How to display all possible options

Без рандома, должны вывести все возможные значения

 

Через регулярки

 

s = 'ABCDEF'
s = 'ABCDEF'
d = '0123456789'

res = list('YXYYYXXX')


with open('codes.txt', 'a') as f:

for i in s:

res[0] = i

for j in d:

res [1] = j

for q in s:

res[2] = q

for w in s:
res[3] =w

for r in s:
res[4] = r

for t in d:
res[5] = t

for y in d:
res[6] = y

for u in d:
res[7] = u

f.write(''.join(res))
f.write('\n')

Python

s = 'ABCDEF'

s = 'ABCDEF'

d = '0123456789'



res = list('YXYYYXXX')





with open('codes.txt', 'a') as f:



    for i in s:

       

        res[0] = i



        for j in d:



            res [1] = j



            for q in s: 



                res[2] = q



                for w in s:

                    res[3] =w 



                    for r in s:

                        res[4] = r 

                       

                        for t in d:

                            res[5] = t



                            for y in d:

                                res[6] = y 



                                for u in d:

                                    res[7] = u



                                    f.write(''.join(res))

                                    f.write('\n')

 

s = 'ABCDEF' s = 'ABCDEF' d = '0123456789' res = list('YXYYYXXX') with open('codes.txt', 'a') as f: for i in s: res[0] = i for j in d: res [1] = j for q in s: res[2] = q for w in s: res[3] =w for r in s: res[4] = r for t in d: res[5] = t for y in d: res[6] = y for u in d: res[7] = u f.write(''.join(res)) f.write('\n')

 

from itertools import permutations
with open("file.txt",'w+') as f:
for i1 in permutations('ABCDEF', 1):
for i2 in permutations('0123456789', 1):
for i3 in permutations('ABCDEF', 3):
for i4 in permutations('0123456789', 3):
f.write(f"{''.join(list(i1)) + ''.join(list(i2)) + ''.join(list(i3)) + ''.join(list(i4))}\n")

Python

from itertools import permutations

with open("file.txt",'w+') as f:

    for i1 in permutations('ABCDEF', 1):

        for i2 in permutations('0123456789', 1):

            for i3 in permutations('ABCDEF', 3):

                for i4 in permutations('0123456789', 3):

                    f.write(f"{''.join(list(i1)) + ''.join(list(i2)) + ''.join(list(i3)) + ''.join(list(i4))}\n")

 

всего значений должно быть 12 960 000
(описание шаблона не такое как в примере)
вот вариант для результатов типа 0A628BFD в bash: