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I currently have a dictionary (Duplicate_combos) that has a unique identifying number for the key value and the value is a list with two elements, a company code and then either a yes or no (both of these values are currently stored as strings). I am essentially just trying to see where the company code is equal and the second term is no for both.
So if this was my dictionary:

1234: ['123' , 'No'] , 1235:['123', 'No'], 1236: ['123','Yes'], 1237: [124,'No']

I would only want to return 1234 and 1235. The code below is what I currently have and I really need to optimize it because while it does work when I tested it on a small data set, I will need to use it on a much larger one (43,000 lines) and in early testing, it is taking 45+ minutes with seemingly no sign of ending soon.

def open_file():

 in_file = open("./Data.csv","r")
 blank = in_file.readline()
 titles = in_file.readline()
 titles = titles.strip()
 titles = titles.split(',')

 cost_center = [] # 0
 cost_center_name = []# 1
 management_site = [] # 15
 sub_function = [] #19
 LER = [] #41
 Company_name = [] #3
 Business_group = [] #7
 Value_center = [] #9 
 Performance_center = [] #10
 Profit_center = [] #11

 total_lines = 

 for line in in_file:

 line = line.strip()
 line = line.split(',')
 cost_center.append(line[0])
 cost_center_name.append(line[1])
 management_site.append(line[15])
 sub_function.append(line[19])
 LER.append(line[41])
 Company_name.append(line[3])
 Business_group.append(line[7])
 Value_center.append(line[9])
 Performance_center.append(line[10])
 Profit_center.append(line[11])

 # create a dictionary of all the lines with the key being the unique cost center number (cost_center list)
 total_lines[line[0]] = line[1:]


 return(cost_center, cost_center_name, management_site, sub_function, LER, Company_name, Business_group, total_lines, titles, Value_center, Performance_center, Profit_center)


def find_duplicates(Duplicate_combos):

 Real_duplicates = []
 archive_duplicates = []


 # loop through the dictionary of duplicate combos by the keys 
 for key in Duplicate_combos:
 code = Duplicate_combos[key][0]
 for key2 in Duplicate_combos:
 # if the two keys are equal to each other, it means you are comparing the key to itself, which we don't want to do so we continue
 if key == key2:
 continue
 # if the company codes are the same and they are BOTH NOT going to be consolidated, we have found a real duplicate 
 elif Duplicate_combos[key2][0] == code and Duplicate_combos[key2][1] == 'No' and Duplicate_combos[key][1] == 'No':
 # make sure that we haven't already dealt with this key before
 if key not in archive_duplicates:
 Real_duplicates.append(key)
 archive_duplicates.append(key)

 if key2 not in archive_duplicates:
 Real_duplicates.append(key2)
 archive_duplicates.append(key2)
 continue 
 return(Real_duplicates)

1. 
   

   It's easier to read code that tuple unpacks the values in the for from dict.items().

   
   
   

   for key1, (code1, option1) in Duplicate_combos.items():
   

   
   


2. 
   archive_duplicates is a duplicate of Real_duplicates. There's no need for it.


3. 
   

   It doesn't seem like the output needs to be ordered, and so you can just make Real_duplicates a set. This means it won't have duplicates, and you don't have to loop through it twice each time you want to add a value.

   
   
   

   This alone speeds up your program from $O(n^3)$ to $O(n^2)$.

   
   



4. Your variable names are quite poor, and don't adhere to PEP8. I have changed them to somewhat generic names, but it'd be better if you replace, say, items with what it actually is.

def find_duplicates(items):
 duplicates = set()
 for key1, (code1, option1) in items.items():
 for key2, (code2, option2) in items.items():
 if key1 == key2:
 continue
 elif code1 == code2 and option1 == option2 == 'No':
 duplicates.add(key1)
 duplicates.add(key2)
 return list(duplicates)

5. 
   

   You don't need to loop over Duplicate_combos twice.

   
   
   

   To do this you need to make a new dictionary grouping by the code. And only adding to it if the option is 'No'.

   
   
   

   After building the new dictionary you can iterate over it's values and return ones where the length of values is greater or equal to two.

   
   

def find_duplicates(items):
 by_code = 
 for key, (code, option) in items.items():
 if option == 'No':
 by_code.setdefault(code, []).append(key)

 return [
 key
 for keys in by_code.values()
 if len(keys) >= 2
 for key in keys
 ]

This now runs in $O(n)$ time rather than $O(n^3)$ time.

>>> find_duplicates(
 101: ['1', 'No'], 102: ['1', 'No'],
 103: ['1','Yes'], 104: ['1', 'No'],
 201: ['2', 'No'], 202: ['2', 'No'],
 301: ['3', 'No'], 401: ['4', 'No'],
)
[101, 102, 104, 201, 202]

When reading your data, you open a file but never .close() it. You should take the habit to use the with keyword to avoid this issue.

You should also benefit from the csv module to read this file as it will remove boilerplate and handle special cases for you:

def open_file(filename='./Data.csv'):
 cost_center = [] # 0
 cost_center_name = []# 1
 management_site = [] # 15
 sub_function = [] #19
 LER = [] #41
 Company_name = [] #3
 Business_group = [] #7
 Value_center = [] #9
 Performance_center = [] #10
 Profit_center = [] #11
 total_lines = 

 with open(filename) as in_file:
 next(in_file) # skip blank line
 reader = csv.reader(in_file, delimiter=',')

 for line in reader:
 cost_center.append(line[0])
 cost_center_name.append(line[1])
 management_site.append(line[15])
 sub_function.append(line[19])
 LER.append(line[41])
 Company_name.append(line[3])
 Business_group.append(line[7])
 Value_center.append(line[9])
 Performance_center.append(line[10])
 Profit_center.append(line[11])

 # create a dictionary of all the lines with the key being the unique cost center number (cost_center list)
 total_lines[line[0]] = line[1:]

 return cost_center, cost_center_name, management_site, sub_function, LER, Company_name, Business_group, total_lines, titles, Value_center, Performance_center, Profit_center

Doing

def get_dupes(df):
 if sum(df.loc[1]=='No')<2:
 return None
 else:
 return list(df.loc[:,df.loc[1]=='No'].columns)
df.groupby(axis=1,by=df.loc[0]).apply(get_dupes)

Got me

 0
 124 None
 123 [1234, 1235]
 dtype: object

Your question wasn't quite clear on what you want the output to be if there are multiple company values with duplicate values (e.g. if the input is 1234: ['123' , 'No'] , 1235:['123', 'No'], 1236: ['123','Yes'], 1237: [124,'No'],1238: [124,'No']
do you want [1234, 1235, 1237, 1238] or [[1234, 1235], [1237, 1238]]), so you can modify this code accordingly.