Consider this simple creation of list of list of list
x = [[[None]*2 for n in range(4)] for j in range(9)]; x
it printed the object and then suddenly doing the same exact code doesn't print anything. Tried del x and doing the code again, but no luck. doing type(x) doesn't print anything neither.
There's no error message.
I actually have the same problem with some other lines of code, including some that print different results with the same code executed at short notice.
Is this a problem of the worksheet or a Sage problem? It is very annoying as you never know if you did something right or not.
Related
I am having a problem running a spiking-neuron simulator. I keep getting the error message, "operation +: Warning adding a matrix with the empty matrix will give an empty matrix result." Now I'm writing this program in "Scilab," but I'm hoping the problem I am having will be clear for the educated eye regardess. What I am doing is converting an existing MATLAB program to Scilab. The original MATLAB program and an explanation can be found here: https://www.izhikevich.org/publications/spikes.pdf
What happens in my Scilab version is that the first pass through the loop produces all the expected values. I Know this becuase I hit pause at the end of the first run, right before "end," and check all the values and matrix elements. However, if I run the program proper, which includes a loop of 20 iterations, I get the error message above, and all of the matrix values are empty! I cannot figure out what the problem is. I am fairly new to programming so the answer may be very simple as far as I know. Here is the Scilab version of the program:
Ne=8; Ni=2;
re=rand(Ne,1); ri=rand(Ni,1);
a=[0.02*ones(Ne,1); 0.02+0.08*ri];
b=[0.2*ones(Ne,1); 0.25-0.05*ri];
c=[-65+15*re.^2; -65*ones(Ni,1)];
d=[8-6*re.^2; 2*ones(Ni,1)];
S=[0.5*rand(Ne+Ni,Ne), -rand(Ne+Ni,Ni)];
v=60*rand(10,1)
v2=v
u=b.*v;
firings=[];
for t=1:20
I=[5*rand(Ne,1,"normal");2*rand(Ni,1,"normal")];
fired=find(v>=30);
j = length(fired);
h = t*ones(j,1);
k=[h,fired'];
firings=[firings;k];
v(fired)=c(fired);
u(fired)=u(fired)+d(fired);
I=I+sum(S(:,fired),"c");
v=v+0.5*(0.04*v.^2+5*v+140-u+I);
v=v+0.5*(0.04*v.^2+5*v+140-u+I);
u=u+a.*(b.*v-u);
end
plot(firings(:,1), firings(:,2),".");
I tried everything to no avail. The program should run through 20 iterations and produce a "raster plot" of dots representing the fired neurons at each of the 20 time steps.
You can add the following line
oldEmptyBehaviour("on")
at the beginning of your script in order to prevent the default Scilab rule (any algebraic operation with an empty matrix yields an empty matrix). However you will still have some warnings (despite the result will be OK). As a definitive fix I recommend testing the emptyness of fired in your code, like this:
Ne=8; Ni=2;
re=rand(Ne,1); ri=rand(Ni,1);
a=[0.02*ones(Ne,1); 0.02+0.08*ri];
b=[0.2*ones(Ne,1); 0.25-0.05*ri];
c=[-65+15*re.^2; -65*ones(Ni,1)];
d=[8-6*re.^2; 2*ones(Ni,1)];
S=[0.5*rand(Ne+Ni,Ne), -rand(Ne+Ni,Ni)];
v=60*rand(10,1)
v2=v
u=b.*v;
firings=[];
for t=1:20
I=[5*rand(Ne,1,"normal");2*rand(Ni,1,"normal")];
fired=find(v>=30);
if ~isempty(fired)
j = length(fired);
h = t*ones(j,1);
k=[h,fired'];
firings=[firings;k];
v(fired)=c(fired);
u(fired)=u(fired)+d(fired);
I=I+sum(S(:,fired),"c");
end
v=v+0.5*(0.04*v.^2+5*v+140-u+I);
v=v+0.5*(0.04*v.^2+5*v+140-u+I);
u=u+a.*(b.*v-u);
end
plot(firings(:,1), firings(:,2),".");
The [] + 1 is not really defined in a mathematical sense. The operation might fail or produce different results depending on the software you use. For example:
Scilab 5 [] + 1 produces 1
Scilab 6 [] + 1 produces [] and a warning
Julia 1.8 [] .+ 1 produces [] but [] + 1 an error.
