How to find the largest number(s) in a list of elements, possibly non-unique?

Question

Here is my program,

item_no = []
max_no = 0
for i in range(5):
    input_no = int(input("Enter an item number: "))
    item_no.append(input_no)
for no in item_no:
    if no > max_no:
       max_no = no
high = item_no.index(max_no)
print (item_no[high])

Example input: [5, 6, 7, 8, 8]

Example output: 8

How can I change my program to output the same highest numbers in an array?

Expected output: [8, 8]

Answer 1

Just get the maximum using max and then its count and combine the two in a list-comprehension.

item_no = [5, 6, 7, 8, 8]

max_no = max(item_no)
highest = [max_no for _ in range(item_no.count(max_no))]
print(highest)  # -> [8, 8]

Note that this will return a list of a single item in case your maximum value appears only once.

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A solution closer to your current programming style would be the following:

item_no = [5, 6, 7, 8, 8]
max_no = 0  # Note 1 
for i in item_no:
    if i > max_no:
        max_no = i
        high = [i]
    elif i == max_no:
        high.append(i)

with the same results as above of course.

Notes

1. I am assuming that you are dealing with N* (1, 2, ...) numbers only. If that is not the case, initializing with -math.inf should be used instead.

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Note that the second code snippet is less efficient than the first by quite a margin. Python allows you to be more efficient than these explicit, fortran-like loops and it is more efficient itself when you use it properly.

Answer 2

You can do it even shorter:

item_no = [5, 6, 7, 8, 8]
#compute once - use many times
max_item = max(item_no)
print(item_no.count(max_item) * [max_item])

Output:

[8, 8]