How to print a particular constraint in pulp for large problems - printing

I am running a OR problem in Pulp with 600*159 decision variables and large set of constraints . every time if i add new constraint to problem to check the correctness of constraint, i put print(model) . my program not able to display due to high memory to show in console. so i use 'writelp' to download file and verify which is cumbersome. can someone tell how to print only the constraint ,i need to see in console or any other option available.

When I want to do something like this, it's easy to just store the constraint to a temporary variable, print that, and then add it to the problem:
>>> x = pl.LpVariable(f"x", lowBound=0, upBound=None)
>>> constraint = x <= 3
>>> print(constraint)
x <= 3
>>> prob += constraint
A constraint is an instance of pulp.LpConstraint, which can be printed, since it has __str__() and __repr__() methods. In other words, it's just a python object.

Related

Provide custom gradient to drake::MathematicalProgram

Drake has an interface where you can give it a generic function as a constraint and it can set up the nonlinearly-constrained mathematical program automatically (as long as it supports AutoDiff). I have a situation where my constraint does not support AutoDiff (the constraint function conducts a line search to approximate the maximum value of some function), but I have a closed-form expression for the gradient of the constraint. In my case, the math works out so that it's difficult to find a point on this function, but once you have that point it's easy to linearize around it.
I know many optimization libraries will allow you to provide your own analytical gradient when available; can you do this with Drake's MathematicalProgram as well? I could not find mention of it in the MathematicalProgram class documentation.
Any help is appreciated!
It's definitely possible, but I admit we haven't provided helper functions that make it pretty yet. Please let me know if/how this helps; I will plan to tidy it up and add it as an example or code snippet that we can reference in drake.
Consider the following code:
from pydrake.all import AutoDiffXd, MathematicalProgram, Solve
prog = MathematicalProgram()
x = prog.NewContinuousVariables(1, 'x')
def cost(x):
return (x[0]-1.)*(x[0]-1.)
def constraint(x):
if isinstance(x[0], AutoDiffXd):
print(x[0].value())
print(x[0].derivatives())
return x
cost_binding = prog.AddCost(cost, vars=x)
constraint_binding = prog.AddConstraint(
constraint, lb=[0.], ub=[2.], vars=x)
result = Solve(prog)
When we register the cost or constraint with MathematicalProgram in this way, we are allowing that it can get called with either x being a float, or x being an AutoDiffXd -- which is simply a wrapping of Eigen's AutoDiffScalar (with dynamically allocated derivatives of type double). The snippet above shows you roughly how it works -- every scalar value has a vector of (partial) derivatives associated with it. On entry to the function, you are passed x with the derivatives of x set to dx/dx (which will be 1 or zero).
Your job is to return a value, call it y, with the value set to the value of your cost/constraint, and the derivatives set to dy/dx. Normally, all of this happens magically for you. But it sounds like you get to do it yourself.
Here's a very simple code snippet that, I hope, gets you started:
from pydrake.all import AutoDiffXd, MathematicalProgram, Solve
prog = MathematicalProgram()
x = prog.NewContinuousVariables(1, 'x')
def cost(x):
return (x[0]-1.)*(x[0]-1.)
def constraint(x):
if isinstance(x[0], AutoDiffXd):
y = AutoDiffXd(2*x[0].value(), 2*x[0].derivatives())
return [y]
return 2*x
cost_binding = prog.AddCost(cost, vars=x)
constraint_binding = prog.AddConstraint(
constraint, lb=[0.], ub=[2.], vars=x)
result = Solve(prog)
Let me know?

Other ways to call/eval dynamic strings in Lua?

