Hello is there a way to generate params in GMPL as example io have a funcion
min:c[i]*x[i] and constrains that looks like A[i][j]*x[i]=b[i]. Where A[i][j]=1/(i+j-1) and i,j=1,2....,n. c[i]=b[i]=sum(j=1,...,n)1/(i+j-1) where i=1,...,n.
So there is a question is there a way to generate matrix A from equation? or do i need to make this matrix manual in data section ? and one more question is there a good way to find (n) maximum size of this problem when with precision of 2 numbers without modifying objective function ?
param n := 3;
set I := 1..n;
param A{i in I, j in I} := 1/(i+j-1);
param c{i in I} := sum{j in I} 1/(i+j-1);
# or better (may be):
# param c{i in I} := sum{j in I} A[i,j];
display A,c;
end;
The output should look like:
Reading model section from x.mod...
9 lines were read
Display statement at line 8
A[1,1] = 1
A[1,2] = 0.5
A[1,3] = 0.333333333333333
A[2,1] = 0.5
A[2,2] = 0.333333333333333
A[2,3] = 0.25
A[3,1] = 0.333333333333333
A[3,2] = 0.25
A[3,3] = 0.2
c[1] = 1.83333333333333
c[2] = 1.08333333333333
c[3] = 0.783333333333333
GMPL is a subset of AMPL. You may want to read the AMPL book to understand more about the syntax.
I am afraid I don't understand your second question.
Related
I have a list like below and need to firs add items in each list and then multiply all results 2+4 = 6 , 3+ (-2)=1, 2+3+2=7, -7+1=-6 then 6*1*7*(-6) = -252 I know how to do it by accessing indexes and it works (as below) but I also need to do it in a way that it will work no matter how many sublist there is
nested_lst = [[2,4], [3,-2],[2,3,2], [-7,1]]
a= nested_lst[0][0] + nested_lst[0][1]
b= nested_lst[1][0] + nested_lst[1][1]
c= nested_lst[2][0] + nested_lst[2][1] + nested_lst[2][2]
d= nested_lst[3][0] + nested_lst[3][1]
def sum_then_product(list):
multip= a*b*c*d
return multip
print sum_then_product(nested_lst)
I have tried with for loop which gives me addition but I don't know how to perform here multiplication. I am new to it. Please, help
nested_lst = [[2,4], [3,-2],[2,3,2], [-7,1]]
for i in nested_lst:
print sum(i)
Is this what you are looking for?
nested_lst = [[2,4], [3,-2],[2,3,2], [-7,1]] # your list
output = 1 # this will generate your eventual output
for sublist in nested_lst:
sublst_out = 0
for x in sublist:
sublst_out += x # your addition of the sublist elements
output *= sublst_out # multiply the sublist-addition with the other sublists
print(output)
Edit: Here is the code I have so far for generating the Patient Oct-Tuples.
(thanks Anon for giving me the bost on how to calculate weighted probability/setting the seed)
fun genPatients(x:int) =
let
val seed=let
val m=Date.minute(Date.fromTimeLocal(Time.now()))
val s=Date.second(Date.fromTimeLocal(Time.now()))
in Random.rand(m,s)
end;
val survivalrate = ref(1)
val numl = ref(1)
val td = ref(1)
val xray = ref(false)
val count= ref(0)
val emnum= ref(1000)
val ageList = [1, 2, 3, 3];
val xrayList=[false,true];
val age = Random.randRange (0, 3) seed;(* random age*)
val nextInt1 = Random.randRange(0, 1)(* random xray*)
val r1 = Random.rand(1,1)
val nextInt2 = Random.randRange(1, 10000000)(* random td*)
val r2 = Random.rand(1,1)
val r1hold= ref(1);
in
while !count < x do
(
count:= !count + 1;
List.nth(ageList, age);
r1hold:= nextInt1 r1;
td:= nextInt2 r2;
(!emnum,age,survivalrate,numl,[],[],xray,td);
emnum:= !emnum + 1
)
end;
My question now is now how to go about indexing a boolean list?
So I was looking for some help defining my Oct-tuple to finish up my project and lo and behold I find someone posting the entirety of my project hoping for a handout answer. Not only that, but I'm almost certain we're in the same class, and you think posting this the night before the morning the project is due is what a responsible student does? Pretty sure nobody on SO is gonna do your homework for you anyway, in fact I'm not even sure it's allowed.
Maybe do some work and then ask for help when you've actually done anything. Or maybe in the next phase try a little harder.
EDIT: I'll give you something to get you started.
To calculate weighted probability you need a seed.
val seed=let
val m=Date.minute(Date.fromTimeLocal(Time.now()))
val s=Date.second(Date.fromTimeLocal(Time.now()))
in Random.rand(m,s)
end;
Here's one. Then you can calculate probability, at least for the age, like this:
val ageList = [1, 2, 3, 3];
val ageInt = Random.randRange (0, 3) seed;
List.nth(ageList, ageInt)
This was how I decided to do the weighted probability portion, you can equate this to the other weighted sections if you're creative. Good luck.
I would like to experiment the weights initialization recommended by Karpathy in his lecture notes,
the recommended heuristic is to initialize each neuron's weight vector
as: w = np.random.randn(n) / sqrt(n), where n is the number of its
inputs
source: http://cs231n.github.io/neural-networks-2/#init
I'm beginner in python, and I don"t know how to implement this :/
weights = tf.Variable(??)
Please help? ...
