Python – Elements with same index
Last Updated :
28 Apr, 2023
Given a List, get all elements that are at their index value.
Input : test_list = [3, 1, 8, 5, 4, 10, 6, 9]
Output : [1, 4, 6]
Explanation : These elements are at same position as its number.
Input : test_list = [3, 10, 8, 5, 14, 10, 16, 9]
Output : []
Explanation : No number at its index.
Method #1: Using loop
In this, we check for each element, if it equates to its index, it’s added to the result list.
Python3
test_list = [3, 1, 2, 5, 4, 10, 6, 9]
print("The original list is : " + str(test_list))
res = []
for idx, ele in enumerate(test_list):
if idx == ele:
res.append(ele)
print("Filtered elements : " + str(res))
|
Output
The original list is : [3, 1, 2, 5, 4, 10, 6, 9]
Filtered elements : [1, 2, 4, 6]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #2 : Using list comprehension + enumerate()
In this, we perform a similar function as the above method, the change being we use list comprehension to make the solution compact.
Python3
test_list = [3, 1, 2, 5, 4, 10, 6, 9]
print("The original list is : " + str(test_list))
res = [ele for idx, ele in enumerate(test_list) if idx == ele]
print("Filtered elements : " + str(res))
|
Output
The original list is : [3, 1, 2, 5, 4, 10, 6, 9]
Filtered elements : [1, 2, 4, 6]
Time Complexity: O(n) where n is the number of elements in the list “test_list”. This is because we’re using the list comprehension + enumerate() function which all has a time complexity of O(n) in the worst case.
Auxiliary Space: O(1), no extra space is required.
Method #3: Using for loop
Python3
test_list = [3, 1, 2, 5, 4, 10, 6, 9]
print("The original list is : " + str(test_list))
res = []
for i in range(0,len(test_list)):
if(i==test_list[i]):
res.append(test_list[i])
print("Filtered elements : " + str(res))
|
Output
The original list is : [3, 1, 2, 5, 4, 10, 6, 9]
Filtered elements : [1, 2, 4, 6]
Method #4: Using numpy library:
- Create a numpy array from the given list using np.array() function.
- Create another numpy array using np.arange(len(test_list)), which creates an array with values from 0 to len(test_list)-1.
- Compare the two numpy arrays element-wise using the equality operator (==). This returns a boolean array with True at positions where the corresponding elements are equal, and False otherwise.
- Use np.where() function to get the indices of True values in the boolean array.
- Convert the resulting numpy array of indices to a Python list using list() function.
- Print the filtered elements using the list of indices.
Python3
import numpy as np
test_list = [3, 1, 2, 5, 4, 10, 6, 9]
print("The original list is : " + str(test_list))
res = np.where(np.array(test_list) == np.arange(len(test_list)))[0]
print("Filtered elements : " + str(list(res)))
|
Output
The original list is : [3, 1, 2, 5, 4, 10, 6, 9]
Filtered elements : [1, 2, 4, 6]
Time complexity: O(n), where n is the length of the input list. This is because creating numpy arrays from a list, comparing them element-wise, and finding indices of True values in the boolean array all take O(n) time.
Auxiliary Space: O(n), where n is the length of the input list. This is because two numpy arrays of size n are created, and a list of filtered indices is created which can have at most n elements.
Method #5 using the filter() function:
Python3
test_list = [3, 1, 2, 5, 4, 10, 6, 9]
print("The original list is : " + str(test_list))
res = list(filter(lambda x: x[0] == x[1], enumerate(test_list)))
res = [ele for idx, ele in res]
print("Filtered elements : " + str(res))
|
Output
The original list is : [3, 1, 2, 5, 4, 10, 6, 9]
Filtered elements : [1, 2, 4, 6]
Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(m), where m is the number of elements in the input list that have the same index and value.
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