Many times we deal with the lists having identical numbers as a sequence and we wish to just keep the 1st occurrence of element and substituting all the occurrences with the different number. Let’s discuss certain ways in which this can be done.
Method #1 : Using list comprehension + enumerate() This task can be achieved using the list comprehension for traversal and checking for element occurrence and index checking can be done using enumerate function.
Python3
# Python3 code to demonstrate# Altering duplicated# using list comprehension + enumerate()# initializing listtest_list = [2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5]# printing original listprint("The original list : " + str(test_list))# using list comprehension + enumerate()# Altering duplicatedres = [False if (ele in test_list[ :idx]) else ele for idx, ele in enumerate(test_list)]# print resultprint("The altered duplicate list is : " + str(res)) |
The original list : [2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5] The altered duplicate list is : [2, False, 3, False, False, False, 4, False, 5, False, False]
Time Complexity: O(n*n), where n is the length of the input list. This is because we’re using list comprehension + enumerate() which has a time complexity of O(n*n) in the worst case.
Auxiliary Space: O(n), as we’re using additional space res other than the input list itself with the same size of input list
Method #2 : Using itertools.groupby() + list comprehension This particular task can also be performed using a combination of above function, using groupby function to get the groups of different elements and list comprehension to alter duplicates.
Python3
# Python3 code to demonstrate# Altering duplicated# using itertools.groupby() + list comprehensionfrom itertools import groupby# initializing listtest_list = [2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5]# printing original listprint("The original list : " + str(test_list))# using itertools.groupby() + list comprehension# Altering duplicatedres = [val for key, grp in groupby(test_list) for val in [key] + [False] * (len(list(grp))-1)]# print resultprint("The altered duplicate list is : " + str(res)) |
The original list : [2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5] The altered duplicate list is : [2, False, 3, False, False, False, 4, False, 5, False, False]
Method #3 : Using dictionary
Initialize an empty result list and a dictionary to track seen values. If the value has been seen before, append False to the result list
Python3
# Initialize the list of valuestest_list = [2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5]# Print the original listprint("Original list:", test_list)# Initialize an empty result list and a dictionary to track seen valuesres = []seen = {}# Iterate through the list of valuesfor ele in test_list: # If the value has not been seen before, append it to the result list # and add it to the dictionary if ele not in seen: res.append(ele) seen[ele] = True # If the value has been seen before, append False to the result list else: res.append(False)# Print the altered listprint("Altered list:", res)#This code is contributed by Edula Vinay Kumar Reddy |
Original list: [2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5] Altered list: [2, False, 3, False, False, False, 4, False, 5, False, False]
Time complexity: O(n)
Auxiliary space: O(n)
Using numpy:
Algorithm:
- Convert the input list into a numpy array using np.array() function.
- Get the unique values and their indices from the array using np.unique() function with return_index=True.
- Create a boolean array with the same size as the input array using np.zeros_like() function with dtype=bool.
- Set the first occurrence of each unique value in the boolean array to True using the unique indices.
- Use np.where() function to replace the duplicate values with False in the original array using the boolean array.
- Convert the numpy array back to a list using .tolist() method and print the altered duplicate list.
Python3
import numpy as np# Initializing listtest_list = [2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5]# Printing original listprint("The original list : " + str(test_list))# Creating numpy arrayarr = np.array(test_list)# Getting unique values and their indicesunique, indices = np.unique(arr, return_index=True)# Creating a boolean array with the same size as input listres = np.zeros_like(arr, dtype=bool)# Setting the first occurrence of each unique value to Trueres[indices] = Trueres = np.where(res, arr, False)res = [False if val == 0 else val for val in res] # replace 0 with False# Printing the altered duplicate listprint("The altered duplicate list is : ", res) |
Output:
The original list : [2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5] The altered duplicate list is : [2, False, 3, False, False, False, 4, False, 5, False, False]
Time complexity: O(nlogn), where n is the length of the input list. This is because np.unique() function has a time complexity of O(nlogn) in the worst case.
Auxiliary space: O(n), as we’re using additional space for the numpy array and the boolean array, both with the same size as the input list.
