153980² = 1
[7, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1]

How to plot 2 functions with Julia

I am struggling to plot evaluated function and Cbebyshev approximation.
I am using Julia 1.2.0.
EDIT: Sorry, added completed code.
using Plots
pyplot()
mutable struct Cheb_struct
c::Vector{Float64}
min::Float64
max::Float64
end
function cheb_coeff(min::Float64, max::Float64, n::Int, fn::Function)::Cheb_struct
struc = Cheb_struct(Vector{Float64}(undef,n), min, max)
f = Vector{Float64}(undef,n)
p = Vector{Float64}(undef,n)
max_plus_min = (max + min) / 2
max_minus_min = (max - min) / 2
for k in 0:n-1
p[k+1] = pi * ((k+1) - 0.5) / n
f[k+1] = fn(max_plus_min + cos(p[k+1])*max_minus_min)
end
n2 = 2 / n
for j in 0:n-1
s = 0
for i in 0:n-1
s += f[i+1]*cos(j*p[i+1])
struc.c[j+1] = s * n2
end
end
return struc
end
function approximate(struc::Cheb_struct, x::Float64)::Float64
x1 = (2*x - struc.max - struc.min) / (struc.max - struc.min)
x2 = 2*x1
t = s = 0
for j in length(struc.c):-1:2
pom = s
s = x2 * s - t + struc.c[j]
t = pom
end
return (x1 * s - t + struc.c[1] / 2)
end
fn = sin
struc = cheb_coeff(0.0, 1.0, 10, fn)
println("coeff:")
for x in struc.c
#printf("% .15f\n", x)
end
println("\n x eval approx eval-approx")
for x in struc.min:0.1:struc.max
eval = fn(x)
approx = approximate(struc, x)
#printf("%11.8f %12.8f %12.8f % .3e\n", x,eval, approx, eval - approx)
display(plot(x=eval,y=approx))
end
I am getting empty plot window.
I would be very grateful if someone coould how to plot these two functions.
You should provide a working code as an example.
However the code below can show you how to plot:
using Plots
pyplot()
fn = sin
approxf(x) = sin(x)+rand()/10
x = 0:0.1:1
evalv = fn.(x)
approxv = approxf.(x)
p = plot(evalv,approxv)
using PyPlot
PyPlot.display_figs() #needed when running in IDE such as Atom

Time-dependent events in ODE

I recently started with Julia and wanted to implement one of my usual problems - implement time-depended events.
For now I have:
# Packages
using Plots
using DifferentialEquations
# Parameters
k21 = 0.14*24
k12 = 0.06*24
ke = 1.14*24
α = 0.5
β = 0.05
η = 0.477
μ = 0.218
k1 = 0.5
V1 = 6
# Time
maxtime = 10
tspan = (0.0, maxtime)
# Dose
stim = 100
# Initial conditions
x0 = [0 0 2e11 8e11]
# Model equations
function system(dy, y, p, t)
dy[1] = k21*y[2] - (k12 + ke)*y[1]
dy[2] = k12*y[1] - k21*y[2]
dy[3] = (α - μ - η)*y[3] + β*y[4] - k1/V1*y[1]*y[3]
dy[4] = μ*y[3] - β*y[4]
end
# Events
eventtimes = [2, 5]
function condition(y, t, integrator)
t - eventtimes
end
function affect!(integrator)
x0[1] = stim
end
cb = ContinuousCallback(condition, affect!)
# Solve
prob = ODEProblem(system, x0, tspan)
sol = solve(prob, Rodas4(), callback = cb)
# Plotting
plot(sol, layout = (2, 2))
But the output that is give is not correct. More specifically, the events are not taken into account and the initial condition doesn't seems to be 0 for y1 but stim.
Any help would be greatly appreciated.
t - eventtimes doesn't work because one's a scalar and the other is a vector. But for this case, it's much easier to just use a DiscreteCallback. When you make it a DiscreteCallback you should pre-set the stop times so that to it hits 2 and 5 for the callback. Here's an example:
# Packages
using Plots
using DifferentialEquations
# Parameters
k21 = 0.14*24
k12 = 0.06*24
ke = 1.14*24
α = 0.5
β = 0.05
η = 0.477
μ = 0.218
k1 = 0.5
V1 = 6
# Time
maxtime = 10
tspan = (0.0, maxtime)
# Dose
stim = 100
# Initial conditions
x0 = [0 0 2e11 8e11]
# Model equations
function system(dy, y, p, t)
dy[1] = k21*y[2] - (k12 + ke)*y[1]
dy[2] = k12*y[1] - k21*y[2]
dy[3] = (α - μ - η)*y[3] + β*y[4] - k1/V1*y[1]*y[3]
dy[4] = μ*y[3] - β*y[4]
end
# Events
eventtimes = [2.0, 5.0]
function condition(y, t, integrator)
t ∈ eventtimes
end
function affect!(integrator)
integrator.u[1] = stim
end
cb = DiscreteCallback(condition, affect!)
# Solve
prob = ODEProblem(system, x0, tspan)
sol = solve(prob, Rodas4(), callback = cb, tstops = eventtimes)
# Plotting
plot(sol, layout = (2, 2))
This avoids rootfinding altogether so it should be a much nicer solution that hacking time choices into a rootfinding system.
Either way, notice that the affect was changed to
function affect!(integrator)
integrator.u[1] = stim
end
It needs to be modifying the current u value otherwise it won't do anything.

