How to find the largest number(s) in a list of ele

问题:

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]

回答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.

---

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.

---

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.

回答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]

回答3:

You could use list comprehension for that task following way:

numbers = [5, 6, 7, 8, 8]
maxnumbers = [i for i in numbers if i==max(numbers)]
print(*maxnumbers,sep=',')

output:

8,8

* operator in print is used to unpack values, sep is used to inform print what seperator to use: , in this case.

EDIT: If you want to get indices of biggest value and call max only once then do:

numbers = [5, 6, 7, 8, 8]
biggest = max(numbers)
positions = [inx for inx,i in enumerate(numbers) if i==biggest]
print(*positions,sep=',')

Output:

3,4

As you might check numbers[3] is equal to biggest and numbers[4] is equal to biggest.

回答4:

1. Count the occurrence of max number

2. iterate over the list to print the max number for the range of the count (1)

Hence:

item_no = [5, 6, 7, 8, 8]
counter = item_no.count(max(item_no))      # 2
print([max(item_no) for x in range(counter)])   

OUTPUT:

[8, 8]

回答5:

This issue can be solved in one line, by finding an item which is equal to the maximum value: To improve performance store max in var Mvalue=max(item_no) [i for i in item_no if i==Mvalue]

回答6:

I think it would be better if we evaluate the max in the array and its count in one iteration

def maxs(iterable):
    max = None
    count = 0
    for index, value in enumerate(iterable):
        if index == 0 or value >= max:
            if value != max:
                count = 0
            max = value
            count += 1
    return count * [max]


print (maxs([5, 6, 7, 8, 8]))   # [8, 8]
print (maxs([3, 2, 4, 5, 1, 2, 4, 5, 2, 5, 0])) # [5, 5, 5]
print (maxs([])) # []

Give it a Try!!