I am trying to plot a 3D graph between 2 scalars and one matrix for each of its entries. On compiling it is giving me "Submatrix incorrectly defined" error on line 11. The code:
i_max= 3;
u = zeros(4,5);
a1 = 1;
a2 = 1;
a3 = 1;
b1 = 1;
hx = linspace(1D-6,1D6,13);
ht = linspace(1D-6,1D6,13);
for i = 1:i_max
for j = 2:4
u(i+1,j)=u(i,j)+(ht*(a1*u(i,j))+b1+(((a2*u(i,j+1))-(2*a2*u(i,j))+(a2*u(i,j-1)))*(hx^-2))+(((a3*u(i,j+1))-(a3*u(i,j-1)))*(0.5*hx^-1)));
plot(ht,hx,u(i+1,j));
end
end
Full error message:
-->exec('C:\Users\deba123\Documents\assignments and lecture notes\Seventh Semester\UGP\Scilab\Simulation1_Plot.sce', -1)
+(((a3*u(i,j+1))-(a3*u(i,j-1)))*(0.5*hx^-1)))
!--error 15
Submatrix incorrectly defined.
at line 11 of exec file called by :
emester\UGP\Scilab\Simulation1_Plot.sce', -1
Please help.
For a 3-dimensional figure, you need 2 argument vectors and a matrix for the function values. So I expanded u to a tensor.
At every operation in your code, I added the current dimension of the term. Now, a transparent handling of you calculation is given. For plotting you have to use the plot3d (single values) or surf (surface) command.
In a 3-dim plot, you want two map 2 vectors (hx,ht) with dim n and m to an scalar z. Therefore you reach a (nxm)-matrix with your results. Is this, what you want to do? Currently, you have 13 values for each u(i,j,:) - entry, but you want (13x13) for every figure. Maybe the eval3d-function can help you.
i_max= 3;
u = zeros(4,5,13);
a1 = 1;
a2 = 1;
a3 = 1;
b1 = 1;
hx = linspace(1D-6,1D6,13); // 1 x 13
ht = linspace(1D-6,1D6,13); // 1 x 13
for i = 1:i_max
for j = 2:4
u(i+1,j,:)= u(i,j)...
+ ht*(a1*u(i,j))*b1... // 1 x 13
+(((a2*u(i,j+1)) -(2*a2*u(i,j)) +(a2*u(i,j-1)))*(hx.^-2))... // 1 x 13
+(((a3*u(i,j+1))-(a3*u(i,j-1)))*(0.5*hx.^-1)) ... // 1 x 13
+ hx*ones(13,1)*ht; // added to get non-zero values
z = squeeze( u(i+1,j, : ))'; // 1x13
// for a 3d-plot: (1x13, 1x13, 13x13)
figure()
plot3d(ht,hx, z'* z ,'*' ); //
end
end
Related
I'm wondering if anyone knows any tools which allow me to count the frequency of amino acids at any specific position in a multiple-sequence alignment.
For example if I had three sequences:
Species 1 - MMRSA
Species 2 - MMLSA
Species 3 - MMRTA
I'd like for a way to search by position for the following output:
Position 1 - M = 3;
Position 2 - M = 3;
Position 3 - R = 2, L = 1;
Position 4 - S = 2, T = 1;
Position 5 - A = 3.
Thanks! I'm familiar with R and Linux, but if there's any other software that can do this I'm sure I can learn.
Using R:
x <- read.table(text = "Species 1 - MMRSA
Species 2 - MMLSA
Species 3 - MMRTA")
ixCol = 1
table(sapply(strsplit(x$V4, ""), "[", ixCol))
# M
# 3
ixCol = 4
table(sapply(strsplit(x$V4, ""), "[", ixCol))
# S T
# 2 1
Depending input file format, there are likely a purpose built bioconductor packages/functions.
That is really easy to parse, you can use any language of choice.
Here is an example in Python using a dict and Counter to assemble the data in a simple object.
from collections import defaultdict, Counter
msa = '''
Species 1 - MMRSA
Species 2 - MMLSA
Species 3 - MMRTA
'''
r = defaultdict(list) #dictionary having the sequences index as key and the list of aa found at that index as value
for line in msa.split('\n'):
line = line.strip()
if line:
sequence = line.split(' ')[-1]
for i, aa in enumerate(list(sequence)):
r[i].append(aa)
count = {k:Counter(v) for k,v in r.items()}
print(count)
#{0: Counter({'M': 3}), 1: Counter({'M': 3}), 2: Counter({'R': 2, 'L': 1}), 3: Counter({'S': 2, 'T': 1}), 4: Counter({'A': 3})}
To print the output as you specified:
for k, v in count.items():
print(f'Position {k+1} :', end=' ') #add 1 to start counting from 1 instead of 0
for aa, c in v.items():
print(f'{aa} = {c};', end=' ')
print()
It prints:
Position 1 : M = 3;
Position 2 : M = 3;
Position 3 : R = 2; L = 1;
Position 4 : S = 2; T = 1;
Position 5 : A = 3;
I'm trying to translate a C routine from an old sound synthesis program into R, but have indexing issues which I'm struggling to understand (I'm a beginner when it comes to using loops).