Direved and integral problème [SCILAB]

I'm working on a project that calculates the derivative of the speed and the integral of the acceleration.
my problem is that I have a lot of acceleration points and speed over time and I can not find the right program for that.
example:
acceleration from 0 km / h to 40 km / h in 5 seconds
from 5 to 10 seconds, the speed is constant 40km/h;
from 10 to 17 seconds there is a deceleration from 40 km / h to 20 km / h
So dv/dt = (v2-v1)/(t2-t1) but I don't know how to declare multiple variables for v1 v2 t1 t2
function a=acc(v1,v2,t1,t2)
a= (v2-v1)/(t2-t1)
endfunction
v1=
v2=
t1=
t2=
disp(acc(v1,v2,t1,t2),'acc = ')
and the same for the integral of (dv/dt)*dt
please help me guys
V(1:5) = linspace(0,40,5);
V(6:10) = 40;
V(11:17) = linspace(40,20,7);
Acc = diff(V);
First we populate a array V with your values for the speed.
Then we create an array Acc with the acceleration a each seconds with diffsince there's only 1s between two values of V.
Another solution based on what you wrote
function a=acc_2(v1,v2,t1,t2)
a= (v2-v1)./(t2-t1) // since v,t are vectors, we need './' and not '/' !
endfunction
V(1:5) = linspace(0,40,5);
V(6:10) = 40;
V(11:17) = linspace(40,20,7);
v1 = V(1:$-1);
v2 = V(2:$);
t1 = 1:length(V)-1;
t2 = 2:length(V);
Acc_2 = acc_2(v1,v2,t1,t2)
And if you want to have h(x) = int_t0^x dv/dt dt then use cumsum
H = cumsum(Acc)
i put this code
V(1:5) = linspace(0,40,5);
V(6:10) = 40;
V(11:17) = linspace(40,20,7);
function a = acc(V)
a=diff(V)
endfunction
function aa = acc_2(v1,v2,t1,t2)
aa = (v2-v1)/(t2-t1)
endfunction
v1 = V(1:$-1);
v2 = V(2:$);
t1 = 1:length(V)-1;
t2 = 2:length(V);
Acc_2 = acc_2(v1,v2,t1,t2)
but he gives me the result of a single variable of Acc_2 ?

