Hello @josef-pkt! Thanks for opening this PR. We checked the lines you've touched for PEP 8 issues, and found:

• In the file statsmodels/stats/tests/test_multicomp_tools.py:

Line 61:19: E124 closing bracket does not match visual indentation

funny detail with current code

When specifying the distance matrix, and a diagonal element is nonzero (1), then it does not form a clique with itself. No letter is assigned to group (if it also has nonzero distance to the other groups).
Found because I made a mistake in a distance matrix example.
Is this a bug or a feature?

Insufficient input validation?

As aside:
I used frozensets, because I used a set of sets, which requires hashable frozenset inside.

problem with pandas html in notebook
print looks fine, but html strips all the leading and trailing whitespace in the letters, so that it's just a single column of letters.

print(group_frame)
      name      mean  ci_sim_low  ci_sim_upp letters
0   France  3.983286    3.835050    4.131523   a    
1  Germany  4.867173    4.726966    5.007380     b  
2    Japan  4.102950    3.955065    4.250834   a    
3   Sweden  4.113893    3.979916    4.247869   a    
4      USA  6.317059    6.187845    6.446273       c

We need to add "-" for empty letter position, optionally

(The above is from an adjusted version of the cylinder example where also Germany differs from rest.)

sorted by mean:

frame_s = group_frame.sort_values("mean")
print(frame_s)
      name      mean  ci_sim_low  ci_sim_upp letters
0   France  3.983286    3.835050    4.131523   a    
2    Japan  4.102950    3.955065    4.250834   a    
3   Sweden  4.113893    3.979916    4.247869   a    
1  Germany  4.867173    4.726966    5.007380     b  
4      USA  6.317059    6.187845    6.446273       c

It looks like relabeling integer indices is quite easy.
uses standard method to get group indicators to and from dummy matrix

e.g. change label of first two groups

a = np.arange(3)
b = np.array([1, 0, 2])
xx = np.repeat(a, 3)
xx, a, b
(array([0, 0, 0, 1, 1, 1, 2, 2, 2]), array([0, 1, 2]), array([1, 0, 2]))

(xx[:, None] == a) @ b
array([1, 1, 1, 0, 0, 0, 2, 2, 2])
# or (which should also work if `b` is a sting array)
b[np.nonzero(xx[:, None] == a)[1]]
array([1, 1, 1, 0, 0, 0, 2, 2, 2])

I did not manage to get a linear algebra row permutation matrix to work.

update
seems to work also for 2-dim instead of 1-dim:
So, we could convert both pair indices at the same time, but I think we are not really gaining anything.

xx2 = np.column_stack((xx, xx[::-1])) 
(xx2[..., None] == a) @ b
array([[1, 2],
       [1, 2],
       [1, 2],
       [0, 0],
       [0, 0],
       [0, 0],
       [2, 1],
       [2, 1],
       [2, 1]])