Replace 1d list as a column of a 2d list-CodePudding

pt_list = [3, 4, 32, 20, 20, 10, 10, 13, 30, 10]

pt = 0

points = [
    ['1.', pt],
    ['2.', pt],
    ['3.', pt],
    ['4.', pt],
    ['5.', pt],
    ['6.', pt],
    ['7.', pt],
    ['8.', pt],
    ['9.', pt],
    ['10.', pt]
]

for i in range(0,10,1):
    for j in points:
        points[i][1] = j

I tried some different ways and I can't find anything on the internet for this specific example. I would like to have the pt_list inside of each pt, so the first one would be ['1.', 3] the second one ['2.', 4] and the third one ['3.'32'] and so on.

How can I achieve this?

CodePudding user response：

You can use zip, which allows you to iterate the two lists in tandem, taking the 0-th column from each row of the 2d list points:

points = [[x[0], y] for x, y in zip(points, pt_list)]

The nested loop you've attempted won't be effective for this sort of problem (that approach would give you a Cartesian Product of the two lists -- for each element of list 1, pair it with each element from list 2), but you can fix your code as follows:

for i in range(len(points)):
    points[i][1] = pt_list[i]

Note that I've used len(points) rather than the hardcoded 10.

These solutions assume your lists are the same length.

CodePudding user response：

You can try this syntax to initialize this particular 2D list:

ls = [[str(i 1) '.',pt_list[i]] for i in range(0,10)]

This is known as list comprehension, and it basically means that we are generating i lists where i is in range from 0 to 10, and that each element would be:

[str(i 1) '.',pt_list[i]]

You can also understand this by manuallly substituting values of i from 0..10 and see what values it gives.

You can also change the value 10 with len(pt_list)..