Get the indices of all occurrences of an element in a Python list
Given a list x and an element target, the goal is to find all the occurrences of the target element in the list. For example, for the given input:
getIndices([1,2,3,4,1,5,1], 1)
The output should be as follows:
[0, 4, 6]
Naive solution
Let's try to implement a naive solution first:
def getIndices(x, target):# list to store indicesindices = []# loop to iterate over the list elementsfor i in range(len(x)):# if the element matches the target element, store its indexif x[i] == target:indices.append(i)# return indicesreturn indicesprint(getIndices([1,2,3,4,1,5,1], 1))
Explanation
Line 1: We define our function, which takes two parameters: a list
xand a target elementtarget.Line 3: We create a list to store the indices of the target element.
Lines 5–8: We iterate over the list of elements and check if an element matches the
targetelement. If yes, we store the index of that element in theindiceslist.Line 11: We return the
indiceslist.
Optimized solution
Our naive solution has a time complexity of O(n), which isn't bad. However, we can optimize the performance by using list comprehension, which is generally faster than using a for loop.
The following image demonstrates list comprehension.
Here is the complete code:
def getIndices2(x, target):indices = [i for i in range(len(x)) if x[i]==target]return indicesprint(getIndices2([1,2,3,4,1,5,1], 1))
Explanation
Line 1: We define our function as before.
Line 2: We use list comprehension and get the indices of the elements in the list
xthat match thetargetelement.Line 4: We return the
indiceslist.
Our optimized solution still has the time complexity of O(n); however, list comprehension may offer faster execution time, especially for larger lists.
Note: While list comprehension has its performance benefits, using it with multiple
ifconditions can make our code harder to read and debug. In that case, using aforloop might be a better option.
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