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C({}.fromkeys(ws, 0)).u(C) solution in Creative category for Popular Words by flpo
from collections import Counter as C
C.u = lambda c, o: c.update(o) or c
popular_words = lambda t, ws: C({}.fromkeys(ws, 0)).u(C(w for w in t.lower().split() if w in ws))
May 15, 2018
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