I've been given the task of translating a piece of MATLAB code into IDL and have
hit a roadblock when I came across the MATLAB function accumarry(). The
function, described here
is used to sum elements in one array, based on indices given in another. Example
1 perhaps explains this better than the actual function description at the top
of the page. In trying to reproduce Example 1 in IDL, I haven't been able to avoid a for loop, but I'm confident that it is possible. My best attempt is the following:
vals = [101,102,103,104,105]
subs = [0,1,3,1,3]
n = max(subs)+1
accum = make_array(n)
for i = 0, n-1 do begin
wVals = where(subs eq i,count)
accum[i] = count eq 0 ? 0 : total(vals[wVals])
endfor
print,accum
; 101.000 206.000 0.00000 208.000
Any advice on improving this would be greatly appreciated! I expected IDL to have a similar built-in function, but haven't been able to track one down. Perhaps some magic with histogram binning?
I found a number of possible solutions to this problem on Coyote's IDL site (not surprisingly.)
http://www.idlcoyote.com/code_tips/drizzling.html
I ended up using the following, as a compromise between performance and readability:
function accumarray,data,subs
mx = max(subs)
accum = fltarr(mx+1)
h = histogram(subs,reverse_indices=ri,OMIN=om)
for j=0L,n_elements(h)-1 do if ri[j+1] gt ri[j] then $
accum[j+om] = total(vals[ri[ri[j]:ri[j+1]-1]])
return,accum
Related
I'm coding the matrix for the 2-dimensional graph, now.
Although it's so simple equation, it takes a lot of time for performing. I think it could get faster.
especially, "for - command term" could be simplified I think.
How can I simplify this?
q=1:1:30
x(q)=330+q*0.3
F=1:30:8970
T=x(1)-0.3:0.001:x(30)+0.3
n=size(T,2)
k=1:1:n
for a=1:1:30
I(a,k)=F(a)*exp(-2.*(T(:,k)))
end
happy=sum(I)
plot(k,I)
I would say that the time is used to print results. Try to use ; at the end of each line, it will fasten computation.
You can also replace the for loop by the following element by element computation:
a = (1:1:30).';
aux = repmat(exp(-2.*(T(:,k))), length(a), 1);
a = repmat(a, 1, length(k));
I = a.'.*aux.';
I'm trying to make Maple solve a complex equation, but it produces an incorrect result.
The following images tells it all :
At (3) I would expect to get something close to 1 (as (2) shows), yet it gives me something that doesn't make any sense. Is it that the || (to express the complex number modulus) operator has another significance in the solve() function?
The more appropriate function here is fsolve.
Example 1
restart:
G:=(w,L)->(5+I*L*2*Pi*w)/(150+I*L*2*Pi*w);
evalf(5*abs(G(10,1)));
fsolve(5*abs(G(10,L))=%,L=0..10)
Example 2
As above, you need to specify the interval L=0..1 where the solution might be.
G:=(f,L)->(256.4+I*L*2*Pi*f)/(256.4+9845+I*L*2*Pi*f);
evalf(5*abs(G(20000,0.03602197444)));
fsolve(5*abs(G(20000,L))=%,L=0..1);
If you are facing difficulties to specify the interval then you should plot it first, it will give you an idea about it?
plot(5*abs(G(20000,L)),L=0..1)
Restrict the values of L in the solve command with the assuming command.
Example 1:
G:= (w,L) -> (50+I*L*2*Pi*w)/(150+I*L*2*Pi*w);
result := evalf(5*abs(G(10,1)));
solve({5*abs(G(10,L)) = result},L) assuming L::real;
{L = 1.000000000}, {L = -1.000000000}
Example 2:
G:=(f,L) -> (256.4+I*2*Pi*L*f)/(256.4+9845+I*2*Pi*L*f);
result := 5*abs(G(20000,0.03602197444));
solve({5*abs(G(20000,L)) = result},L) assuming L::real;
{L = 0.03602197445}, {L = -0.03602197445}
L=1; Nx=51; PeriodicFlag=1; T=15; Nt=51;
spacedef='Pade6FirstDeriv'; casedef='LinearAdvection';
if (spacedef == 'Pade6FirstDeriv')
D1 = sparse(Pade6(Nx,dx,PeriodicFlag));
elseif (spacedef == 'Upwind3FirstDeriv')
D1 = sparse(Upwind3(Nx,dx,PeriodicFlag));
elseif (spacedef == 'Central4FirstDeriv')
D1 = sparse(Central4(Nx,dx,PeriodicFlag));
elseif (spacedef == 'Central2FirstDeriv')
D1 = sparse(Central2(Nx,dx,PeriodicFlag));
else
error(sprintf('Unknown spacedef = %s',spacedef));
end
In the above code, the if section is a small segment from a function I've constructed. I'm trying to get the function to know which methods to use based on my input (spacedef). Central2, Central4, Upwind3, and Pade6 are other functions I've written. The weird thing is that when spacedef =/= to 'Pade6FirstDeriv', I would get an error stating Error using ==, Matrix dimensions must agree. I've tried swapping the order of the if loop (by placing Central4, Central2, Pade6, and Upwind3 in the first line of the loop), and it seems like only the top line of the loop will work (the elseifs are not working). I'd greatly appreciate it if anybody can help me out. Thanks!
As has been noted in the comments, this is a common error when people first start comparing strings in MATLAB, and strcmp or strcmpi is generally the solution.
