通过迭代列表追加项目

Question

I've designed a following function that takes a list as a parameter and returns another list.

def function(list):
    A = []
    B = []
    for num in list:
        A.append(num)
        B.append(A)
    return B

In this function, I assumed that the function would append A to A depending on the current state of A. Working under that assumption. I called the function

>>> test([1,2,3,4,5])

expected the output to be

[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 3, 4, 5]]

However, the result that I get is

[[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]]

Why is this? Is there any solution around this problem?

Answer 1

You need to copy it by value. In your current situation, you are passing by reference. Hence, every index gets updated as A changes. Should be:

    B.append(A[:])

Notice the [:] for copying by value.

To clarify, running your code here we can examine the result of calling function on [1,2] :

Now, consider what happens when we make a copy:

def function(list):
    A = []
    B = []
    for num in list:
        A.append(num)
        B.append(A[:])
    return B

print function([1,2])

Also, as a side note: you should change the names of your variables. For example, by declaring the parameter as list you shadow the list method.

Answer 2

You can append the copy of A to B to avoid using the same reference of A :

for num in list:
      A.append(num)
      B.append(A[:])

Answer 3

Lists are passed by reference. To copy them use the syntax B.append(A[:])

def function(list):
    A = []
    B = []
    for num in list:
        A.append(num)
        B.append(A[:])
    return B

Answer 4

In your code you sending reference instead of data.

Working code

def function(list):
   A = []
   B = []
   for num in list:
      A.append(num)
      B.append(A[:])
   return B