Hello everyone I am working on CIFAR10 dataset using pytorch. I have developed a model which works absolutely fine but the main problem occurrs while runing the following code:
import time
start_time=time.time()
epochs=5
train_losses=[]
test_losses=[]
train_correct=[]
test_correct=[]
for i in range(epochs):
tsn_corr=0
tst_corr=0
for b, (X_train,y_train) in enumerate(train_loader):
b+=1
y_pred=model(X_train)
loss=criterion(y_pred,y_train)
#Tally the number of correct predictions
predicted= torch.max(y_pred.data, 1)[1]
batch_corr=(predicted==y_train).sum()
tsn_corr += batch_corr
#optimize paramters
optimizer.zero_grad()
loss.backward()
optimizer.step()
#print interim results
if b%600 == 0:
print(f"epochs: {i}, batch: {b}, loss: {loss.item():10.8f}")
loss=loss.detach().numpy()
train_losses.append(loss)
train_correct.append(tsn_corr)
#Running the test_batches
with torch.no_grad():
for b, (X_test,y_test) in enumerate(test_loader):
b+=1
y_val=model(X_test)
#TALLY THE NUMBER OF CORRECT PREDICTIONS
predicted=torch.max(y_val.data, 1)[1]
batch_corr= (predicted==y_test).sum()
tst_corr += batch_corr
loss=criterion(y_val,y_test)
loss=loss.detach().numpy()
test_losses.append(loss)
test_correct.append(tst_corr)
the following error occurrs while running the following code:
NotImplementedError Traceback (most recent call last)
<ipython-input-43-48e21e83e9f7> in <module>
15 b+=1
16
---> 17 y_pred=model(X_train)
18 loss=criterion(y_pred,y_train)
19
~\Anaconda3\lib\site-packages\torch\nn\modules\module.py in _call_impl(self, *input, **kwargs)
887 result = self._slow_forward(*input, **kwargs)
888 else:
--> 889 result = self.forward(*input, **kwargs)
890 for hook in itertools.chain(
891 _global_forward_hooks.values(),
~\Anaconda3\lib\site-packages\torch\nn\modules\module.py in _forward_unimplemented(self, *input)
199 registered hooks while the latter silently ignores them.
200 """
--> 201 raise NotImplementedError
202
203
NotImplementedError:
Can someone tell me what can I do to fix this code. Apart from this previous all codes work fine and the model which I made using ConvolutionalNeural-Networks also runs successfully meaning that there is no problem with the model. I guess this detail might help. It might be noted that this code works just fine on MNIST dataset. I dont know what is the problem with CIFAR datasets
Your model class needs to implement a forward method. See the PyTorch Example on Subclassing to see an example.
everyone,
I'm trying to do some calculations and plot the results, but it seems that these are too heavy for Maxima. When I try to calculate N1 and N2 the program crashes when parameter j is too high or when I try to plot them, the program displays the following error message: "Heap exhausted, game over." What should I do? I've seen some people saying to try to compile Maxima with ccl, but I don't know how to do it or if it will work.
I usually receive error messages like:
Message from maxima's stderr stream: Heap exhausted during garbage collection: 0 bytes available, 16 requested.
Gen Boxed Unboxed LgBox LgUnbox Pin Alloc Waste Trig WP GCs Mem-age
0 0 0 0 0 0 0 0 20971520 0 0 0,0000
1 0 0 0 0 0 0 0 20971520 0 0 0,0000
2 0 0 0 0 0 0 0 20971520 0 0 0,0000
3 16417 2 0 0 43 1075328496 707088 293986768 16419 1 0,8032
4 13432 21 0 1141 70 955593760 838624 2000000 14594 0 0,2673
5 0 0 0 0 0 0 0 2000000 0 0 0,0000
6 741 184 34 28 0 63259792 1424240 2000000 987 0 0,0000
7 0 0 0 0 0 0 0 2000000 0 0 0,0000
Total bytes allocated = 2094182048
Dynamic-space-size bytes = 2097152000
GC control variables:
*GC-INHIBIT* = true
*GC-PENDING* = true
*STOP-FOR-GC-PENDING* = false
fatal error encountered in SBCL pid 13884(tid 0000000001236360):
Heap exhausted, game over.
