Getting all combinations of a string with subscript and specif format

Question

I have seen various combination formats. But I have to try something new. Lets say, we have three lists:

A = ('Li', 'B', 'C', 'N')
B = ('Li', 'B', 'C', 'N')  
X = ('O', 'Br', 'Cl')

All the element of A are same as B. Now I need to create all possible combinations in the following formats:

# (i)   ABX3
# (ii)  AA'B2X6
# (iii) A2B2X6

Where A2 (is A_{2}) is 'A' with 2 in subscript and A!=A' in the combination.

Here is what I have tried so far:

import itertools
elem_list = ['Li','B','C','N']
CCA_combos1 = list(itertools.combinations(elem_list, 2)) 
display(CCA_combos1, len(CCA_combos1))
   
listX1 = ('O','Cl', 'Br')
CCA_combosX = []
for x in range(len(CCA_combos1)):
  for y in range(0,len(listX1)):
    CCA_combosX += [tuple(CCA_combos1[x]) + tuple(listX1[y])]

resX1 = [''.join(tups) for tups in CCA_combosX]

The above code not working well for Br and Cl I have no idea yet to create combination in A2BB'X6.

Can you clarify what you mean by `A!=A' in the combination.`? — ChrisOram, May 17 '22 at 06:48
`"Where A2 (is A_{2}) is 'A' with 2 in subscript and A!=A' in the combination"` hard to understand. Can you explain that a more in detail ? — Rabinzel, May 17 '22 at 06:49
I don't want to speak for OP but `A != A'` looks like it means they are distinct elements. — ChrisOram, May 17 '22 at 06:50
If I'd have to guess I'd say, `A2` means "take element of group A twice ('C2' or 'N2')" while `AA'` means "pick two elements of group A but the one can not be equal to the other" — Rabinzel, May 17 '22 at 07:14

Robin Thibaut · Accepted Answer · 2022-05-19T08:53:20.957

You could use itertools.product. I hope I understood your question correctly.

from itertools import product

A = ('Li', 'B', 'C', 'N')
B = ('Li', 'B', 'C', 'N')
X = ('O', 'Br', 'Cl')

list1, list2, list3 = [], [], []

# (i)   ABX3
for a, b, x in product(A, B, X):
    print(a, b, x, 3)
    list1.append((a, b, x, 3))

# (ii)  AA'B2X6
for a, ap, b, x in product(A, A, B, X):
    if a != ap:
        print(a, ap, b, 2, x, 6)
        list2.append((a, ap, b, 2, x, 6))

# (iii) A2B2X6
for a, b, x in product(A, B, X):
    print(a, 2, b, 2, x, 6)
    list3.append((a, 2, b, 2, x, 6))
    
    
# print the number of elements in each list
print('\n(i)   ABX3:', len(list1))
print('(ii)  AA\'B2X6:', len(list2))
print('(iii) A2B2X6:', len(list3))

Outputs:

# (i)
Li Li O 3
Li Li Br 3
Li Li Cl 3
Li B O 3
Li B Br 3
Li B Cl 3
Li C O 3
Li C Br 3
Li C Cl 3
Li N O 3
Li N Br 3
Li N Cl 3
B Li O 3
B Li Br 3
B Li Cl 3
B B O 3
B B Br 3
B B Cl 3
B C O 3
B C Br 3
B C Cl 3
B N O 3
B N Br 3
B N Cl 3
C Li O 3
C Li Br 3
C Li Cl 3
C B O 3
C B Br 3
C B Cl 3
C C O 3
C C Br 3
C C Cl 3
C N O 3
C N Br 3
C N Cl 3
N Li O 3
N Li Br 3
N Li Cl 3
N B O 3
N B Br 3
N B Cl 3
N C O 3
N C Br 3
N C Cl 3
N N O 3
N N Br 3
N N Cl 3

