获取一个打印语句两次打印一个值的不同输出，python

Question

I have a code I'm trying to run. But after running the code I get two print value for one print statement and a single test case.

The output I am getting:

PS D:\VS Code\Python> & C:/Users/zyx/AppData/Local/Programs/Python/Python310/python.exe "d:/VS Code/Python/dump/dump 3.py"
lo: 0 , hi: 7 , mid: 3 , mid_number: 7
lo: 4 , hi: 7 , mid: 5 , mid_number: 1
True
PS D:\VS Code\Python> 

The only output I should be getting is

PS D:\VS Code\Python> & C:/Users/zyx/AppData/Local/Programs/Python/Python310/python.exe "d:/VS Code/Python/dump/dump 3.py"
lo: 0 , hi: 7 , mid: 3 , mid_number: 7
True
PS D:\VS Code\Python> 

How can I get remove this extra print statement error? Here's the code:

test1 = {
    'input': {
        'nums': [4, 5, 6, 7, 8, 1, 2, 3]
    },
    'output': 5
}

def count_rotations_binary(nums):
    lo = 0
    hi = len(nums)-1
    
    while lo<=hi:
        mid = (lo+hi)//2
        mid_number = nums[mid]
        
        print("lo:", lo, ", hi:", hi, ", mid:", mid, ", mid_number:", mid_number)

        if mid > 0 and nums[mid]<nums[mid-1]:
            return mid
        
        elif nums[mid] > nums[hi]:
            lo = mid + 1  
        
        else:
            hi = mid - 1
    
    return 0

print(count_rotations_binary(**test1['input']) == test1['output'])

Answer 1

Its a while loop, so it can run multiple times. Therefore your code must be making tthe while loop multiple times therefor printing multiple times.

Answer 2

This isn't an error. For the input you have provided, and your binary search logic to find the starting position of the number sequence, 2 loops (and therefore 2 print statements) is expected.

During the first pass through the loop we have the following values:
lo: 0 , hi: 7 , mid: 3 , mid_number: 7

Then we get to the binary search logic:

        if mid > 0 and nums[mid]<nums[mid-1]:
            return mid
        
        elif nums[mid] > nums[hi]:
            lo = mid + 1  
        
        else:
            hi = mid - 1

The if statement is False as 7 < 6 is False.
Then the elif statement nums[mid] > nums[hi] evaluates to True, as 7 > 3 . So lo is set to mid+1 (4), and the loop starts again.

On the second loop we have
lo: 4 , hi: 7 , mid: 5 , mid_number: 1
The if statement is evaluated as True:
mid > 0 and nums[mid]<nums[mid-1]
evaluates as 5 > 0 and 1 < 8
So return mid is called and the function returns 5.

Answer 3

Use If instead of while that's it. use this.

if lo<=hi:

instead of this.

while lo<=hi:

Reason: It can check conditions a single time and give an output no matter how long your data is present here:

'nums': [4, 5, 6, 7, 8, 1, 2, 3]

Fact: if you cannot change your program and add some extra data here:

'nums': [4, 5, 6, 7, 8, 1, 2, 3,5,6,8,4]

you will get some extra lines because of your algorithm until while loop does not matches your given condition.