Python+Numpy 1.23 np.zeros((0,0)) + 1 produces [].
I suggest checking with size() or a comparison to the empty matrix to avoid such strange behaviour.
I'm in the process of working on a coding project in a pluto notebook. It's early days but I'm curious why the value of 1000 doesn't show beside quadorder in the screenshot above. It seems to output for everything else!
It is the ; in your code that triggers that. However, I don't think it is supposed to be expected behavior. It is supposed to be a comment, but the parser probably sees the ; and thinks that it is part of the code.
If you put it between quoting marks then this doesn't happen.
I.e. this should work as expected:
quadorder = 1000 # like python Julia doesn't require ';'
Otherwise, it would usually also work if you just don't put it at the end of the line.
The Julia REPL evaluates the semicolon wrongly and it is a know issue github.com/JuliaLang/julia/issues/28743.
You can actually trick it into some weird behavior that suppresses output. For example this returns no output:
julia> ";#"
julia> a = ["; #", "hi", 3]
julia> a = "223" #;
The reason is that the REPL parser looks for the last semicolon in the line and if there is any # or if it is at the end of the line (whitespaces don't matter), then it suppresses any output.
I've written a prime-generating function generatePrimes (full code here) that takes input bound::Int64 and returns a Vector{Int64} of all primes up to bound. After the function definition, I have the following code:
println("Generating primes...")
println("Last prime: ", generatePrimes(10^7)[end])
println("Primes generated.")
which prints, unexpectedly,
Generating primes...
9999991
Primes generated.
This output misses the "Last prime: " segment of the second print statement. The output does work as expected for smaller inputs; any input at least up to 10^6, but somehow fails for 10^7. I've tried several workarounds for this (e.g. assigning the returned value or converting it to a string before calling it in a print statement, combining the print statements, et cetera) and discovered some other weird behaviour: if the "Last prime", is removed from the second print statement, for input 10^7, the last prime doesn't print at all and all I get is a blank line between the first and third print statements. These issues are probably related, and I can't seem to find anything online about why some print statements wouldn't work in Julia.
Thanks so much for any clarification!
Edit: Per DNF's suggestion, following are some reductions to this issue:
Removing the first and last print statements doesn't change anything -- a blank line is always printed in the case I outlined and each of the cases below.
println(generatePrimes(10^7)[end]) # output: empty line
Calling the function and storing the last index in a variable before calling println doesn't change anything either; the cases below work exactly the same either way.
lastPrime::Int = generatePrimes(10^7)[end]
println(lastPrime) # output: empty line
If I call the function in whatever form immediately before a println, an empty line is printed regardless of what's inside the println.
lastPrime::Int = generatePrimes(10^7)[end]
println("This doesn't print") # output: empty line
println("This does print") # output: This does print
If I call the function (or print the pre-generated-and-stored function result) inside a println, anything before the function call (that's also inside the println) isn't printed. The 9999991 and anything else there may be after the function call is printed only if there is something else inside the println before the function call.
# Example 1
println(generatePrimes(10^7)[end]) # output: empty line
# Example 2
println("This first part doesn't print", generatePrimes(10^7)[end]) # output: 9999991
# Example 3
println("This first part doesn't print", generatePrimes(10^7)[end], " prints") # output: 9999991 prints
# Example 4
println(generatePrimes(10^7)[end], "prime doesn't print") # output: prime doesn't print
I could probably list twenty different variations of this same thing, but that probably wouldn't make things any clearer. In every single case version of this issue I've seen so far, the issue only manifests if there's that function call somewhere; println prints large integers just fine. That said, please let me know if anyone feels like they need more info. Thanks so much!
Most likely you are running this code from Atom Juno which recently has some issues with buffering standard output (already reported by others and I also sometimes have this problem).
One thing you can try to do is to flush your standard output
flush(stdout)
Like with any unstable bug restarting Atom Juno also seems to help.
