程序在 break 语句后继续循环

Question

I'm new to python and I was going through a few problem to practice. The question is:

#Given an array, return the first recurring character
#Example1 : array = [2,1,4,2,6,5,1,4]
#It should return 2
#Example 2 : array = [2,6,4,6,1,3,8,1,2]
#It should return 6


lsts = [2,5,1,2,3,5,1,2,4]
    
def findDouble(arrs):
  repeats = dict()
  for arr in arrs:
    repeats[arr] = repeats.get(arr, 0) + 1
    if repeats[arr] == 2: break
    print(arr)
        
        
    
findDouble(lsts)
    
#0(n)

My understanding is that after the "break" it should end the loop, so I should just get 2. Instead it goes through the entire thing and I get 2, 5, and 1. What am I not getting?

Answer 1

Perhaps it is easier to understand if you put a print(repeats) right after assigning repeats[arr] = ...

Iteration 1: arr == 2

{2: 1} # key `2` was created and assigned `0 + 1`

Iteration 2: arr == 5

{2: 1, 5: 1} # key `5` created and assigned  `0 + 1`

Iteration 3: arr == 1

{2: 1, 5: 1, 1: 1} # key `1` created and assigned `0 + 1`

Iteration 4: arr == 2

{2: 2, 5: 1, 1: 1} # key `2` was already present, assigned `1 + 1`
repeat[arr] == 2: # evaluates to True, so it breaks

Answer 2

The first time through the loop, arrs is 2. That key does not yet exist in the dictionary, so repeats[2] is created with a value of 1, and the program prints 2 .

The second time through the loop, arrs is 5. That key does not yet exist in the dictionary, so repeats[5] is created with a value of 1, and the program prints 5 .

The third time through the loop, arrs is 1. That key does not yet exist in the dictionary, so repeats[1] is created with a value of 1, and the program prints 1 .

The fourth time through the loop, arrs is 2. That key already exists in the dictionary with a value of 1, so repeats[2] is assigned a new value of 2, and the loop breaks.