I am working with a third party device which has some implementation of Lua, and communicates in BACnet. The documentation is pretty janky, not providing any sort of help for any more advanced programming ideas. It's simply, "This is how you set variables...". So, I am trying to just figure it out, and hoping you all can help.
I need to set a long list of variables to certain values. I have a userdata 'ME', with a bunch of variables named MVXX (e.g. - MV21, MV98, MV56, etc).
(This is all kind of background for BACnet.) Variables in BACnet all have 17 'priorities', i.e., every BACnet variable is actually a sort of list of 17 values, with priority 16 being the default. So, typically, if I were to say ME.MV12 = 23, that would set MV12's priority-16 to the desired value of 23.
However, I need to set priority 17. I can do this in the provided Lua implementation, by saying ME.MV12_PV[17] = 23. I can set any of the priorities I want by indexing that PV. (Corollaries - what is PV? What is the underscore? How do I get to these objects? Or are they just interpreted from Lua to some function in C on the backend?)
All this being said, I need to make that variable name dynamic, so that i can set whichever value I need to set, based on some other code. I have made several attempts.
This tells me the object(MV12_PV[17]) does not exist:
x = 12
ME["MV" .. x .. "_PV[17]"] = 23
But this works fine, setting priority 16 to 23:
x = 12
ME["MV" .. x] = 23
I was trying to attempt some sort of what I think is called an evaluation, or eval. But, this just prints out function followed by some random 8 digit number:
x = 12
test = assert(loadstring("MV" .. x .. "_PV[17] = 23"))
print(test)
Any help? Apologies if I am unclear - tbh, I am so far behind the 8-ball I am pretty much grabbing at straws.
Underscores can be part of Lua identifiers (variable and function names). They are just part of the variable name (like letters are) and aren't a special Lua operator like [ and ] are.
In the expression ME.MV12_PV[17] we have ME being an object with a bunch of fields, ME.MV12_PV being an array stored in the "MV12_PV" field of that object and ME.MV12_PV[17] is the 17th slot in that array.
If you want to access fields dynamically, the thing to know is that accessing a field with dot notation in Lua is equivalent to using bracket notation and passing in the field name as a string:
-- The following are all equivalent:
x.foo
x["foo"]
local fieldname = "foo"
x[fieldname]
So in your case you might want to try doing something like this:
local n = 12
ME["MV"..n.."_PV"][17] = 23
BACnet "Commmandable" Objects (e.g. Binary Output, Analog Output, and o[tionally Binary Value, Analog Value and a handful of others) actually have 16 priorities (1-16). The "17th" you are referring to may be the "Relinquish Default", a value that is used if all 16 priorities are set to NULL or "Relinquished".
Perhaps your system will allow you to write to a BACnet Property called "Relinquish Default".

Modify values programmatically SPSS

I have a file with more than 250 variables and more than 100 cases. Some of these variables have an error in decimal dot (20445.12 should be 2.044512).
I want to modify programatically these data, I found a possible way in a Visual Basic editor provided by SPSS (I show you a screen shot below), but I have an absolute lack of knowledge.
How can I select a range of cells in this language?
How can I store the cell once modified its data?
--- EDITED NEW DATA ----
Thank you for your fast reply.
The problem now its the number of digits that number has. For example, error data could have the following format:
Case A) 43998 (five digits) ---> 4.3998 as correct value.
Case B) 4399 (four digits) ---> 4.3990 as correct value, but parsed as 0.4399 because 0 has been removed when file was created.
Is there any way, like:
IF (NUM < 10000) THEN NUM = NUM / 1000 ELSE NUM = NUM / 10000
Or something like IF (Number_of_digits(NUM)) THEN ...
Thank you.
there's no need for VB script, go this way:
open a syntax window, paste the following code:
do repeat vr=var1 var2 var3 var4.
compute vr=vr/10000.
end repeat.
save outfile="filepath\My corrected data.sav".
exe.
Replace var1 var2 var3 var4 with the names of the actual variables you need to change. For variables that are contiguous in the file you may use var1 to var4.
Replace vr=vr/10000 with whatever mathematical calculation you would like to use to correct the data.
Replace "filepath\My corrected data.sav" with your path and file name.
WARNING: this syntax will change the data in your file. You should make sure to create a backup of your original in addition to saving the corrected data to a new file.

