Find 3 maximum elements of list with python

Question

How can I search for the three maximum elements of a list and replace it at same index with its result when divided by 2.

Please what am I doing wrong:

input is: 2 5 8 19 1 15 7 20 11
output should be : 2 5 8 9.5 1 7.5 7 10 11

Index out of range is the result displayed

def numberInput(line):
  lineInput = [int(i) for i in line.split()]

  min1 = min(lineInput)

  for j in range(len(lineInput)):

     if lineInput[j]>min1 and lineInput[j] > lineInput[j+1]:

           max1 = lineInput[j]/float(2)
     else:
           max1 = lineInput[j]/float(2)
           lineInput[j] = max1

     lineInput[j] = max1
  return(lineInput)

number = '2 5 8 19 1 15 7 20 11' 
print(numberInput(number))

When you get to the last element of the list, lineInput[j+1] is outside the list. You need to use range(len(lineInput)-1) — Barmar
– Barmar, Commented Feb 13, 2020 at 1:57
@MateenUlhaq: heapq.nlargest should be somewhat faster; it's O(n log k) (where k is 3 in this case) rather than the O(n log n) you pay on a complete sort (whether it's faster in practice would require testing; sorted is implemented in C, nlargest is Python with C-accelerated helpers, so it's slower than big-O might lead you to believe). — ShadowRanger
– ShadowRanger, Commented Feb 13, 2020 at 2:10

Swati Srivastava · Accepted Answer · 2020-02-13 06:45:25Z

1

If the order of list isn't important, you can simply sort the list in descending order and then replace the first 3 elements as

a = [2, 5, 8, 19, 1, 15, 7, 20, 11]
a.sort(reverse = True)
a[0] = a[0] / 2
a[1] = a[1] / 2
a[2] = a[2] / 2

Output

[10.0, 9.5, 7.5, 11, 8, 7, 5, 2, 1]

If the order is important,

import heapq
largest = heapq.nlargest(3, a)
for i in range(len(a)):
    if a[i] in largest:
        a[i] = a[i] / 2

Output

[2, 5, 8, 9.5, 1, 7.5, 7, 10.0, 11]

heapq.nlargest() is a function of heapq which can give you n largest numbers from a list. Since lists in Python do not have a replace function, the list had to be traversed once, and then replaced manually. Hope this resolves your issue.

answered Feb 13, 2020 at 6:45

Swati Srivastava

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Comments

mattyx17 · Accepted Answer · 2020-02-13 02:30:47Z

0

Why wouldn't you just walk through the list once (it maybe a bit longer code, but definitely efficient)

def numberInput(line):
    lineInput = [int(i) for i in line.split()]
    n = len(lineInput)
    if n <= 3:
        return [i / 2 for i in lineInput]
    max_pairs = [(lineInput[i], i) for i in range(3)] # keep track of 3 max's and their indices
    max_pairs.sort(key = lambda x: -x) # sort in descending order
    for i in range(3, n):
        if lineInput[i] >= max_pairs[0][0]: # greater than the largest element
            max_pairs = [(lineInput[i], i)] + max_pairs[:2]
        elif lineInput[i] >= max_pairs[1][0]: # greater than second element
            max_pairs = [max_pairs[0], (lineInput[i], i), max_pairs[1]]
        elif lineInput[i] >= max_pairs[2][0]: # greater than third element
            max_pairs = max_pairs[:2] + [(lineInput[i], i)]
     for pair in max_pairs:
         lineInput[pair[1]] = lineInput[pair[0]] / 2
     return lineInput

Explanation: max_pairs is a set of three tuples, containing the maximum three elements and their indices

Note: I think the above is easiest to understand, but you can do it in a loop if you don't like all those ifs

answered Feb 13, 2020 at 2:30

mattyx17

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1 Comment

mattyx17 Over a year ago

Also, as @AMC commented above, python convention is to use underscores rather than camelCase so you usually see lineInput and numberInput as line_input and number_input respectively

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