I had the same issue. For me, changing the terminal renderer (File -> Settings -> Packages -> julia-client -> Terminal Options) from webgl to canvas (see pic below) seems to solve the issue.
change terminal renderer
I've also encountered this problem many times. (First time, it was triggered after using the debugger. It is probably unrelated but I have been using Julia+Juno for 2 weeks prior to this issue.)
In my case, the code before the println statement needed to have multiple dictionary assignation (with new keys) in order to trigger the behavior.
I also confirmed that the same code ran in Command Prompt (with same Julia interpreter) prints fine. Any hints about how to further investigate this will be appreciated.
I temporarily solve this issue by printing to stderr, thinking that this stream has more stringent flush mechanism: println(stderr, "hello!")
I'm trying to create a schedule with regular expressions and given this piece of code, I don't understand the functionality of "next" here and how it could be compared to "continue", "pass" or "break", etc.
for j, line in enumerate(lines):
m = reg_expr_w.match(line)
if (m != None):
week_variabel = m.group(4)
next
This is a very odd thing to have. next is a function. So that line doesn't do anything. Could you be missing part of the code? Otherwise this is a mistake of some kind, either a bug or a remnant of some older piece of code that wasn't fully cleaned up. Where did you get it?
I am using the Kernel Density Estimator toolbox form http://www.ics.uci.edu/~ihler/code/kde.html . But I am getting the following error when I try to execute the demo files -
>> demo_kde_3
KDE Example #3 : Product sampling methods (single, anecdotal run)
Attempt to reference field of non-structure array.
Error in double (line 10)
if (npd.N > 0) d = 1; % return 1 if the density exists
Error in repmat (line 49)
nelems = prod(double(siz));
Error in kde (line 39)
if (size(ks,1) == 1) ks = repmat(ks,[size(points,1),1]); end;
Error in demo_kde_3 (line 8)
p = kde([.1,.45,.55,.8],.05); % create a mixture of 4 gaussians for
testing
Can anyone suggest what might be wrong? I am new to Matlab and having a hard time to figure out the problem.
Thank You,
Try changing your current directory away from the #kde folder; you may have to add the #kde folder to your path when you do this. For example run:
cd('c:\');
addpath('full\path\to\the\folder\#kde');
You may also need to add
addpath('full\path\to\the\folder\#kde\examples');
Then see if it works.
It looks like function repmat (a mathworks function) is picking up the #kde class's version of the double function, causing an error. Usually, only objects of the class #kde can invoke that functions which are in the #kde folder.
I rarely use the #folder form of class definitions, so I'm not completely sure of the semantics; I'm curious if this has any effect on the error.
In general, I would not recommend using the #folder class format for any development that you do. The mathworks overhauled their OO paradigm a few versions ago to a much more familiar (and useful) format. Use help classdef to see more. This #kde code seems to predate this upgrade.
MATLAB gives you the code line where the error occurs. As double and repmat belong to MATLAB, the bug probably is in kde.m line 39. Open that file in MATLAB debugger, set a breakpoint on that line (so the execution stops immediately before the execution of that specific line), and then when the code is stopped there, check the situation. Try the entire code line in console (copy-paste or type it, do not single-step, as causing an uncatched error while single-stepping ends the execution of code in debugger), it should give you an error (but doesn't stop execution). Then try pieces of the code of that code line, what works as it should and what not, eg. does the result of size(points, 1) make any sense.
However, debugging unfamiliar code is not an easy task, especially if you're a beginner in MATLAB. But if you learn and understand the essential datatypes of MATLAB (arrays, cell arrays and structs) and the different ways they can be addressed, and apply that knowledge to the situation on the line 39 of kde.m, hopefully you can fix the bug.
Repmat calls double and expects the built-in double to be called.
However I would guess that this is not part of that code:
if (npd.N > 0) d = 1; % return 1 if the density exists
So if all is correct this means that the buil-tin function double has been overloaded, and that this is the reason why the code crashes.
EDIT:
I see that #Pursuit has already addressed the issue but I will leave my answer in place as it describes the method of detection a bit more.