Move One List Element to Another List - Python

The task of moving one list element to another in Python involves locating a specific element in the source list, removing it, and inserting it into the target list at a desired position. For example, if a = [4, 5, 6, 7, 3, 8] and b = [7, 6, 3, 8, 10, 12], moving 10 from b to index 4 in a results in a = [4, 5, 6, 7, 10, 3, 8] and b = [7, 6, 3, 8, 12].

Using list comprehension

This method efficiently moves an element from one list to another in a single step using list comprehension. It minimizes function calls and optimizes performance by finding and inserting the element in a compact and readable way. This approach is ideal for quick modifications in small to medium-sized lists.

a = [4, 5, 6, 7, 3, 8]
b = [7, 6, 3, 8, 10, 12]

[a.insert(4, b.pop(idx)) for idx in [b.index(10)]]

print(a)  # After insert
print(b)  # After removal

[4, 5, 6, 7, 10, 3, 8]
[7, 6, 3, 8, 12]

Explanation: b.pop(idx) removes the element 10 from list b and a.insert(4, b.pop(idx)) inserts this removed element at index 4 in list a, effectively moving 10 from b to a.

Using pop() and insert()

A direct method where pop is used to remove an element from one list and insert places it into another. This approach ensures clarity while maintaining efficiency. It is widely used for moving elements between lists without unnecessary complexity.

a = [4, 5, 6, 7, 3, 8]
b = [7, 6, 3, 8, 10, 12]

idx = b.index(10)
val = b.pop(idx)
a.insert(4, val)

print(a)
print(b)

[4, 5, 6, 7, 10, 3, 8]
[7, 6, 3, 8, 12]

Explanation: This code first finds the index of the element 10 in list b using b.index(10), storing it in idx. Then, b.pop(idx) removes 10 from b and stores it in val. Finally, a.insert(4, val) inserts 10 at index 4 in list a, effectively moving the element from b to a.

Using filter()

filter() method quickly finds the element’s index before removing and inserting it into the target list. By avoiding unnecessary iterations, it enhances efficiency, making it a good choice for handling large lists where performance is a concern.

a = [4, 5, 6, 7, 3, 8]
b = [7, 6, 3, 8, 10, 12]

idx = next(filter(lambda i: b[i] == 10, range(len(b))))
a.insert(4, b.pop(idx))

print(a)
print(b)

[4, 5, 6, 7, 10, 3, 8]
[7, 6, 3, 8, 12]

Explanation: filter() iterates over b's indices, finding where b[i] == 10 and next retrieves the first matching index as idx. Then, b.pop(idx) removes 10 from b and a.insert(4, b.pop(idx)) places it at index 4 in a, completing the transfer.

Using loop

A simple but less efficient method where a loop finds the element, removes it and inserts it into another list. While not the fastest, it offers flexibility for handling custom conditions when moving elements between lists.

a = [4, 5, 6, 7, 3, 8]
b = [7, 6, 3, 8, 10, 12]

for i in range(len(b)):
    if b[i] == 10:
        a.insert(4, b.pop(i))
        break

print(a)
print(b)

[4, 5, 6, 7, 10, 3, 8]
[7, 6, 3, 8, 12]

Explanation: loop iterates through b, finds 10, removes it using b.pop(i) and inserts it at index 4 in a. The break ensures only the first occurrence is moved efficiently.