Many times we might have problem in which we need to find indices rather than the actual numbers and more often, the result is conditioned. First approach coming to mind can be a simple index function and get indices less than or equal than particular number, but this approach fails in case of duplicate numbers. Let’s discuss certain ways in which this problem can be successfully solved.
Method #1 : Using loop
This problem can easily be solved using loop with a brute force approach in which we can just check for the index as we iterate and append it in a new list as we proceed forward.
# Python3 code to demonstrate# Atmost K element indices# using loop # initializing list test_list = [12, 11, 7, 15, 8, 18] # printing original listprint("The original list : " + str(test_list)) # using loop# Atmost K element indicesres = []for idx in range(0, len(test_list)) : if test_list[idx] <= 11: res.append(idx) # print resultprint("The list of indices at most 11 : " + str(res)) |
The original list : [12, 11, 7, 15, 8, 18] The list of indices at most 11 : [1, 2, 4]
Method #2 : Using list comprehension + enumerate()
The combination of these two function can also perform this particular task efficiently and in one line. The enumerate function is used to get element and its index simultaneously.
# Python3 code to demonstrate# Atmost K element indices# using list comprehension + enumerate() # initializing list test_list = [12, 11, 7, 15, 8, 18] # printing original listprint("The original list : " + str(test_list)) # using list comprehension + enumerate()# Atmost K element indicesres = [idx for idx, val in enumerate(test_list) if val <= 11] # print resultprint("The list of indices at most 11 : " + str(res)) |
The original list : [12, 11, 7, 15, 8, 18] The list of indices at most 11 : [1, 2, 4]
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