I am trying to create a model with 'sequence.gather' operator, but getting an error "Where operation can only operate on scalar input" when calling 'train_minibatch'.
input_seq_axis = Axis('inputAxis')
input_sequence = sequence.input_variable(shape=vocab_dim, sequence_axis=input_seq_axis)
vowel_mask_sequence = sequence.input_variable(shape=2, sequence_axis=input_seq_axis)
a = Sequential([
C.layers.Recurrence(C.layers.LSTM(hidden_dim)),
])
b=C.sequence.gather(a(input_sequence),vowel_mask_sequence)
z=Dense(3)(b)
label_sequence = sequence.input_variable(3, sequence_axis=z.dynamic_axes[1])
How can I fix the error ? I even dont use 'where' operator.
For sequence.gather(x, y), y has to be a scalar, that is to say:
assert y.shape == (1,)
The values of y must be either 0 or 1, and also with the same exact dynamic axis as x.
An example on how to use sequence.gather from a library i maintain.
Related
Is it possible (and if so how) to use an objective function that has a conditional expression?
Changing the example from the docs, I would like an expression like:
def objective_function(model):
return model.x[0] if model.x[1] < const else model.x[2]
model.Obj = Objective(rule=objective_function, sense=maximize)
Can this be modelled directly like this or do I have to consider some sort of transformation (and if so how would this look like)?
Just executing the above gives an error message like:
Evaluating Pyomo variables in a Boolean context, e.g.
if expression <= 5:
is generally invalid. If you want to obtain the Boolean value of the
expression based on the current variable values, explicitly evaluate the
expression using the value() function:
if value(expression) <= 5:
or
if value(expression <= 5):
which I think is because Pyomo thinks I'd like to obtain a value, instead of an expression with the variable.
One way to formulate that is by using a logical disjunction. You can look into the Pyomo.GDP documentation for usage, but it would look like:
m.helper_var = Var()
m.obj = Objective(expr=m.helper_var)
m.lessthan = Disjunct()
m.lessthan.linker = Constraint(expr=m.helper_var == m.x[0])
m.lessthan.constr = Constraint(expr=m.x[1] < const)
m.greaterthan = Disjunct()
m.greaterthan.linker = Constraint(expr=m.helper_var == m.x[2])
m.greaterthan.constr = Constraint(expr=m.x[1] >= const)
m.lessthanorgreaterthan = Disjunction(expr=[m.lessthan, m.greaterthan])
# some kind of transformation (convex hull or big-M)
You can also formulate this using complementarity constraints.
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}
From the documentation, we see the following example:
g = gallery('integerdata',3,[15,1],1);
x = gallery('uniformdata',[15,1],9);
y = gallery('uniformdata',[15,1],2);
A = table(g,x,y)
func = #(x, y) (x - y);
B = rowfun(func,A,...
'GroupingVariable','g',...
'OutputVariableName','MeanDiff')
When the function func is applied to A in rowfun how does it know that there are variables in A called x and y?
EDIT: I feel that my last statement must not be true, as you do not get the same result if you did A = table(g, y, x).
I am still very confused by how rowfun can use a function that does not actually use any variables defined within the calling environment.
Unless you specify the rows (and their order) with the Name/Value argument InputVariables, Matlab will simply take column 1 as first input, column 2 as second input etc, ignoring eventual grouping columns.
Consequently, for better readability and maintainability of your code, I consider it good practice to always specify InputVariables explicitly.
I have a problem with If statement in OpenScad.
I have 4 variables
a=20;
b=14;
w=1;
c=16;
I want to check witch number is bigger a or b.
And after depending who is smaller to take the value of smaller variable(in our case b < a) and to make a simple operation with c variable ( c=b-w).
I tried like this but it doesn't work.
a=20;
b=14;
w=1;
c=16;
if(a>b)
{
c=b-w;
}
if (a<b)
{
c=a-w;
}
if (a==b)
{
c=a-w;
}
It seems logic, but in openscad as I understood you can't change the value of variable inside a If statement. What trick can I use in order to get my goal.
Thank you!
in the 3. leg you confused the assignment-operator „=“ with the equal-operator „==“ (correct if (a==b)).
in your 3. leg you do the same as in the 2., so you could handle both as an „else“-leg.
Correct: assignment is not allowed in if-statement. In openscad you can use the ? operator instead:
c = a > b ? b-w : a-w;
After = follows the condition. The statement after the ? becomes the value if the condition is true, and the statement after the : becomes the value if the condition is false. Nested conditions are possible, e.g. your conditions:
c = a > b ? b-w : (a < b ? a-w : a-w);
More information in the documentation.
OpenSCAD's variable assignment is different. You can only assign variables inside a bracket. So c = b - w will only be assigned inside the if bracket. Outside if this bracket it will still be 16. Don't ask me why. You can read more in the Documentation of OpenSCAD.
c = min(c,min(a,b)/2-w);
this also solve the problem )
temp(i,1) = rand(1)*(pb(1,num).pos(i,1) - pw(1,num).pos(i,1));
This line gives the following error:
Error using ==> minus
Not enough input arguments.
The following are the definitions of pb and pw.
pw=struct('fitness',[],'pos',{});
pb=struct('fitness',[],'pos',{});
pos is a 2 x 1 array.
When tracking down errors like this, I break the problem up into smaller bits. Especially when the logic isn't readily apparent. Not only does it provide a path that can be used to step through your function using the debugger, but it also makes it more readable.
I've taken liberty with the intermediate variable names.
thisPb = pb(1,num);
thisPw = pw(1,num);
initialPos= pw.pos(i,1);
finalPos = pb.pos(i,1);
whos initialPos finalPos
temp(i,1) = rand(1) * (finalPos - initialPos);
The line with whos will print out the values. Make sure that finalPos and initialPos are both numbers.
One way that you can get this error is when num is an empty matrix.
The expression
>> s(x).a
can return a variable number of outputs, depending on the size of x.
If x = [1,2,3] for example, it will return three values (as long as s has at least three elements).
If x = [] on the other hand, then s(x).a will return no outputs, so the expression
>> disp(s(x).a)
will give you a Not enough input arguments error, which is almost certainly what you're seeing. You should check that num is not empty.
Are you sure, that all values are really initialised? Try to check this before your codeline.
disp(pb(1,num).pos(i,1))
disp(pw(1,num).pos(i,1))
temp(i,1) = rand(1)*(pb(1,num).pos(i,1) - pw(1,num).pos(i,1));