Here goes the code:
enter code here
a: 80$;
b: 6*a$;
h1: 80$;
t: 2$;
j: 5$;
carga: 250$;
sig: -carga/2$;
n: 2*q*%pi/b$;
m: i*%pi/a$;
i: 2*p-1$;
i1: 2*p1-1$;
/*i1: p1$;*/
Φ: a/b$;
τ: cosh(x) - (x/sinh(x))$;
σ: sinh(x) - (x/cosh(x))$;
Ψ: sinh(x)/τ$;
Χ: cosh(x)/σ$;
Λ0: 1/(((i/2)^2+Φ^2*q^2)^2)$;
Λ1: sum((((i/2)^3*subst([x=(i*%pi/(2*Φ))],Ψ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ0, p, 1, j)$;
Λ2: sum(((q1^3*subst([x=(q1*%pi*Φ)],Χ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ1, q1, 1, j)$;
Λ3: sum((((i/2)^3*subst([x=(i*%pi/(2*Φ))],Ψ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ2, p, 1, j)$;
Λ4: sum(((q1^3*subst([x=(q1*%pi*Φ)],Χ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ3, q1, 1, j)$;
Λ5: sum((((i/2)^3*subst([x=(i*%pi/(2*Φ))],Ψ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ4, p, 1, j)$;
Ζ0: sum(((q^3*subst([x=(q*%pi*Φ)],Χ))/(((i1/2)^2+Φ^2*q^2)^2))*Λ0, q, 1, j)$;
Ζ2: sum(((q^3*subst([x=(q*%pi*Φ)],Χ))/(((i1/2)^2+Φ^2*q^2)^2))*Λ2, q, 1, j)$;
Ζ4: sum(((q^3*subst([x=(q*%pi*Φ)],Χ))/(((i1/2)^2+Φ^2*q^2)^2))*Λ4, q, 1, j)$;
E: 200000$;
ν: 0.3$;
λ: (ν*E)/((1+ν)*(1-2*ν))$;
μ: E/(2*(1+ν))$;
a0: float(1/(b/2)*integrate(0, y, -(b/2), -h1/2)+1/b*integrate(sig, y, -h1/2, h1/2)+1/(b/2)*integrate(0, y, h1/2, (b/2)))$;
aq: float(1/(b/2)*integrate(0*cos(q*y*%pi/(b/2)), y, -(b/2), - h1/2)+1/(b/2)*integrate(sig*cos(q*y*%pi/(b/2)), y, -h1/2, h1/2)+1/(b/2)*integrate(0*cos(q*y*%pi/(b/2)), y, h1/2, (b/2)))$;
aq1: float(1/(b/2)*integrate(0*cos(q1*y*%pi/(b/2)), y, -(b/2), - h1/2)+1/(b/2)*integrate(sig*cos(q1*y*%pi/(b/2)), y, -h1/2, h1/2)+1/(b/2)*integrate(0*cos(q1*y*%pi/(b/2)), y, h1/2, (b/2)))$;
Bq: aq/((λ+μ)*subst([x=q*%pi*Φ],σ))+((16*Φ^4*q^2*(-1)^q)/((λ+μ)*%pi^2*subst([x=q*%pi*Φ],σ)))*sum(q1*aq1*(-1) ^q1*subst([x=q1*%pi*Φ],Χ)*(Λ1+(16*Φ^4/(%pi^2))*Λ3+((16*Φ^4/(%pi^2))^2)*Λ5), q1, 1, j)+(8*λ*Φ^3*q^2*(-1)^q*a0)/((λ+μ)*(λ+2*μ)*(%pi^3)*subst([x=q*%pi*Φ],σ))*sum(subst([x=i*%pi/(2*Φ)],Ψ)/(i/ 2)*(Λ0+(16*Φ^4/(%pi^2))*Λ2+((16*Φ^4/(%pi^2))^2)*Λ4), p, 1, j)$;
