How to remove a single number from a list with multiples of that number - python-3.7

Question

As I'm a beginner in coding I wanted to try to find the first three repeated numbers in a list. My problem is that in my code when there is a number repeated three, the code breaks.
The usual, remove, pop, and del, don't work as they delete one element in the list.
import random
r = random.randint
string = ""
def first_repeat(myList):
myList = sorted(list(myList))
print(myList)
number = 0
final_numbers = []
loop = 0
while loop < 2:
try:
if number == 0:
number += 1
else:
if myList[loop] == myList[loop-1]:
final_numbers.append(myList[loop])
else:
myList.pop(loop)
myList.pop (loop-1)
number = 0
if loop == 0 :
loop += 1
else:
loop -= 1
if len(final_numbers) > 3:
return final_numbers[0], final_numbers[1], final_numbers[2]
if len(myList) <=1:
loop += 2
except:
continue
return final_numbers
for n in range(20):
string = string+str(r(0,9))
print(first_repeat(string))
the expected result should be at the first three repeated numbers.

Answer 1

I added some print statements so you can go through your program and find out where the logic is wrong with your code.
import random
r = random.randint
string = ""
def first_repeat(myList):
myList = sorted(list(myList))
print(myList)
number = 0
final_numbers = []
loop = 0
while loop < 2:
print( 'inside while loop: loop = {}'.format( loop ))
try:
if number == 0:
number += 1
else:
if myList[loop] == myList[loop-1]:
print( 'in -> if myList[loop] == myList[loop-1]' )
final_numbers.append(myList[loop])
print( 'final_numbers: [{}]'.format( ','.join( final_numbers )))
else:
print( 'in first -> else' )
myList.pop(loop)
myList.pop (loop-1)
number = 0
print( 'myList: [{}]'.format( ','.join( myList ) ))
if loop == 0 :
loop += 1
else:
loop -= 1
if len(final_numbers) > 3:
print( 'returning final numbers' )
print( final_numbers )
return final_numbers[0], final_numbers[1], final_numbers[2]
if len(myList) <=1:
loop += 2
except:
continue
print( 'at end of this loop final numbers is: [{}]'.format( ','.join( final_numbers)))
print( 'press any key to continue loop: ')
input()
return final_numbers
for n in range(20):
string = string+str(r(0,9))
print(first_repeat(string))
Following is a method to do it taking advantage of pythons defaultdict
https://docs.python.org/2/library/collections.html#collections.defaultdict
#import defaultdict to keep track of number counts
from collections import defaultdict
#changed parameter name since you are passing in a string, not a list
def first_repeat( numbers_string ):
#create a dictionary - defaulddict( int ) is a dictionary with keys
#instantiated to 0 - (instead of throwing a key error)
number_count = defaultdict( int )
#convert your string to a list of integers - look up list iterations
numbers = [ int( s ) for s in list( numbers )]
# to store the repeated numbers
first_three_repeats = []
for number in numbers:
# for each number in the list, increment when it is seen
number_count[number] += 1
#at first occurence of 3 numbers, return the number
if number_count[number] == 2:
first_three_repeats.append( number )
if len( first_three_repeats ) == 3:
return first_three_repeats
#if here - not three occurrences of repeated numbers
return []
for n in range(20):
string = string+str(r(0,9))
print( findFirstThreeNumbers( string ))