However, one solution the questions links in the comments don't present, and a solution I think looks much nicer, is the switch statement:
switch (spacedef)
case('Pade6FirstDeriv')
D1 = sparse(Pade6(Nx,dx,PeriodicFlag));
case('Upwind3FirstDeriv')
D1 = sparse(Upwind3(Nx,dx,PeriodicFlag));
case('Central4FirstDeriv')
D1 = sparse(Central4(Nx,dx,PeriodicFlag));
case('Central2FirstDeriv')
D1 = sparse(Central2(Nx,dx,PeriodicFlag));
otherwise
error(sprintf('Unknown spacedef = %s',spacedef));
end
Note: if I expect others to use my code with string comparisons, I usually lower the input such that the comparison is case-insensitive, although I have not done that here.
I have a huge cell vector cc (size: 1xN) of the form:
cc{1} = {'indexString1', 'str_row1col1', 'str_row1col2' }
cc{2} = {'indexString2', 'str_row2col1', 'bighello', 'str_row1col3' }
cc{3} = {'indexString3','str_row3col1'}
cc{4} = {'indexString4','str_row3col1', 'helloWorld'}
I want to traverse each cell and remove specific cells that contain the word "hello", e.g c{4}{2}. Can we do that without for loops keeping the final structure of cc?
Best,
Thoth.
EDIT: From the answers and comments I have seen that the structure of the cell impose some limitations. So any other suggestion to store my data are welcome. I just want to keep together all the cells (e.g. 'str_row1col1', 'str_row1col2') that correspond to the same indexString*n* (e.g. indexString1). I made this edit in case it helps some final reshape.
Using regular expressions, you can obtain a logical array in which zeros represent occurences of the word 'hello' somewhere in the nested cell. As #LuisMendo pointed out, this would be much easier to delete the unwanted cells if they were not nested:
clc
clear
cc{1} = {'str_row1col1', 'str_row1col2' };
cc{2} = {'str_row2col1', 'bighello', 'str_row1col3' };
cc{3} = {'str_row3col1'};
cc{4} = {'str_row3col1', 'helloWorld'};
A = (cellfun(#isempty,regexp([cc{:}],'(\w*hello|hello\w*)','match')))
Gives the following array:
A =
1 1 1 0 1 1 1 0
For the rest I think you would need a loop since the nested cells are not all of the same size. Anyhow I hope it helps you a bit.
EDIT Here is what you can do using a for loop. In order to identify words of interest (earth and water as in your comment below), simply add them to the argument in the call to regexp. This character: | is used to make some sort of list so that Matlab checks all the expressions in the brackets.
Please refer to this page for more infos on regular expressions. There is also a possibility to look for regular expressions with case-sensitivity.
Sample code, in which I added strings containing earth and water:
cc{1} = {'str_row1col1', 'earth!superman' 'str_row1col2' 'DummyString'};
cc{2} = {'str_row2col1', 'bighello', 'str_row1col3' };
cc{3} = {'str_row3col1' 'str_row3col3' 'water_batman'};
cc{4} = {'str_row3col1' 'str_row4col2' 'helloWorld'};
cc{5} = {'str_row5_LegoMan' 'str_row5col2' 'AnotherDummyString' 'Useless String' 'BonjourWorld'};
% With a for loop, for example:
FinalCell = cell(size(cc,2),1);
for k = 1:size(cc,2)
DummyCell = cc{k}; % Use dummy cell for easier indexing
% This is where you tell Matlab what words/expressions you are looking for
A = cellfun(#isempty,regexp(cc{k},'(\w*hello|hello\w*|earth|water)','match'));
DummyCell(~A) = []; % Remove the cells containing the strings/words of interest
FinalCell{k} = DummyCell;
end
Then you're good to go. Hope that helps!
The closest thing possible I found is:
clear all
cc{1} = {'str_row1col1', 'str_row1col2' };
cc{2} = {'str_row2col1', 'bighello', 'str_row1col3' };
cc{3} = {'str_row3col1'};
cc{4} = {'str_row3col1', 'helloWorld'};
cc1 = [cc{:}];
cc1 = cc1(~strcmp('bighello',cc1));
This reorganize your array into a one dimensional array and it cannot match regular expression, but only whole words.
For a better job I am afraid you have to use for loops.
I want to create a CoffeeScript range (like [4...496]) but using a length instead of an end range. This can be done with a loop like
myNum = getBigNumber()
newArray = ( n + myNum for n in [0...50] )
but I'm wondering if there is range-related shortcut that I'm missing. Is there something like
[getBigNumber()...].length(50) available in CoffeeScript?
You can just do
range = [myNum...myNum + 50]
Edit: As mu points out in the comments, CoffeeScript will add some complexity whether you use the snippet above or the original code. If performance is an issue, it might be better to drop down to plain JS for the loop (using backticks in the CoffeeScript code).
Assuming you want an ascending (i.e. low to high) range, you can do:
myNum = getBigNumber()
length = 50
range = new Array length
i = 0
`for(; i < length ; i++) { range[i] = i + myNum }` # raw, escaped JS
It's a lot faster than CoffeeScript's way of doing things, but note that CoffeeScript's range syntax also supports creating descending ranges by just flipping the boundary values. So CoffeeScript is (as always) easier on the eyes and simpler to work with, but raw JS is 3.5x faster in my test.