βp: -(2*λ*a0*(-1)^((i-1)/2))/((λ+μ)*(λ+2*μ)*(i/2)^2*%pi^2*subst([x=i*%pi/(2*Φ)],τ))-((32*λ*Φ^4*(i/2)^2*a0*(-1)^((i-1)/2))/((λ+μ)*(λ+2*μ)*%pi^2*subst([x=i*%pi/(2*Φ)],τ)))*sum(((subst([x=i1*%pi/(2*Φ)],Ψ))/(i1/2))*(Ζ0+Ζ2*((16*Φ^4)/%pi^2)+Ζ4*(((16*Φ^4)/%pi^2)^2)),p1,1,j)-((4*Φ*(i/2)^2*(-1)^((i-1)/2))/((λ+μ)*%pi*subst([x=i*%pi/(2*Φ)],τ)))*sum(q*aq*(-1)^q*subst([x=q*%pi*Φ],Χ)*(Λ0+Λ2*(16*Φ^4/%pi^2)+Λ4*(16*Φ^4/%pi^2)^2),q,1,j)$;
N1: (2*a0/a)*x+(λ+μ)*sum(Bq*((1+((n*a*sinh(n*a/2))/(2*cosh(n*a/2))))*sinh(n*x)-n*x*cosh(n*x))*cos(n*y),q,1,j)+(λ+μ)*sum(βp*((1-((m*b*cosh(m*b/2))/(2*sinh(m*b/2))))*cosh(m*y)+m*y*sinh(m*y))*sin(m*x),p,1,j)$;
N2: ((2*λ*a0)/(a*(λ+2*μ)))*x+(λ+μ)*sum(Bq*((1-((n*a*sinh(n*a/2))/(2*cosh(n*a/2))))*sinh(n*x)+n*x*cosh(n*x))*cos(n*y),q,1,j)+(λ+μ)*sum(βp*((1+((m*b*cosh(m*b/2))/(2*sinh(m*b/2))))*cosh(m*y)-m*y*sinh(m*y))*sin(m*x),p,1,j);
wxplot3d(N1, [x,-a/2,a/2], [y,-b/2,b/2])$;
wxplot3d(N2, [x,-a/2,a/2], [y,-b/2,b/2])$;
This is not a complete answer, since I don't know how this should work with wxMaxima: I would suggest that you ask the developers. However it's too long for a comment and I think might be useful to people, and it does answer the question of how you solve the heap-size limit for Maxima itself when using SBCL, at least when run on Linux or some other platform with a command-line.
As a note, I suspect that the underlying problem is not the heap size, but that the calculation is blowing up in some horrible way: the best fix is probably to understand what's blowing up and fix that. See Robert Dodier's answer, which is probably going to be a lot more helpful. However, if heap size is the problem, this is how you deal with it for Maxima.
The trick is that you can tell SBCL what the heap limit should be by passing it the --dynamic-space-size <MB> argument, and you can pas arguments through the maxima wrapper to do this.
Here is a transcript of Maxima, being run on Linux, with SBCL as a back end (this is a version built from source: the packaged version will I assume be the same):
$ maxima
Maxima 5.43.2 http://maxima.sourceforge.net
using Lisp SBCL 2.0.0
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) :lisp (sb-ext:dynamic-space-size)
1073741824
So, on this system the defaule heap limit is 1GB (this is SBCL's default limit on the platform).
