如果列表中的元素数大于 2，则将列表前半部分中的 n 个元素与列表另一半中的元素组合

Question

I have a problem with dealing with elements in a list. To be precise in a list of lists. For example, I have list of elements that are read from a file:

list_1 = [['void', None], ['uint8', 'f_MbistTestType_u8'], ['uint8', 'uint32', 'f_MbistTestType_u8', 'f_chip_id_u32'], ['void', None], ['void', None], ['void', None], ['void', None]]

In this case third element has more than two elements. I want to switch element 2 with element 3. So it would look like this:

list_1[2] = ['uint8', 'f_MbistTestType_u8', 'uint32', 'f_chip_id_u32']

If there would be 6 elements ie

list_example = ['uint8', 'uint32', 'void', 'f_chip_id_u32', 'f_MbistTestType_u8', None]

After the operation it should look like this:

list_example_sorted = ['uint8', 'f_chip_id_u32', 'uint32', 'f_MbistTestType_u8', 'void', None]

Right now I know how to get those elements in case I have only one occurrence of more than 2 elements, but don't know how to switch their places and also what to do in case I have more than one occurrence:

for elements in list_1:
print(elements)
if len(elements) > 2:
    list_el = elements
    print(list_el)

I tried to pop them out and append, but it won't scale well with more than 4 elements.

I tried to use swap function, but it seems that it doesn't work or I used it wrong?

Answer 1

Going by an input of [1, 1, 1, 2, 2, 2] with the desired output [1, 2, 1, 2, 1, 2] , ie you want the first element of the left half followed by the first element of the right half and so forth. To make it more obvious:

• input = [1, 2, 3, 4, 5, 6]
• output = [1, 4, 2, 5, 3, 6]

Define a function combine_inplace that combines the ith element of the left half with the ith element of the right half of l :

def combine_inplace(l):
    mid = len(l) // 2
    ptr = 0
    for left, right in zip(l[:mid], l[mid:]):
        l[ptr], l[ptr+1] = left, right
        # Increment pointer ptr by 2 for the next combination
        ptr += 2

• combine_inplace mutates the passed list l
• left half and right half are created using slice operator
• use zip to iterate over both list
• increment ptr by 2 to get to the next list indices for l

If you don't want to mutate the list itself you can instead create a new list combined that is returned by the function combine :

def combine(l):
    mid = len(l) // 2
    combined = []
    for left, right in zip(l[:mid], l[mid:]):
        combined.extend((left, right))
    return combined

• Does not mutate the passed list l
• Initialise empty list combined to store the combined values
• use zip to iterate over both list halves
• Returns the list combined

This uses the same logic as combine_inplace but you keep the original list intact.

Both functions combine the elements of the left half with the right half of a given list. The only difference is that with combine you have to store the returned list in a variable to access it.

>> l = [1, 1, 1, 2, 2, 2]
>> combine_inplace(l)
>> print(l)
[1, 2, 1, 2, 1, 2]
>> input_list = [1, 2, 3, 4, 5, 6]
>> output_list = combine(input_list)
>> print(output_list)
[1, 4, 2, 5, 3, 6]

Now using either combine or combine_inplace to combine elements of lists with a length > 2 inside a list:

ll = [[1, 2], [1, 2], [1, 1, 2, 2], [1, 2], [1, 2, 3, 4, 5, 6]]
# Non-destructive way using combine to create a new list comb_ll
comb_ll = []
for el in ll:
    if len(el) > 2:
        el = combine(el)
    comb_ll.append(el)

# Mutates the original list
for i in range(len(ll)):
    if len(ll[i]) > 2:
        combine_inplace(ll[i])

In both cases you'll get the same result:

>> print(comb_ll)
[[1, 2], [1, 2], [1, 2, 1, 2], [1, 2], [1, 4, 2, 5, 3, 6]]
>> print(ll)
[[1, 2], [1, 2], [1, 2, 1, 2], [1, 2], [1, 4, 2, 5, 3, 6]]