I want to generate x number of variants of an address, but with a list of predefined symbols.
adress = "RUE JEAN ARGENTIN"
a_variants = ["À","Á","Â","Ä","Ã","Å"]
e_variants = ["É","È","Ê","Ë"]
i_variants = ["Ì","Í","Î","Ï"]
u_variants = ["Ù","Ú","Û","Ü"]
r_variants = ["Ŕ","Ŗ"]
o_variants = ["Ò","Ó","Ô","Ö","Õ"]
s_variants = ["Ś","Š","Ş"]
n_variants = ["Ń","Ň"]
d_variants = ["D","Đ"]
Example of output
# RUE JEAN ARGENTINS
# ŔUE JEAN ARGENTINS
# ŖUE JEAN ARGENTINS
# RÙE JEAN ARGENTINS
# RÚE JEAN ARGENTINS
# RÛE JEAN ARGENTINS
# RÙE JEAN ARGENTINS
# RÜE JEAN ARGENTINS
# ....
# ŖUE JEAN ARGENTÌNS
# ....
# ....
# ŖÛÉ JÉÀŇ ÀŖGÈNTÏŇŞ
# ....
I don't especially want a code answer in python or anything, but mostly an algorithmic idea to achieve this. Or even a mathematical calculation to find the number of possible variations.
Thank you very much
It should be something like:
R
has 3
versions (r_variants
+ original R
)U
has 5
versions (u_variants
+ original U
)
etc.and you have to multipicate all values
(number of items in r_variants + 1) * (number of items in u_variants + 1) * ....
If you count repeated chars in text then you can reduce it to
((number of items in a_variants + 1) * number of chars A in text) * ((number of items in e_variants + 1) * number of chars E in text) * ....
But in code I will use first method:
address = "RUE JEAN ARGENTIN"
variation = {
'A': ["À","Á","Â","Ä","Ã","Å"],
'E': ["É","È","Ê","Ë"],
'I': ["Ì","Í","Î","Ï"],
'U': ["Ù","Ú","Û","Ü"],
'R': ["Ŕ","Ŗ"],
'O': ["Ò","Ó","Ô","Ö","Õ"],
'S': ["Ś","Š","Ş"],
'N': ["Ń","Ň"],
'D': ["D","Đ"],
}
result = 1
for char in address:
if char in variation:
result *= (len(variation[char]) + 1)
print(result)
Result:
37209375 # 37 209 375
EDIT:
You can substract 1
to skip original version "RUE JEAN ARGENTIN"
37209374 # 37 209 374
EDIT:
If you use shorter text and smaller number of variation then you can easly generate all version on paper (or display on screen) and count them - and check if calculation is correct.
For example for RN
it gives 8 variation (9 versions - original)
You can use itertools.product()
to generate all versions
import itertools
for number, chars in enumerate(itertools.product(['R', "Ŕ","Ŗ"], ['N', "Ń","Ň"]), 1):
print(f'{number} | {"".join(chars)}')
Result:
1 | RN
2 | RŃ
3 | RŇ
4 | ŔN
5 | ŔŃ
6 | ŔŇ
7 | ŖN
8 | ŖŃ
9 | ŖŇ