In Maple, I have a matrix N and its elements N[i,j], If I modify the elements of this matrix as follows for example
>for j from 1 to 4 do
>print(F[i,j]=(diff(N[i,j],x)));
>od;od;
where the matrix elements are functions of x.
I've wanted to define new matrix elements
>BA[i,j]:=(diff(N[i,j],x)));
but I can't do this with Maple, through the above command. Can someone help me ?
Better than using a loop is simply BA:= diff~(N,x). The ~ can be appended to any operator to mean "apply the operator to each member of the container and return a new container containing the modified members."
Also, be careful about using print. Its only purpose is to print stuff on the screen from the middle (not the end) of a computation. It can't be used to change any stored values. Good programs use print very sparingly, if at all. The end result of a computation is displayed automatically, without needing a print command.
Related
I am a complete noob in Julia and its syntax. I am trying to follow this article on semi-definite-programming on Julia.
I would apprecieate if someone can help me figure out what the for loop in In[4] actually does:
for i in 1:m
A[:, (i-1)*n+1:i*n] .= random_mat_create(n)
b[i] = tr(A[:, (i-1)*n+1:i*n]*X_test)
end
To my understanding it should create a vector of matrices A (m of those) as well as an m-dimensional vector b. I am totally confused though on the indexing of A and the indexing of b.
I would like an explanation of the :, (i-1)*n+1:i*n part of this code. The reason I ask here is because I also dont know what to Google or what to search for in Julia documentation.
(i-1)*n+1:i*n creates a range from (i-1)*n + 1 to i*n. For example, if i=2 and n=10, this range becomes 11:20, so A[:, (i-1)*n+1:i*n] will grab all the rows of A (that's what : does), and columns 11-20.
There are two operations there that are not clear to you:
: operator. Consider a Matrix a = zeros(3,3). You could use array slicing operator (similarly to numpy or Matlab) to select the entire second columns as: a[1:end,2]. However, when selecting everything from start to the end those two values can be omitted and hence you can write a[:,2] (this always looked the easiest way for me to remember that)
. (dot) operator. Julia is very careful about what gets vectorized and what not. In numpy or R, vectorizing operations happens kind of always automatically. In Julia you have the control - but with the control comes the responsibility. Hence trying to assign values to the second column by writing a[:, 2] = 5.0 will throw an error because there is vector on the right and a scalar on the left. If you want to vectorize you need to tell that to Julia. Hence the dot operator .= means "perform element-wise assignment". Note that any Julia function or operator, even your own functions can be decorated by such dot .. Since this is a very important language feature have a look at https://docs.julialang.org/en/v1/manual/arrays/#Broadcasting
Is there a way I can use a command that selects a matrix to use based on a variable?
Need in this /
:If (way to select a matrix based on what variable L equals) (E,F)=1:Output E,F,"O
I don't want to make a specific go-to for every single matrix I need.
This is for creating maps with the matrix in case anyone has a better way.
If i understand correctly you want to get the value from a certain matrix, chosen dynamically depending on the value of a variable. You can kinda do this by putting the names of the matrices into a string, then grab a substring of the string, using sub(, at a dynamic offset, based on L, and then feeding that string into expr( to get a reference to the matrix, ie
:"[A][B][C]"->Str1, sub(Str1,2,1) yields "[B]", expr("[B]") yields Matrix B...so 2 maps to [B]. TI considers the symbol [A] (and all the other matrix vars) to be a single character, so "[A][B][C]" is a 3 character string.
Note that all of the matrix vars need to be input from the MATRIX menu (including inside the string). Typing in the individual [ A ] chracters will not work.
Also note you can't grab indexes off of a matrix returned with expr (ie expr("[A]")(1,2) so you need an extra matrix (I used [J]) to store the result.
For example
:"MAKE SOME MATRICES"
:[[1,2][3,4]]->[A]
:[[5,6][7,8]]->[B]
:[[9,10],[11,12]]->[C]
:"SAMPLE L VALUE"
:2->L
:"STORE REFERENCES TO THE"
:"MATRICES IN A STRING"
:"[A][B][C]"->Str1
:expr(sub(Str1, L, 1))->[J]
:"SHOWS 6"
:[J](1,2)
so then proceed normally with [J]
:If [J](E,F)
: "DO WHATEVER
Tested on an 84 SE, I assume it would work the same for anything in that family, except IIRC some older models only have matrices A-F
I have a matrix A which has a size of 54x100. For some specific condition I perform an operation on each row of A. I need to save the output of this for loop. I've tried the following but it did not work.
