# Find the most frequently chosen number

Asked 1 weeks ago, Updated 1 weeks ago, 2 views

``````import numpy as np
import xlwings as xw
import random

for i in range(100):
F1AA = random.randint(0, 52)
F1BB = random.randint(0, 52)
F1CC = random.randint(0, 156)
F2AA = random.randint(0, 26)
F2BB = random.randint(0, 52)
F2CC = random.randint(0, 104)
F3AA = random.randint(0, 52)
F3BB = random.randint(0, 52)
F3CC = random.randint(0, 156)
P2AA = random.randint(0, 51)
P2BB = random.randint(0, 68)
P2CC = random.randint(0, 136)
P3AA = random.randint(0, 51)
P3BB = random.randint(0, 51)
P3CC = random.randint(0, 153)
S1OO = random.randint(0, 108)
S2AA = random.randint(0, 6)
S2BB = random.randint(0, 12)
S2CC = random.randint(0, 36)
S3AA = random.randint(0, 6)
S3BB = random.randint(0, 12)
S3CC = random.randint(0, 36)

print(f'F1-AA-{F1AA}',f'F1-BB-{F1BB}',f'F1-CC-{F1CC}',f'F2-AA-{F2AA}',f'F2-BB-{F2BB}',f'F2-CC-{F2CC}',f'F3-AA-{F3AA}',f'F3-BB-{F3BB}',f'F3-CC-{F3CC}', f'P2-AA-{P2AA}', f'P2-BB-{P2BB}', f'P2-CC-{P2CC}', f'P3-AA-{P3AA}', f'P3-BB-{P3BB}', f'P3-CC-{P3CC}', f'S1-OO-{S1OO}',f'S2-AA-{S2AA}', f'S2-BB-{S2BB}', f'S2-CC-{S2CC}', f'S3-AA-{S3AA}', f'S3-BB-{S3BB}', }', f'S3-CC-{S3CC}' )
``````

It seems like it's simple coding, but it's getting longer because I'm trying to do something without learning it.

To give you a rough explanation, I want to find the number that comes out the most per code (F1AA, F1BB, etc.) by spinning a lot randomly, but I can only implement it yet.

Is there any way to simplify it or code for finding the numbers that come out a lot here?

I'm sorry I didn't explain too much.

python

2022-09-20 14:44

I'm not sure. If you have learned probability and statistics, you will know that if you perform `random.randomint (0, 52)` very often, `F1AA` will be assigned very evenly with a probability of 1/53 each of the integers from `0` to `52`.
For `random.randomint(0,156)`, `F1CC` would be from `0` to `156`. If so, literally finding the "number that came out the most for each code" is a little overshadowed, as long as the law of large numbers is valid.

Looking at the comments again, I guess I just need to get a snapshot of 100 times. Then you can just use the code roughly.

``````# This makes F1AA a list of 100 random numbers between 0 and 52.
# Perm: https://stackoverflow.com/a/30386371/8680764
from random import randint
F1AA = [randint(0, 52) for _ in range(100)]

# If you give a list, it's a function that tells you the most frequently appearing element.
# Perm: https://www.geeksforgeeks.org/python-find-most-frequent-element-in-a-list/
def most_frequent(List):
counter = 0
num = List
for i in List:
curr_frequency = List.count(i)
if (curr_frequency> counter):
counter = curr_frequency
num = i
return num

# Please use a moderate combination.
print(F1AA)
print(most_frequent(F1AA))

# By the way.
# If you take a random number from 0 to 52,
# All of those numbers are equally qualified to be the "most likely" numbers.
# That's always the case, whether it's 100 or 1000000.
# Thanks to this mathematical fact, the above code is simplified into the next line.

# A function that obtains the most frequently generated integer when the integer from start to end is randomly selected times
def get_most_frequent(start, end, times) :
return landint(start, end) # No need to use times...!
``````

2022-09-20 14:44

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