Filling a vector with a varying amount of variables

Question

Hey currently I’m trying to program a quadratic programming algorithm with Python.
But I’m struggling with creating a vector which indicates the dimensions of the problem.

My problem: I am trying to create a vector x = [x_1, x_2, x_3, …, x_N] which containts N elements. N varies.The elements ‘x_N’ should be variables / the solution of my optimization problem.

My code so far: (‘NoE’… is equal to N+1, thats why Im substracting 1 in my while statement)

 #filling the x-vector according to the dimension of the given problem
    temp_2 = 0
    while temp_2 <= (NoE-1):   
        x[temp_2]= 'x_' + temp_2
        print(x)

Thanks for helping! 🙂

Asked By: Lalaaar

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Source

Answer 1

Your aim is to create a vector that has n elemnts formmated x_N ?..If so..Simply take a list and append to it inside the loop.

le =[]
num = 7

for x in range(num):

    a = f'x_{x}'
    le.append(a)

print(le)

Gives #

['x_0', 'x_1', 'x_2', 'x_3', 'x_4', 'x_5', 'x_6']

Answered By: Bhargav

Answer 2

You have to create an empty vector x and append to it 'x_' and convert temp_2 into a string in each iteration and concatenate it to 'x_', also the incrementation is missing in the code.

The following is the correct code:

temp_2 = 0

NoE=10

x=[] # Create an empty vector

while temp_2 <= (NoE-1):
   
    x.append('x_' + str(temp_2))

    temp_2 += 1 

print(x)

Output

['x_0', 'x_1', 'x_2', 'x_3', 'x_4', 'x_5', 'x_6', 'x_7', 'x_8', 'x_9']

Answered By: Khaled DELLAL

Filling a vector with a varying amount of variables

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