Collapsing a 10 period curve to 4 periods

I have a 10 period cost curve table below. How do I programmatically collapse/condense/shrink this to 4 periods. I'm using VBA but I should be able to follow other languages. The routine should work for whatever period you pass to it. For example, if I pass it a 7 it should condense the percentages to 7 periods. If I pass it 24 then expand the percentages to 24 periods, spreading the percentages based on the original curve. Any help or example will be appreciated. Thanks...
ORIGINAL
Period Pct
1 10.60%
2 19.00%
3 18.30%
4 14.50%
5 10.70%
6 8.90%
7 6.50%
8 3.10%
9 3.00%
10 5.40%
COLLAPSED
Period Pct
1 38.75%
2 34.35%
3 16.95%
4 9.95%
EDITED: I've added sample code below as to what I have so far. It only works for periods 1, 2, 3, 5, 9, 10. Maybe someone can help modify it to work for any period. Disclaimer, I'm not a programmer so my coding is bad. Plus, I have no clue as to what I'm doing.
Sub Collapse_Periods()
Dim aPct As Variant
Dim aPer As Variant
aPct = Array(0.106, 0.19, 0.183, 0.145, 0.107, 0.089, 0.065, 0.031, 0.03, 0.054)
aPer = Array(1, 2, 3, 5, 9, 10)
For i = 0 To UBound(aPer)
pm = 10 / aPer(i)
pct1 = 1
p = 0
ttl = 0
For j = 1 To aPer(i)
pct = 0
k = 1
Do While k <= pm
pct = pct + aPct(p) * pct1
pct1 = 1
p = p + 1
If k <> pm And k = Int(pm) Then
pct1 = (pm - Int(pm)) * j
pct = pct + (pct1 * aPct(p))
pct1 = 1 - pct1
End If
k = k + 1
Loop
Debug.Print aPer(i) & " : " & j & " : " & pct
ttl = ttl + pct
Next j
Debug.Print "Total: " & ttl
Next i
End Sub
I would like to know how this is done also using an Integral? This is how I would have done it - perhaps it's a longhand/longwinded method but I'd like to see some better suggestions.
It's probably easier to see the method in Excel first using the LINEST function and Named ranges. I've assumed the function is logarithmic. I've outlined steps [1.] - [5.]
This VBA code then essentially replicates the Excel method using a function to pass 2 arrays, periods and a return array that can be written to a range
Sub CallingProc()
Dim Periods As Long, returnArray() As Variant
Dim X_Values() As Variant, Y_Values() As Variant
Periods = 4
ReDim returnArray(1 To Periods, 1 To 2)
With Sheet1
X_Values = Application.Transpose(.Range("A2:A11"))
Y_Values = Application.Transpose(.Range("B2:B11"))
End With
FGraph X_Values, Y_Values, Periods, returnArray 'pass 1D array of X, 1D array of Y, Periods, Empty ReturnArray
End Sub
Function FGraph(ByVal x As Variant, ByVal y As Variant, ByVal P As Long, ByRef returnArray As Variant)
Dim i As Long, mConstant As Double, cConstant As Double
'calc cumulative Y and take Ln (Assumes Form of Graph is logarithmic!!)
For i = LBound(y) To UBound(y)
If i = LBound(y) Then
y(i) = y(i)
Else
y(i) = y(i) + y(i - 1)
End If
x(i) = Log(x(i))
Next i
'calc line of best fit
With Application.WorksheetFunction
mConstant = .LinEst(y, x)(1)
cConstant = .LinEst(y, x)(2)
End With
'redim array to fill for new Periods
ReDim returnArray(1 To P, 1 To 2)
'Calc new periods based on line of best fit
For i = LBound(returnArray, 1) To UBound(returnArray, 1)
returnArray(i, 1) = UBound(y) / P * i
If i = LBound(returnArray, 1) Then
returnArray(i, 2) = (Log(returnArray(i, 1)) * mConstant) + cConstant
Else
returnArray(i, 2) = ((Log(returnArray(i, 1)) * mConstant) + cConstant) - _
((Log(returnArray(i - 1, 1)) * mConstant) + cConstant)
End If
Next i
'returnArray can be written to range
End Function
EDIT:
This VBA code now calculates the linear trend of the points either side of the new period reduction. The data is returned in a 2dimension array named returnArray
Sub CallingProc()
Dim Periods As Long, returnArray() As Variant
Dim X_Values() As Variant, Y_Values() As Variant
Periods = 4
ReDim returnArray(1 To Periods, 1 To 2)
With Sheet1
X_Values = Application.Transpose(.Range("A2:A11"))
Y_Values = Application.Transpose(.Range("B2:B11"))
End With
FGraph X_Values, Y_Values, returnArray 'pass 1D array of X, 1D array of Y, Dimensioned ReturnArray
End Sub
Function FGraph(ByVal x As Variant, ByVal y As Variant, ByRef returnArray As Variant)
Dim i As Long, j As Long, mConstant As Double, cConstant As Double, Period As Long
Period = UBound(returnArray, 1)
'calc cumulative Y
For i = LBound(y) + 1 To UBound(y)
y(i) = y(i) + y(i - 1)
Next i
'Calc new periods based on line of best fit
For i = LBound(returnArray, 1) To UBound(returnArray, 1)
returnArray(i, 1) = UBound(y) / Period * i
'find position of new period to return adjacent original data points
For j = LBound(x) To UBound(x)
If returnArray(i, 1) <= x(j) Then Exit For
Next j
'calc linear line of best fit between existing data points
With Application.WorksheetFunction
mConstant = .LinEst(Array(y(j), y(j - 1)), Array(x(j), x(j - 1)))(1)
cConstant = .LinEst(Array(y(j), y(j - 1)), Array(x(j), x(j - 1)))(2)
End With
returnArray(i, 2) = (returnArray(i, 1) * mConstant) + cConstant
Next i
'returnarray holds cumulative % so calc period only %
For i = UBound(returnArray, 1) To LBound(returnArray, 1) + 1 Step -1
returnArray(i, 2) = returnArray(i, 2) - returnArray(i - 1, 2)
Next i
'returnArray now holds your data
End Function
Returns:
COLLAPSED
1 38.75%
2 34.35%
3 16.95%
4 9.95%

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