S=zeros(54,100);
for i=1:54;
Ri=A(i,:);
answer=mean(reshape(Ri,5,20),1);
S(i)=answer;
end
Firstly, judging by your question I'd recommend some basic Matlab tutorials like this or just detailed documentation like this.
To actually help you with your issue though; you can do this:
%% Make up A (since I don't know what it actually is)
n = 54; m = 100;
A = randn(n,m); % N x m matrix of random numbers
%% Loop over each row of A
S = cell(n,1);
for j = 1:n;
Rj = A(j,:); % j'th row
answer = mean(reshape(Rj,5,20),1); % some operation
S{j} = answer; % store the answer in cell S
end
The problem was that your answer was not a single number (1x1 matrix) but a vector and so you got a dimension mismatch error. Above I'm putting the answers into a cell object of size n. The result of your operation on j'th row can then be retrieved by calling S{j}.
Also:
Do not using i as an iterator since it also represents the imaginary unit.
Do not hard-code values but reference the existing ones. For example here I referenced n in the for-loop declaration as opposed to just writing for j = 1:54 because otherwise, if I got struck by a fancy to use my code for a 53x100 array it would not work anymore.
When you post your code I reccomend adding a minimal working example - a pece of code which people can just copy and paste into their Matlab (or whatever interpreter of whatever language) and run to reproduce your problem. Here you have not included anything which tells the code what A is, for example.
This is quite a good read in general and should help you in the future
I was reading Parallel Computing docs of Julia, and having never done any parallel coding, I was left wanting a gentler intro. So, I thought of a (probably) simple problem that I couldn't figure out how to code in parallel Julia paradigm.
Let's say I have a matrix/dataframe df from some experiment. Its N rows are variables, and M columns are samples. I have a method pwCorr(..) that calculates pairwise correlation of rows. If I wanted an NxN matrix of all the pairwise correlations, I'd probably run a for-loop that'd iterate for N*N/2 (upper or lower triangle of the matrix) and fill in the values; however, this seems like a perfect thing to parallelize since each of the pwCorr() calls are independent of others. (Am I correct in thinking this way about what can be parallelized, and what cannot?)
To do this, I feel like I'd have to create a DArray that gets filled by a #parallel for loop. And if so, I'm not sure how this can be achieved in Julia. If that's not the right approach, I guess I don't even know where to begin.
This should work, first you need to propagate the top level variable (data) to all the workers:
for pid in workers()
remotecall(pid, x->(global data; data=x; nothing), data)
end
then perform the computation in chunks using the DArray constructor with some fancy indexing:
corrs = DArray((20,20)) do I
out=zeros(length(I[1]),length(I[2]))
for i=I[1], j=I[2]
if i<j
out[i-minimum(I[1])+1,j-minimum(I[2])+1]= 0.0
else
out[i-minimum(I[1])+1,j-minimum(I[2])+1] = cor(vec(data[i,:]), vec(data[j,:]))
end
end
out
end
In more detail, the DArray constructor takes a function which takes a tuple of index ranges and returns a chunk of the resulting matrix which corresponds to those index ranges. In the code above, I is the tuple of ranges with I[1] being the first range. You can see this more clearly with:
julia> DArray((10,10)) do I
println(I)
return zeros(length(I[1]),length(I[2]))
end
From worker 2: (1:10,1:5)
From worker 3: (1:10,6:10)
where you can see it split the array into two chunks on the second axis.
The trickiest part of the example was converting from these 'global' index ranges to local index ranges by subtracting off the minimum element and then adding back 1 for the 1 based indexing of Julia.
Hope that helps!
I was trying to perform a descend loop in Scilab
like:
for i=7: 1
disp(i);
end
But apparently it didn't work and there are no errors too.
Is this possible? And if yes, how to do it?
Syntax
As the for-loop documentation states, the for loop's syntax is:
for variable=expression do instruction, ,instruction,end
If expression is a matrix or a row vector, variable takes as values the values of each column of the matrix.
Your example
Your for-loop uses 7:1 as expression, what will result in a valid but empty vector:
-->7:1
ans =
[]
The for-loop then takes each value in the vector and then finishes, but since your vector is empty it finishes directly. So it is valid syntax.
Working example
The documentation gives the following example of a working decreasing loop:
for j=4:-1:1
disp(j);
end // decreasing loop