# How to determine the maximum number from a list of numbers?

### Question

In this exercise, the function is passed three numbers as parameters and determines the maximum value. How can the function return the maximum value for a list of numbers?

Python includes a `max()` function which does this task. If you pass the `max()` function a list of numbers, it will return the largest number in the list as shown in this example. If the exercise function was re-written to accept a list of numbers as a parameter, it could just call `max()` to find the largest value.

``````num = (100, 2, 5, 99, 24, 300, 88, 72)
print(max(num))
#Output: 300
``````
15 Likes

But with max you can’t handle the tie cases, it just returns the max number indepently of how many cases it encounters.

4 Likes

you could find the largest number with max() but you can’t find the second largest or generate it’s a tie.

I ended up using the below code after checking this post on stackoverflow

``````def max_num(num1,num2,num3):
l = [num1,num2,num3]
sort = sorted(l,reverse=True)
if sort == sort:
return "It's a tie!"
else:
return sort
``````

This adds the numbers to a list and sorts by largest.
I then check to see if the largest and second largest are equal, if they are return the string.
else return the largest number.

15 Likes

def max_num(num1, num2, num3):
if (num2 == num3) | (num1 == num3) | (num1 == num2):
return “It’s a tie!”
else:
all_num = [num1, num2, num3]
return max(all_num)

More of a long hand answer but might be simpler for someone learning at this stage

6 Likes

What if your list is [2, 1, 1] ?

2 Likes
``````def max_num(num1, num2, num3):
if (num2 == num3) | (num1 == num3) | (num1 == num2):
return "It's a tie!"
else:
all_num = [num1, num2, num3]
return max(all_num)
``````

it returns 2 as the max. I added indentation here to make it pretty

The indentation makes it Python!

Does it?

``````def max_num(num1, num2, num3):
if (num2 == num3) | (num1 == num3) | (num1 == num2):
return "It's a tie!"
else:
all_num = [num1, num2, num3]
return max(all_num)

print(max_num(2,1,1))
``````

Output:

``````It's a tie!
``````
1 Like

No, it doesn’t. It returns 1, because the if statement will be evaluated first and you will exit the function once the first True condition is met, which is that 1 and 1 are equal

If you actually run the function, you’ll see that it prints “It’s a tie!”

It does so because or (just as bitwise or as used here for some reason) stops processing (“is short-circuited”) when the first True-returning expression is reached, and in this case, that is the that is the first one, (num2 == num3). That causes the if block to run and return “It’s a tie!”

1 Like

But that’s not the behavior we want. The instructions say: The function should return the largest of these three numbers. If any of two numbers tie as the largest, you should return `"It's a tie!"`, so you would want this to return 2, because that’s the highest number and you’re evaluating for equality first. So if you have two numbers that are equal, but lower than the max, you’re going to get the wrong result

1 Like

Another (specific to the example) option would be:

def max_num(num1, num2, num3):
maximum = max(num1, num2, num3)
count = 0

if num1 == maximum:
count += 1
if num2 == maximum:
count += 1
if num3 == maximum:
count += 1

if count > 1:
return “It’s a tie!”
else:
return maximum

1 Like
``````def max_num(num1, num2, num3):
maximum = max(num1, num2, num3)
if (maximum == num1 or maximum == num2) and maximum == num3:
return "It's a tie!"
elif maximum == num1 and maximum == num2:
return "It's a tie!"
else:
return maximum
``````

Just posting this here for people struggling with this as I did. The real issue with this task is just testing the max_num against the other numbers to find if and two are equal to the max. This was as concise as I could formulate it with my current knowledge.

3 Likes

There’s another list command that counts the number of times that one value appears in a list. You can use it to identify Ties. The command is .count()

``````def max_num(num1, num2, num3):
num_list = [num1, num2, num3]
if(num_list.count(max(num_list)) == 1):
return max(num_list)
else:
return "It's a tie!"
``````

Here I used the max() function to find que greatest value and then the .count() to see if there’s more than one.

3 Likes

def max_num(list_of_nums):
max = list_of_nums
for i in list_of_nums:
if max <= i:
max = i
return max

can’t format

Try the </> icon in the middle of the menu bar that appears atop the text box when you open it to type.

Also, I think this works.

``````def max_num(num1, num2, num3):
if num1 == num2 or num1 == num3 or num2 == num3:
return "It's a tie!"
else:
return max(num1, num2, num3)
print(max_num(-10, 0, 10))
``````
1 Like

why use pipes instead of or in your first conditional? Seems to throw it off

Here’s my solution:

# Write your max_num function here:

def max_num(num1, num2, num3):
max = num1
if num2 > max:
max = num2
if num3 > max:
max = num3

if (max == num1 and max == num2) or (max == num2 and max == num3) or (max == num1 and max == num3):
return “It’s a tie!”
else:
return max

indeed it is- thank you

It kinda works except for e.g. max_num(5, -10, -10) (as one of the examples from the exercise shows). The problem here is that 5 is clearly the biggest number yet your code will print “It’s a tie!”. (I made the same mistake, therefore I know it)