Now we can pass the -X <lisp options> aka --lisp-options=<lisp options> option to the maxima wrapper to pass the appropriate option through to sbcl:
$ maxima -X '--dynamic-space-size 2000'
Lisp options: (--dynamic-space-size 2000)
Maxima 5.43.2 http://maxima.sourceforge.net
using Lisp SBCL 2.0.0
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) :lisp (sb-ext:dynamic-space-size)
2097152000
As you can see this has doubled the heap size.
If someone knows the answer for wxMaxima then please do add an edit to this answer: I can't experiment it because all my Linux VMs are headless.
Also not a complete answer here, but some more notes and pointers which I hope will help.
To make the problem easier for Maxima to digest, use only exact numbers (integers and ratios), and avoid float and numer. (Plotting functions will apply float and numer automatically.) I changed 0.3 to 3/10 and cut out the calls to float.
Also, try setting j to a smaller number (I tried j equal to 1) to try to work all the way through the problem before increasing it to 5 again.
Also, replace all sum and integrate with 'sum and 'integrate (i.e. noun expressions instead of verb expressions). Take a look at the summands and integrands to see if they look right. You can evaluate the sums and/or integrals or both via ev(expr, sum) or ev(expr, integrate) or ev(expr, nouns) to evaluate 'sum, 'integrate, or all noun expressions, respectively.
With j equal to 1, I get the following expression for N1:
(2500000*((-(13*cosh(%pi/6)
*((8503056*cosh(%pi/6)^2*sinh(3*%pi)^2)
/(9765625*%pi^4
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^2
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+(52488*cosh(%pi/6)*sinh(3*%pi))
/(15625*%pi^2*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))
+324/25))
/(120000*%pi^2*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))))
+(13*sinh(3*%pi)
*((2754990144*cosh(%pi/6)^3*sinh(3*%pi)^2)
/(244140625*%pi^4
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^3
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+(17006112*cosh(%pi/6)^2*sinh(3*%pi))
/(390625*%pi^2
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^2
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))
+(104976*cosh(%pi/6))
/(625*(sinh(%pi/6)-%pi/(6*cosh(%pi/6))))))
/(22680000*%pi^2*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+13/(35000*%pi^2*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))))
*sin((%pi*(2*p-1)*x)/80)
*((%pi*(2*p-1)*y*sinh((%pi*(2*p-1)*y)/80))/80
+(1-(3*%pi*(2*p-1)*cosh(3*%pi*(2*p-1)))
/sinh(3*%pi*(2*p-1)))
*cosh((%pi*(2*p-1)*y)/80)))
/13
+(2500000*((-(13*cosh(%pi/6)
*((344373768*cosh(%pi/6)^2*sinh(3*%pi)^3)
/(244140625*%pi^4
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
^2
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))
^3)
+(2125764*cosh(%pi/6)*sinh(3*%pi)^2)
/(390625*%pi^2
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+(13122*sinh(3*%pi))
/(625*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))))
/(1620000*%pi^3*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^2))
+(13*sinh(3*%pi)
*((8503056*cosh(%pi/6)^2*sinh(3*%pi)^2)
/(9765625*%pi^4
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^2
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+(52488*cosh(%pi/6)*sinh(3*%pi))
/(15625*%pi^2*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))
+324/25))
/(3780000*%pi^3*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))
-13/(20000*%pi*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))))
*(((%pi*sinh(%pi/6))/(6*cosh(%pi/6))+1)
*sinh((%pi*x)/240)
-(%pi*x*cosh((%pi*x)/240))/240)*cos((%pi*y)/240))
/13-(25*x)/48$
Now in order to plot that, it should be a function of x and y only. However listofvars reports that it contains x, y, and p. Hmm. I see that βp has a summation over p1 but it contains Ζ0, which contains Λ0, which contains p. Is the summation over p1 supposed to be over p? Is the summand supposed to contain p1 instead of p?
Likewise it appears that N2, after evaluating the sums and integrals with j equal to 1, contains p.
Maybe you need to rework the formulas somewhat? I don't know what the correct form might be.
