在 python 的列表中附加列表中的项目时出现问题

Question

I think similar questions exist but I can't seem to find them out

x=list(([],)*3)
x[0].append(1)
x[1].append(2)
x[2].append(3)
print(x)

I want the result to look like this:

[[1],[2],[3]]

But instead, I get:

[[1, 2, 3], [1, 2, 3], [1, 2, 3]]

Also I really don't want to use any loops but if it is necessary then I'll try

Answer 1

Just realized you don't want to use a for loop.

Option 1: no for loops:

x = []
x.append([1])
x.append([2])
x.append([3])
print(x)
> [[1], [2], [3]]

Option 2: using loops.

Easiest and cleanest way is to use list comprehension like this:

x = [[i] for i in range(1,4)]
print(x)
> [[1], [2], [3]]

Or you can use a plain for loop and create an empty list before:

result = []
for i in range(1,4):
    result.append([i])
result

Answer 2

Instead of the first line, use

x = [[] for _ in range(3)]

In the current code, x is a list of the empty lists which are actually the same object . That's how [...] * n works; it repeats the same objects n times. So appending an item to one of these empty lists would append the item to the others, because they are the same.

Answer 3

You can create an empty list first and starts appending each element as a list.

x=[]
x.append([1])
x.append([2])
x.append([3])
print(x)

Answer 4

All the lists are copies (or views would be more accurate) of each other. Any changes made to one list will make changes to all the list. You'll have to manually create lists or use list comprehension to make separate lists. @j1-lee has already told you about list comprehension so I won't copy that. Create manual lists and try to append the data. Your code:

x=[[],[],[]]
x[0].append(1)
x[1].append(2)
x[2].append(3)
print(x)