For a single value, use:
weights = tf.Variable(10)
For a vector with random values:
shape = [784, 625]
weights = tf.Variable(tf.random_normal(shape, stddev=0.01)/tf.sqrt(n))
Please note that you need to sess.run to evaluate the variables.
Also, please check out other Random Tensors: https://www.tensorflow.org/versions/r0.8/api_docs/python/constant_op.html#random-tensors
n = 10
init_x = np.random.randn(n)
x = tf.Variable(init_x)
sess = tf.InteractiveSession()
sess.run(tf.initialize_all_variables())
print(sess.run(x))
I do it in the following way:
self.w_full, self.b_full = [], []
n_fc_layers = len(structure)
structure.insert(0, self.n_inputs)
with vs.variable_scope(self.scope):
for lr_idx in range(n_fc_layers):
n_in, n_out = structure[lr_idx], structure[lr_idx+1]
self.w_full.append(
vs.get_variable(
"FullWeights{}".format(lr_idx),
[n_in, n_out],
dtype=tf.float32,
initializer=tf.random_uniform_initializer(
minval=-tf.sqrt(tf.constant(6.0)/(n_in + n_out)),
maxval=tf.sqrt(tf.constant(6.0)/(n_in + n_out))
)
)
)
self.b_full.append(
vs.get_variable(
"FullBiases{}".format(lr_idx),
[n_out],
dtype=tf.float32,
initializer=tf.constant_initializer(0.0)
)
)
after
structure.insert(0, self.n_inputs)
you'll have [n_inputs, 1st FC layer size, 2nd FC layer size ... output layer size]
Hi everyone I'm newbie in prolog and I have such a list: (actually It is output of my predicate not a list )
P = [1/1, 1/3] ;
P = [1/1, 2/3] ;
P = [1/3, 1/1] ;
P = [1/3, 2/1] ;
P = [2/1, 1/3] ;
P = [2/1, 2/3] ;
P = [2/3, 1/1] ;
P = [2/3, 2/1] ;
and I need to remove dublicete terms.For example [1/1,2/3] and [2/3,1/1]is same and I should remove one of them , which one is not important ,How could I do that in prolog ?? Thanks in advance
NOTE I LEARNT THAT findALL should be good way for this but still dont know the answer please help me .
Unless you actually show us your code, it's never going to be possible to give you precise answers.
I assume you have a predicate f/1 such that:
?- f(P).
produces the interactive result you show above. A simple solution is to change your query:
?- f([X,Y]), X < Y.
This will produce the following result:
X = 1/3, Y = 1/1 ;
X = 1/3, Y = 2/1 ;
X = 2/3, Y = 1/1 ;
X = 2/3, Y = 2/1 ;
findall/3 isn't sufficient to solve this particular situation, because you've defined uniqueness in a way that ignores the position in the list. In Prolog (and everything else) [X,Y] and [Y,X] are not equal, so you'd have to find a trick to get this to give you "unique" results.
i was not able to print an a value (float) to a file with the OCaml lenguage.
How can i do?
If you know how, can you show me a little example?
Thank you advance and have a good day!
Printf.fprintf allows direction to an out_channel, in your case a file. Reasonably, you'd open the file for writing first, and pass around that channel.
Printf.fprintf (open_out "file.txt") "Float Value of %f" 1.0
If you want to print the textual representation of a float to a file, perhaps the simplest thing to do is:
output_string outf (string_of_float myfloat)
If you want to print the float to the console, you can use
print_string (string_of_float myfloat)
Of course, Printf.printf can also do that and more, so it is worth knowing it.
If you want to output the binary representation of a float, things are more complicated. Since a float value is represented as an IEEE 754 double, it is 8 bytes long which can be written in different orders depending on the platform. In the case of little-endian order, as is normal in X86, you can use the following:
let output_float_le otch fv =
let bits = ref (Int64.bits_of_float fv) in
for i = 0 to 7 do
let byte = Int64.to_int (Int64.logand !bits 0xffL) in
bits := Int64.shift_right_logical !bits 8;
output_byte otch byte
done
The float value so written can be read back with the following:
let input_float_le inch =
let bits = ref 0L in
for i = 0 to 7 do
let byte = input_byte inch in
bits := Int64.logor !bits (Int64.shift_left (Int64.of_int byte) (8 * i))
done;
Int64.float_of_bits !bits
This has the advantage of being a very compact way to exactly preserve floats in a file, that is, what you write will be read back exactly as it originally was. For example, I did this in the interactive top-level:
# let otch = open_out_bin "Desktop/foo.bin" ;;
val otch : out_channel = <abstr>
# output_float_le otch 0.5 ;;
- : unit = ()
# output_float_le otch 1.5 ;;
- : unit = ()
# output_float_le otch (1. /. 3.) ;;
- : unit = ()
# close_out otch ;;
- : unit = ()
# let inch = open_in_bin "Desktop/foo.bin" ;;
val inch : in_channel = <abstr>
# input_float_le inch ;;
- : float = 0.5
# input_float_le inch ;;
- : float = 1.5
# input_float_le inch ;;
- : float = 0.333333333333333315
# close_in inch ;;
- : unit = ()
and as you can see I got back exactly what I put in the file. The disadvantage of this form of writing floats to files is that the result is not human-readable (indeed, the file is binary by definition) and you lose the possibility to interoperate with other programs, like Excel for instance, which in general exchange data in human-readable textual form (CSV, XML, etc.).
Did you try printf?
let a = 4.0
printf "my float value: %f" a
Cf the doc inria and don't forget to open the module