Currier A/B split is not what we thought it was!

[nablator]

the cho counter sums both cho and sho. Likewise for the che counter: it sums both che and she.

f2r: 17 cho+sho, 6 che+she → r_cho = 0.739 (73.9% cho) → Language A 
f2v: 21 cho+sho, 4 che+she → r_cho = 0.840 (84.0% cho) → Language A

In the RF1b-e.txt file (from your You are not allowed to view links. Register or Login to view.) I have:
f2r: 18 cho+sho, 6 che+she
f2v: 24 cho+sho, 7 che+she

[BTW RenéZ might have updated a few lines without changing the version number, you don't have the latest.]

These counts are produced by the switch.py script.

                c = classify_word_cho(w) 
                if c == 'cho': 
                    cho_c += 1 
                elif c == 'che': 
                    che_c += 1 

...

def classify_word_cho(word): 
... 
    if has_cho and not has_che: 
        return 'cho' 
    elif has_che and not has_cho: 
        return 'che' 
    return None 

For example in You are not allowed to view links. Register or Login to view. I see 7 che+she.

But cheopchor and chotchey are classified as None, so it should be 5 not 4 che+she. I must be missing something...

>>> classify_word_cho('cheor')
'che'
>>> classify_word_cho('sshey')
'che'
>>> classify_word_cho('chees')
'che'
>>> classify_word_cho('cheaiin')
'che'
>>> classify_word_cho('cheol')
'che'

Something is wrong with parse_ivtff. '<->' should be processed as a word separator and what's inside {} should not be skipped. 'chees' is not counted because 'chokeos<->chees' gets converted to a unique word 'chokeoschees' classified as None.

Why not simply count occurrences of [cs]ho and [cs]he and avoid all the complications?