Logic behind COBOL code

I am not able to understand what is the logic behind these lines:
COMPUTE temp = RESULT - 1.843E19.
IF temp IS LESS THAN 1.0E16 THEN
Data definition:
000330 01 VAR1 COMP-1 VALUE 3.4E38. // 3.4 x 10 ^ 38
Here are those lines in context (the sub-program returns a square root):
MOVE VAR1 TO PARM1.
CALL "SQUAREROOT_ROUTINE" USING
BY REFERENCE PARM1,
BY REFERENCE RESULT.
COMPUTE temp = RESULT - 1.843E19.
IF temp IS LESS THAN 1.0E16 THEN
DISPLAY "OK"
ELSE
DISPLAY "False"
END-IF.
These lines are just trying to test if the result returned by the SQUAREROOT_ROUTINE is correct. Since the program is using float-values and rather large numbers this might look a bit complicated. Let's just do the math:
You start with 3.4E38, the squareroot is 1.84390889...E19.
By subtracting 1.843E19 (i.e. the approximate result) and comparing the difference against 1.0E16 the program is testing whether the result is between 1.843E19 and 1.843E19+1.0E16 = 1.844E19.
Not that this test would not catch an error if the result from SQUAREROOT_ROUTINE was too low instead of too high. To catch both types of wrong results you should compare the absolute value of the difference against the tolerance.
You might ask "Why make things so complicated"? The thing is that float-values usually are not exact and depending on the used precision you will get sightly different results due to rounding-errors.
well the logic itself is very straight forward, you are subtracting 1.843*(10^19) from the result you get from the SQUAREROOT_ROUTINE and putting that value in the variable called temp and then If the value of temp is less than 1.0*(10^16) you are going to print a line out to the SYSOUT that says "OK", otherwise you are going to print out "False" (if the value was equal to or greater than).
If you mean the logic as to why this code exists, you will need to talk to the author of the code, but it looks like a debugging display that was left in the program.

Can't modify loop-variable in lua [duplicate]

This question already has answers here:
Lua for loop reduce i? Weird behavior [duplicate]
(3 answers)
Closed 7 years ago.
im trying this in lua:
for i = 1, 10,1 do
print(i)
i = i+2
end
I would expect the following output:
1,4,7,10
However, it seems like i is getting not affected, so it gives me:
1,2,3,4,5,6,7,8,9,10
Can someone tell my a bit about the background concept and what is the right way to modify the counter variable?
As Colonel Thirty Two said, there is no way to modify a loop variable in Lua. Or rather more to the point, the loop counter in Lua is hidden from you. The variable i in your case is merely a copy of the counter's current value. So changing it does nothing; it will be overwritten by the actual hidden counter every time the loop cycles.
When you write a for loop in Lua, it always means exactly what it says. This is good, since it makes it abundantly clear when you're doing looping over a fixed sequence (whether a count or a set of data) and when you're doing something more complicated.
for is for fixed loops; if you want dynamic looping, you must use a while loop. That way, the reader of the code is aware that looping is not fixed; that it's under your control.
When using a Numeric for loop, you can change the increment by the third value, in your example you set it to 1.
To see what I mean:
for i = 1,10,3 do
print(i)
end
However this isn't always a practical solution, because often times you'll only want to modify the loop variable under specific conditions. When you wish to do this, you can use a while loop (or if you want your code to run at least once, a repeat loop):
local i = 1
while i < 10 do
print(i)
i = i + 1
end
Using a while loop you have full control over the condition, and any variables (be they global or upvalues).
All answers / comments so far only suggested while loops; here's two more ways of working around this problem:
If you always have the same step size, which just isn't 1, you can explicitly give the step size as in for i =start,end,stepdo … end, e.g. for i = 1, 10, 3 do … or for i = 10, 1, -1 do …. If you need varying step sizes, that won't work.
A "problem" with while-loops is that you always have to manually increment your counter and forgetting this in a sub-branch easily leads to infinite loops. I've seen the following pattern a few times:
local diff = 0
for i = 1, n do
i = i+diff
if i > n then break end
-- code here
-- and to change i for the next round, do something like
if some_condition then
diff = diff + 1 -- skip 1 forward
end
end
This way, you cannot forget incrementing i, and you still have the adjusted i available in your code. The deltas are also kept in a separate variable, so scanning this for bugs is relatively easy. (i autoincrements so must work, any assignment to i below the loop body's first line is an error, check whether you are/n't assigning diff, check branches, …)

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