The routine creates an exponential lookup table - the vector exptab:
# Define parameters
sinetabsize <- 8192
prop <- 0.8
BP <- 10
BD <- -5
BA <- -1
# Create output vector
exptab <- vector("double", sinetabsize)
# Loop
while(abs(BD) > 0.00001){
BY = (exp(BP) -1) / (exp(BP*prop)-1)
if (BY > 2){
BS = -1
}
else{
BS = 1
}
if (BA != BS){
BD = BD * -0.5
BA = BS
BP = BP + BD
}
if (BP <= 0){
BP = 0.001
}
BQ = 1 / (exp(BP) - 1)
incr = 1 / sinetabsize
x = 0
stabsize = sinetabsize + 1
for (i in (1:(stabsize-1))){
x = x + incr
exptab [[sinetabsize-i]] = 1 - (BQ * (exp(BP * x) - 1))
}
}
Running the code gives the error:
Error in exptab[[sinetabsize - i]] <- 1 - (BQ * (exp(BP * x) - 1)) :
attempt to select less than one element in integerOneIndex
Which, I understand from looking at other posts, indicates an indexing problem. But, I'm finding it difficult to work out the exact issue.
I suspect the error may lie in my translation. The original C code for the last few lines is:
for (i=1; i < stabsize;i++){
x += incr;
exptab[sinetabsize-i] = 1.0 - (float) (BQ*(exp(BP*x) - 1.0));
}
I had thought the R code for (i in (1:(stabsize-1))) was equivalent to the C code for (i=1; i< stabsize;i++) (i.e. the initial value of i is i = 1, the test is whether i < stabsize, and the increment is +1). But now I'm not so sure.
Any suggestions as to where I'm going wrong would be greatly appreciated!
As you say, array indexing in R starts at 1. In C it starts at zero. I reckon that's your problem. Can sinetabsize-i ever get to zero?
I'm trying to implement the following code from here but it won't work correctly.
What I want is the shortest path distances from a source to all nodes and also the predecessors. Also, I want the input of the graph to be an adjacency matrix which contains all of the edge weights.
I'm trying to make it work in just one function so I have to rewrite it. If I'm right the original code calls other functions (from graph.jl for example).
I don't quite understand how to rewrite the for loop which calls the adj() function.
Also, I'm not sure if the input is correct in the way the code is for now.
function dijkstra(graph, source)
node_size = size(graph, 1)
dist = ones(Float64, node_size) * Inf
dist[source] = 0.0
Q = Set{Int64}() # visited nodes
T = Set{Int64}(1:node_size) # unvisited nodes
pred = ones(Int64, node_size) * -1
while condition(T)
# node selection
untraversed_nodes = [(d, k) for (k, d) in enumerate(dist) if k in T]
if minimum(untraversed_nodes)[1] == Inf
break # Break if remaining nodes are disconnected
end
node_ind = untraversed_nodes[argmin(untraversed_nodes)][2]
push!(Q, node_ind)
delete!(T, node_ind)
# distance update
curr_node = graph.nodes[node_ind]
for (neigh, edge) in adj(graph, curr_node)
t_ind = neigh.index
weight = edge.cost
if dist[t_ind] > dist[node_ind] + weight
dist[t_ind] = dist[node_ind] + weight
pred[t_ind] = node_ind
end
end
end
return dist, pred
end
So if I'm trying it with the following matrix
A = [0 2 1 4 5 1; 1 0 4 2 3 4; 2 1 0 1 2 4; 3 5 2 0 3 3; 2 4 3 4 0 1; 3 4 7 3 1 0]
and source 2 i would like to get the distances in a vector dist and the predeccessors in anothe vectore pred.
Right now I'm getting
ERROR: type Array has no field nodes
Stacktrace: [1] getproperty(::Any, ::Symbol) at .\sysimg.jl:18
I guess I have to rewrite it a bit more.
I m thankful for any help.
