Numeric Anagrams - Count Occurances

[monty024]

Greetings,

I am new to Python and messing around with a script with that will read a text file. Each line in the text file contains a three digit number from 000 to 999, the number of lines in the text file can vary. The script I made counts the number of occurrences for each number. However what I want to expand that and get the numeric anagram count.
For example if I have the list as shown below:

123
132
750
213
231
312
321
123
990
570

i would like the output to be:
(123,132,213,231,321,312) = 7
(099,909,990) = 1
(057, 075, 507, 570, 705, 750) = 2


This is the code I have to count the occurrences of a number. I know it can be cleaned up a but I am still just learning and will figure out how to clean it up, but I just need help with the code for the numeric anagram portion. I have searched the forum but only came across word anagram code. Any help would be greatly appreciated.

Regards,
Monty

for occurance in range(000, 100):
    file = open("/home/monty/Documents/Python_Projects/Book1-new.txt", "r")
    data = file.read()
    occurance_fill = f"{occurance:03d}"
    occurances = data.count(str(occurance_fill))

for occurance2 in range(100, 1000):
    file = open("/home/monty/Documents/Python_Projects/Book1-new.txt", "r")
    data = file.read()
    occurance2_fill = f"{occurance2}"
    occurances = data.count(str(occurance2_fill))
    print(str(occurance2_fill) + " : " + str(occurances))

[bowlofred]

Although these are numbers, you can still deal with it as strings. If you alphabetically sort each number, you can collapse each to be the "reference" number for each of the anagrams. So "312" would always reference "123".

Then you can just count 'em up. This code doesn't show all the possible anagrams (that is doable with the more-itertools.distinct_permutations). It just shows the reference one.

from collections import Counter

num_list = """123
132
750
213
231
312
321
123
990
570"""

num_frequency = Counter()
for number in num_list.splitlines():
    num_frequency["".join(sorted(number))] += 1

for number, frequency in num_frequency.items():
    print(f"({number}) = {frequency}")

Output:
(123) = 7
(057) = 2
(099) = 1

[monty024]

Thank you BowloFred! That helps so much.
I am learning Python on my own so sometimes I just need to be pointed in the right direction.