Making a 2 dimensional list/matrix with different number of columns for each row

I would like to make a list or matrix that has a known number of rows(3), but for each row the number of elements will be different.
So it could look something like this:

[[4, 6, 8],
[1, 2, 3, 4],
[0, 2, 3, 4, 8]]


Each row will have a known maximum of elements(8).

So far I have tried the following:

Sets1=np.zeros((3,8))

for j in range(3):
Sets1[0,:]=[i for i, x in enumerate(K[:-1]) if B[x,j]==1 or B[x+1,j]==1]


I want this because I want to have a list for each j in range(3) over which I can do a for loop and add constraints to my ILP.

Any help will be greatly appreciated!

Final expected output for the given sample?
– Divakar
Jun 29 at 8:35

You shouldn't call that a matrix because matrix is a rectangular array of numbers. It's just a 2D array.
– Kasramvd
Jun 29 at 8:36

Edit: Never mind I have figured it out.. I just wanted a list of lists. A got by creating an empty list and then just appending each list with the list I needed. Thanks though!!
– Tine
Jun 29 at 8:37

1 Answer
1

Here is a sample of what your code should look like:

from random import *

myMatrix = [
,
,

]

elementsNums = [1, 2, 3, 4, 5, 6, 7, 8] # possible length of each list in the matrix
elements = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] # random values in each list in the matrix


def addElements(Num, List, Matrix):
count = 0

a = choice(Num) #length of first list in matrix
Num.remove(a) # removing to prevent same length lists in matrix

b = choice(Num) # length of second list in matrix
Num.remove(b)

c = choice(Num) # length of third list in matrix

nums = [a, b, c]

for x in nums: # iterating through each list (a, b, c)
for i in range(x): # placing x number of values in each list
Matrix[count].append(choice(List)) # adding the values to the list
count += 1 # moving to the next list in the matrix


addElements(elementsNums, elements, myMatrix)
print(myMatrix)


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