Let's say I have:
a=[1,2,3]
b=[4,5,6]
Now I want to create a list of list from a and b, I would do it like this:
c=[a,b]=[[1,2,3],[4,5,6]]
As a.append(b) would result in: [1,2,3,b]=[1,2,3,[4,5,6]]
Now let's say there exists anew list which I want to append to c:
d=[7,8,9]
I now have to do c.append(d) to get [[1,2,3],[4,5,6],[7,8,9]]
Because
e=[c,d]=[[[1,2,3],[4,5,6]],[7,8,9]]
How can I get a list of list from individual lists without know how my lists are structured?
Try this:
a = [1,2,3]
b = [4,5,6]
c = []
c.append(a)
c.append(b)
This should work, and only takes 2 simple lines.
A bit roundabout of a way but looks nice, using numpy
a = np.array([[1,2,3]])
b = np.array([[4,5,6]])
c = np.append(a,b,axis=0)
print(c.tolist())
gives you
[[1,2,3],[4,5,6]]
Appending another list in the same way keeps the structure of list of lists, for example
d = np.array([[7,8,9]])
e = np.append(c,d,axis=0)
print(e.tolist())
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Now this is quite roundabout. I would just keep everything in numpy arrays if possible.
EDIT: Figured out how to do this without numpy
Simply state each list as a list of lists to begin with
a = [[1,2,3]]
b = [[4,5,6]]
a.extend(b)
print(a)
[[1,2,3],[4,5,6]]
Furthermore you can do this
d = [[7,8,9]]
a.extend(d)
print(a)
[[1, 2, 3], [4, 5, 6], [4, 5, 6]]
The two actions you are describing are distinctly different. The first is creating the outer list (a.k.a. c) and the second is appending to it.
To make the process more uniform you can just start off with an outer list and append all the child lists to it.
c = []
a=[1,2,3]
b=[4,5,6]
d=[7,8,9]
c.append(a)
c.append(b)
c.append(d)
c is now
[[[1,2,3],[4,5,6]],[7,8,9]]