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May 18, 2024 04:00
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python implementation of vruddhi
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import re | |
vruddhi = {'अ': 'आ', 'इ': 'ऐ', 'ई': 'ऐ', 'उ': 'औ', 'ऊ': 'औ', 'ए': 'ऐ', 'ओ': 'औ', | |
'ि': 'ै', 'ी': 'ै', 'ु': 'ौ', 'ू': 'ौ', 'े': 'ै', 'ो': 'ौ', | |
'क': 'का', 'ख': 'खा', 'ग': 'गा', 'घ': 'घा', 'ङ': 'ङा', | |
'च': 'चा', 'छ': 'छा', 'ज': 'जा', 'झ': 'झा', 'ञ': 'ञा', | |
'ट': 'टा', 'ठ': 'ठा', 'ड': 'डा', 'ढ': 'ढा', 'ण': 'णा', | |
'त': 'ता', 'थ': 'था', 'द': 'दा', 'ध': 'धा', 'न': 'ना', | |
'प': 'पा', 'फ': 'फा', 'ब': 'बा', 'भ': 'भा', 'म': 'मा', | |
'य': 'या', 'र': 'रा', 'ल': 'ला', 'व': 'वा', 'श': 'शा', | |
'ष': 'षा', 'स': 'सा', 'ह': 'हा', 'ळ': 'ळा'} | |
svar = ["ा", "ि", "ी", "ु", "ू", "े", "ै", "ो", "ौ"] | |
def sanskrit_expand(s): | |
try: | |
pattern = re.compile(r"(.्)?.[ािीुूृेैोौ]?") | |
x = pattern.match(s).group(0) | |
myr = x[-1] | |
ns = s.replace(myr, vruddhi[myr], 1) | |
if s[-1] in svar: | |
ns1 = ns[0:-1] + "िक" | |
ns2 = ns[0:-1] + "ी" | |
ns3 = ns[0:-1] + "्य" | |
else: | |
ns1 = ns + "िक" | |
ns2 = ns + "ी" | |
ns3 = ns + "्य" | |
except KeyError: | |
ns = s | |
if s[-1] in svar: | |
ns1 = ns[0:-1] + "िक" | |
ns2 = ns[0:-1] + "ी" | |
ns3 = ns[0:-1] + "्य" | |
else: | |
ns1 = ns + "िक" | |
ns2 = ns + "ी" | |
ns3 = ns + "्य" | |
return (ns1, ns2, ns3) | |
sanskrit_expand('नगर') | |
# returns ('नागरिक', 'नागरी', 'नागर्य') |
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