# (ii)
Li B Li 2 O 6
Li B Li 2 Br 6
Li B Li 2 Cl 6
Li B B 2 O 6
Li B B 2 Br 6
Li B B 2 Cl 6
Li B C 2 O 6
Li B C 2 Br 6
Li B C 2 Cl 6
Li B N 2 O 6
Li B N 2 Br 6
Li B N 2 Cl 6
Li C Li 2 O 6
Li C Li 2 Br 6
Li C Li 2 Cl 6
Li C B 2 O 6
Li C B 2 Br 6
Li C B 2 Cl 6
Li C C 2 O 6
Li C C 2 Br 6
Li C C 2 Cl 6
Li C N 2 O 6
Li C N 2 Br 6
Li C N 2 Cl 6
Li N Li 2 O 6
Li N Li 2 Br 6
Li N Li 2 Cl 6
Li N B 2 O 6
Li N B 2 Br 6
Li N B 2 Cl 6
Li N C 2 O 6
Li N C 2 Br 6
Li N C 2 Cl 6
Li N N 2 O 6
Li N N 2 Br 6
Li N N 2 Cl 6
B Li Li 2 O 6
B Li Li 2 Br 6
B Li Li 2 Cl 6
B Li B 2 O 6
B Li B 2 Br 6
B Li B 2 Cl 6
B Li C 2 O 6
B Li C 2 Br 6
B Li C 2 Cl 6
B Li N 2 O 6
B Li N 2 Br 6
B Li N 2 Cl 6
B C Li 2 O 6
B C Li 2 Br 6
B C Li 2 Cl 6
B C B 2 O 6
B C B 2 Br 6
B C B 2 Cl 6
B C C 2 O 6
B C C 2 Br 6
B C C 2 Cl 6
B C N 2 O 6
B C N 2 Br 6
B C N 2 Cl 6
B N Li 2 O 6
B N Li 2 Br 6
B N Li 2 Cl 6
B N B 2 O 6
B N B 2 Br 6
B N B 2 Cl 6
B N C 2 O 6
B N C 2 Br 6
B N C 2 Cl 6
B N N 2 O 6
B N N 2 Br 6
B N N 2 Cl 6
C Li Li 2 O 6
C Li Li 2 Br 6
C Li Li 2 Cl 6
C Li B 2 O 6
C Li B 2 Br 6
C Li B 2 Cl 6
C Li C 2 O 6
C Li C 2 Br 6
C Li C 2 Cl 6
C Li N 2 O 6
C Li N 2 Br 6
C Li N 2 Cl 6
C B Li 2 O 6
C B Li 2 Br 6
C B Li 2 Cl 6
C B B 2 O 6
C B B 2 Br 6
C B B 2 Cl 6
C B C 2 O 6
C B C 2 Br 6
C B C 2 Cl 6
C B N 2 O 6
C B N 2 Br 6
C B N 2 Cl 6
C N Li 2 O 6
C N Li 2 Br 6
C N Li 2 Cl 6
C N B 2 O 6
C N B 2 Br 6
C N B 2 Cl 6
C N C 2 O 6
C N C 2 Br 6
C N C 2 Cl 6
C N N 2 O 6
C N N 2 Br 6
C N N 2 Cl 6
N Li Li 2 O 6
N Li Li 2 Br 6
N Li Li 2 Cl 6
N Li B 2 O 6
N Li B 2 Br 6
N Li B 2 Cl 6
N Li C 2 O 6
N Li C 2 Br 6
N Li C 2 Cl 6
N Li N 2 O 6
N Li N 2 Br 6
N Li N 2 Cl 6
N B Li 2 O 6
N B Li 2 Br 6
N B Li 2 Cl 6
N B B 2 O 6
N B B 2 Br 6
N B B 2 Cl 6
N B C 2 O 6
N B C 2 Br 6
N B C 2 Cl 6
N B N 2 O 6
N B N 2 Br 6
N B N 2 Cl 6
N C Li 2 O 6
N C Li 2 Br 6
N C Li 2 Cl 6
N C B 2 O 6
N C B 2 Br 6
N C B 2 Cl 6
N C C 2 O 6
N C C 2 Br 6
N C C 2 Cl 6
N C N 2 O 6
N C N 2 Br 6
N C N 2 Cl 6

# (iii)
Li 2 Li 2 O 6
Li 2 Li 2 Br 6
Li 2 Li 2 Cl 6
Li 2 B 2 O 6
Li 2 B 2 Br 6
Li 2 B 2 Cl 6
Li 2 C 2 O 6
Li 2 C 2 Br 6
Li 2 C 2 Cl 6
Li 2 N 2 O 6
Li 2 N 2 Br 6
Li 2 N 2 Cl 6
B 2 Li 2 O 6
B 2 Li 2 Br 6
B 2 Li 2 Cl 6
B 2 B 2 O 6
B 2 B 2 Br 6
B 2 B 2 Cl 6
B 2 C 2 O 6
B 2 C 2 Br 6
B 2 C 2 Cl 6
B 2 N 2 O 6
B 2 N 2 Br 6
B 2 N 2 Cl 6
C 2 Li 2 O 6
C 2 Li 2 Br 6
C 2 Li 2 Cl 6
C 2 B 2 O 6
C 2 B 2 Br 6
C 2 B 2 Cl 6
C 2 C 2 O 6
C 2 C 2 Br 6
C 2 C 2 Cl 6
C 2 N 2 O 6
C 2 N 2 Br 6
C 2 N 2 Cl 6
N 2 Li 2 O 6
N 2 Li 2 Br 6
N 2 Li 2 Cl 6
N 2 B 2 O 6
N 2 B 2 Br 6
N 2 B 2 Cl 6
N 2 C 2 O 6
N 2 C 2 Br 6
N 2 C 2 Cl 6
N 2 N 2 O 6
N 2 N 2 Br 6
N 2 N 2 Cl 6

(i)   ABX3: 48
(ii)  AA'B2X6: 144
(iii) A2B2X6: 48

How do I store 1. print(a, b, x, 3), 2. print(a, ap, b, 2, x, 6) and 3. print(a, 2, b, 2, x, 6) in different variables. So that we can operate them in future. For example I want to merge the combinations (1 and 2) or (1 and 3). — Neelesh Gupta, May 19 '22 at 06:43
same time I want to count the number of combination in each. — Neelesh Gupta, May 19 '22 at 07:18
You can simply use lists. I edited the answer to store the variables into 3 different lists. — Robin Thibaut, May 19 '22 at 08:50
There is huge redundancy in the output. For example ABO3 is similar to BAO3. I am just pointing on one example of output list i.e. LiBO3 and BLiO3 are same — Neelesh Gupta, May 19 '22 at 10:56

Getting all combinations of a string with subscript and specif format

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