greg wrote:
> But I still don't know how to replace all whitespace with
> space
string.join(phrase.split(), " ")
or
re.sub("(?u)\s+", " ", phrase)
not sure which one's faster; I suggest benchmarking.
(if you want to preserve leading/trailing space with the split
approach, use isspace on the start/end of phrase)
> or detect words that end with a lowercase letter.
word[-1].islower()
</F>
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