Assuming that graph[i,j] is a length of path from i to j (your graph is directed looking at your data), and it is a Matrix with non-negative entries, where 0 indicates no edge from i to j, a minimal rewrite of your code should be something like:
function dijkstra(graph, source)
#assert size(graph, 1) == size(graph, 2)
node_size = size(graph, 1)
dist = fill(Inf, node_size)
dist[source] = 0.0
T = Set{Int}(1:node_size) # unvisited nodes
pred = fill(-1, node_size)
while !isempty(T)
min_val, min_idx = minimum((dist[v], v) for v in T)
if isinf(min_val)
break # Break if remaining nodes are disconnected
end
delete!(T, min_idx)
# distance update
for nei in 1:node_size
if graph[min_idx, nei] > 0 && nei in T
possible_dist = dist[min_idx] + graph[min_idx, nei]
if possible_dist < dist[nei]
dist[nei] = possible_dist
pred[nei] = min_idx
end
end
end
end
return dist, pred
end
(I have not tested it extensively, so please report if you find any bugs)
I have a simple question. How do I use the command qpsolve from Scilab if I only want to use the lower bounds and upper bounds limit?
ci <= x <= cs
The command can be used as this:
[x [,iact [,iter [,f]]]] = qpsolve(Q,p,C,b,ci,cs,me)
But I want to use it like this:
x = qpsolve(Q,p,[],[],ci,cs,[])
Only ci and cs should explain the limits for vector x. Unfortunately, the command cannot take empty []. Should I take [] as a row vector of ones or zeros?
https://help.scilab.org/docs/6.0.1/en_US/qpsolve.html
In Scilab 5.5.1 , [] works for C and b but not for me. so C = [];b = [];me = 0; should work.
Why
qpsolve is an interface for qp_solve :
function [x, iact, iter, f]=qpsolve(Q,p,C,b,ci,cs,me)
rhs = argn(2);
if rhs <> 7
error(msprintf(gettext("%s: Wrong number of input argument(s): %d expected.\n"), "qpsolve", 7));
end
C(me+1:$, :) = -C(me+1:$, :);
b(me+1:$) = -b(me+1:$);
// replace boundary contraints by linear constraints
Cb = []; bb = [];
if ci <> [] then
Cb = [Cb; speye(Q)]
bb = [bb; ci]
end
if cs <> [] then
Cb = [Cb; -speye(Q)]
bb = [bb; -cs]
end
C = [C; Cb]; b = [b; bb]
[x, iact, iter, f] = qp_solve(Q, -p, C', b, me)
endfunction
It transform every bound constraints into linear constraints. To begin, it swap the sign of the inequality constraints. To do that, it must know me, ie it must be an integer. Since C and b are empty matrices, is value doesn't matter.
Bonus:
if Q is inversible, you could skip the qpsolve macro and write
x = -Q\p
x(x<ci) = ci(x<ci)
x(x>cs) = cs(x>cs)
Hi have made this code to plot a function.
I need to mark with an red X all the crossings between x = 0 and the blue wave line in the graph.
I have made some tries but with '-xr' in the plot function but it places X marks out of the crossings.
Anyone knows how to do it. Many thanks.
Code:
% entrada
a = input('Introduza o valor de a: ');
% ficheiro fonte para a função
raizes;
% chamada à função
x = 0:.1:50;
or = x;
or(:) = 0;
h = #(x) cos(x);
g = #(x) exp(a*x)-1;
f = #(x) h(x) - g(x);
zeros = fzero(f,0);
plot(x,f(x));
hold on
plot(zeros,f(zeros),'-xr')
hold off
Graph (it only marks one zero, i need all the zero crossings):
As mentioned in the comments above, you need to look for the zeros of your function before you can plot them. You can do this mathematically (in this case set f(x) = g(x) and solve for x) or you can do this analytically with something like fsolve.
If you read the documentation for fsolve, you will see that it searches for the zero closest to the provided x0 if passed a scalar or the first zero if passed an interval. What we can do for a quick attempt at a solution is to pass our x values into fsolve as initial guesses and filter out the unique values.
% Set up sample data
a = .05;
x = 0:.1:50;
% Set up equations
h = #(x) cos(x);
g = #(x) exp(a*x)-1;
f = #(x) h(x) - g(x);
% Find zeros of f(x)
crossingpoints = zeros(length(x), 1); % Initialize array
for ii = 1:length(x) % Use x data points as guesses for fzero
try
crossingpoints(ii) = fzero(f, x(ii)); % Find zero closest to guess
end
end
crossingpoints(crossingpoints < 0) = []; % Throw out zeros where x < 0
% Find unique zeros
tol = 10^-8;
crossingpoints = sort(crossingpoints(:)); % Sort data for easier diff
temp = false(size(crossingpoints)); % Initialize testing array
% Find where the difference between 'zeros' is less than or equal to the
% tolerance and throw them out
temp(1:end-1) = abs(diff(crossingpoints)) <= tol;
crossingpoints(temp) = [];
% Sometimes catches beginning of the data set, filter it out if this happens
if abs(f(crossingpoints(1))) >= (0 + tol)
crossingpoints(1) = [];
end
% Plot data
plot(x, f(x))
hold on
plot(crossingpoints, f(crossingpoints), 'rx')
hold off
grid on
axis([0 20 -2 2]);
Which gives us the following:
Note that due to errors arising from floating point arithmetic we have to utilize a tolerance to filter our zeros rather than utilizing a function like unique.