Python, looping and get the first max value

Question

Let's say I have this list containing these numbers: 1, 4, 8, 3, 5, 9, 2

I want to loop through this and get 8 because it's the first max value I encounter when loop through the list I'm thinking of loop through each [i], and + the if statement:

if [i] < [i+1] and [i+1] > [i+2]:

print(i)

Although this print 8 which is what I want, it also prints 9 because it satisfied the condition. Is there a way to make it stop after printing 8?!

Answer 1

Below is the code you can try though to get the first max value between i and i+1 index through the list

nums = [1, 4, 8, 3, 5, 9, 2]
for i in range(len(nums)-1):
    if nums[i] > nums[i+1]:
        print(nums[i])
        break

Answer 2

You can use the break statement to break out of your loop. You can check out the use of break, continue and pass statements in Python here,

https://www.geeksforgeeks.org/break-continue-and-pass-in-python/

The break statement is what is applicable to you here.

So, your code would look something like this,

for i in range(len(nums)-1):
  if [i] < [i+1]:
    print(i)
    break

Answer 3

By "first max", I'm assuming you mean the point where the immediate next element is lower?

Thats what I understood also going by your example - Want 8 as the output out of [1, 4, 8, 3, 5, 9, 2]

If that is what you're looking for, you can try this -

a = [1, 4, 8, 3, 5, 9, 2]
for idx, val in enumerate(a):
    if val > a[idx+1]:
        # print(f"{idx} : {val}")
        print(val)
        break

I'm checking for the 1st occurrence where an element is higher than its very next element and breaking out of the loop at that point.