Python | Shift sublist in list

Sometimes, while working with list, we can have a problem in which we need to shift some sublist to desired index in same sublist. This problem can occur in day-day programming. Let’s discuss certain ways in which this task can be performed. 

Method #1 : Using insert() + pop() + loop The combination of above functions can be used to perform the particular task. The pop function can be used to remove the sublist and insert function inserts the sublist. This happens for each element in single iteration in loop. 

Python3

# Python3 code to demonstrate working of
# Shift sublist in list
# Using insert() + pop() + loop

# function to perform the task
def shift_sublist(test_list, strt_idx, no_ele, shft_idx):
    for ele in range(no_ele):
        test_list.insert(shft_idx + 1, test_list.pop(strt_idx))
    return test_list

# initializing list
test_list = [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16]

# printing original list
print("The original list is : " +  str(test_list))

# Using insert() + pop() + loop
# Shift sublist in list
res = shift_sublist(test_list, 2, 3, 6)

# Printing result
print("The list after shift of sublist : " +  str(res))

Output : 

The original list is : [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16] The list after shift of sublist : [4, 5, 8, 10, 1, 6, 7, 3, 12, 15, 16]

Method #2 : Using list slicing This task can also be using list slicing technique in which one can just add the different sections of list at required positions. 

Python3

# Python3 code to demonstrate working of
# Shift sublist in list
# Using list slicing

# function to perform the task
def shift_sublist(test_list, strt_idx, no_ele, shft_idx):
    return (test_list[:strt_idx] + test_list[strt_idx + no_ele : no_ele + shft_idx - 1]
            + test_list[strt_idx : strt_idx + no_ele] + test_list[shft_idx + no_ele -1:])

# initializing list
test_list = [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16]

# printing original list
print("The original list is : " +  str(test_list))

# Using list slicing
# Shift sublist in list
res = shift_sublist(test_list, 2, 3, 6)

# Printing result
print("The list after shift of sublist : " +  str(res))

Output : 

The original list is : [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16]
The list after shift of sublist : [4, 5, 8, 10, 1, 6, 7, 3, 12, 15, 16]

Method #3: Using list comprehension and slicing

In this method, extract the sublist to be shifted using list slicing, create a new list by iterating through the original list and excluding the indices of the sublist, and insert the shifted sublist at the desired index using slicing.

Python3

def shift_sublist(test_list, strt_idx, no_ele, shft_idx):
    # Extract sublist to be shifted using list slicing and reverse it
    shifted_sublist = test_list[strt_idx:strt_idx+no_ele][::-1]
    # Create a new list by excluding indices of the sublist
    shifted_list = [test_list[i] for i in range(len(test_list)) if i not in range(strt_idx, strt_idx+no_ele)]
    # Insert the shifted sublist at the desired index using slicing
    shifted_list[shft_idx:no_ele+shft_idx] = shifted_sublist
    return shifted_list

# initializing list
test_list = [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16]

# printing original list
print("The original list is:", test_list)

# Using list comprehension and slicing
# Shift sublist in list
res = shift_sublist(test_list, 2, 3, 6)

# Printing result
print("The list after shift of sublist:", res)

Output

The original list is: [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16]
The list after shift of sublist: [4, 5, 8, 10, 1, 12, 3, 7, 6]

Time complexity: O(n), where n is the length of the original list. This is because we only need to iterate through the original list once to extract the shifted sublist and create a new list.
Auxiliary space: O(n), as we create a new list to hold the shifted sublist and a new list to hold the remaining elements of the original list.