I'm trying to print the outcome of an anova like so:
library(pander)
m.aov = aov(Sepal.Width ~ Species * Sepal.Length, iris)
pander(m.aov, split.table=Inf)
and I get this as expected if I type it into the console:
----------------------------------------------------------------------
Df Sum Sq Mean Sq F value Pr(>F)
-------------------------- ---- -------- --------- --------- ---------
**Species** 2 11.34 5.672 76.48 2.329e-23
**Sepal.Length** 1 4.769 4.769 64.3 3.368e-13
**Species:Sepal.Length** 2 1.513 0.7566 10.2 7.19e-05
**Residuals** 144 10.68 0.07417 NA NA
----------------------------------------------------------------------
Table: Analysis of Variance Model
However, if I embed this into a knitr chunk, I don't get the table:
```{r, results='asis'}
library(pander)
m.aov = aov(Sepal.Width ~ Species * Sepal.Length, iris)
pander(m.aov, split.table=Inf)
```
Knit the above and one obtains
```r
pander(m.aov, split.table=Inf)
```
i.e., the code chunk with no output.
Question: Is this a bug (in knitr? pander?) or something I've overlooked? How can I work around it?
> sessionInfo()
R version 3.0.2 (2013-09-25)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_AU.UTF-8 LC_NUMERIC=C LC_TIME=en_AU.UTF-8
[4] LC_COLLATE=en_AU.UTF-8 LC_MONETARY=en_AU.UTF-8 LC_MESSAGES=en_AU.UTF-8
[7] LC_PAPER=en_AU.UTF-8 LC_NAME=C LC_ADDRESS=C
[10] LC_TELEPHONE=C LC_MEASUREMENT=en_AU.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] knitr_1.8 pander_0.5.1 vimcom_1.0-0 setwidth_1.0-3 colorout_1.0-3
loaded via a namespace (and not attached):
[1] digest_0.6.4 evaluate_0.5.5 formatR_1.0 Rcpp_0.11.2 stringr_0.6.2 tools_3.0.2
To run a certain software I'm using .txt-input files which I need to manipulate with Matlab.
I know how to do it, and I didn't expected problems. As it was not working I reduced my manipulation script to a minimum, so actually nothing is changed. Except some white spaces, and the other software seems to react very sensitive on that.
parts of my file look like that:
...
*CONTROL_TERMINATION
$# endtim endcyc dtmin endeng endmas
1.000000 0 0.000 0.000 0.000
*CONTROL_TIMESTEP
$# dtinit tssfac isdo tslimt dt2ms lctm erode ms1st
0.000 0.900000 0 0.000 -1.000E-4 0 0 0
$# dt2msf dt2mslc imscl
0.000 0 0
...
I'm loading it to Matlab and directly save it again without changes:
% read original file
fid = fopen('filename.txt','r');
param = textscan(fid,'%s','delimiter','\n');
rows = param{1,1};
fclose(fid);
% overwrite to new file
fid = fopen('filename.txt','w');
fprintf(fid, '%s\r\n', rows{:});
fclose(fid);
The output file is lacking of the white spaces at the begin of every line, that seems to be the only difference of input and output file. (at least I hope so)
...
*CONTROL_TERMINATION
$# endtim endcyc dtmin endeng endmas
1.000000 0 0.000 0.000 0.000
*CONTROL_TIMESTEP
$# dtinit tssfac isdo tslimt dt2ms lctm erode ms1st
0.000 0.900000 0 0.000 -1.000E-4 0 0 0
$# dt2msf dt2mslc imscl
0.000 0 0
...
Though it seems weird to me, that this should be the reason - what can I change, that both files look 100% identical? The problem I'm having is that the white spaces have different lengths.
You can use the whitespace option in textscan, and setting it to an empty string.
param = textscan(fid,'%s','delimiter','\n','whitespace','');
By default, textscan does not include leading white-space characters in the processing of any data fields (doc center).