Possible to get a word count in Calibre?

[BetterRed]

Originally Posted by theducks

BR
Didn't Word count (paid) used to be really Character count (not whitespace char) divided by N?

That was before we had markup.
So wouldn't converting to plain Text (removes all tags) and subtracting all the 'Spaces',Tabs, LF/CR get you 'just the letters'?

Wasn't that the algo for typist pay rates. As I recall newspaper and magazine journo's were paid by the column inch. For the stuff I was paid to write (20+ years ago) the rate was in units of 100 words rounded down. The editor's lackeys counted the words manually, they were written (in purple of course) on each page of the galley proofs.

She was deadly on unnecessary and superfluous hyphens. I am pretty sure that some hyphenated words counted as one word (well-worn, thirty-three) whilst others were counted as multiples. The lackeys had those counting gadgets that 'time and motion' engineers had on their clipboards.

Originally Posted by davidfor

BR: At the start of that article, it says:

Hyphens' main purpose is to glue words together.

Whenever I glue two things together, it is to make a single thing. Hence, the hyphen glues two words together to be one word.

And the devil in me wants to mention that your example is for a "compound adjective". Doesn't that mean the non-hyphenated version should be counted as a single word? Yeah, I'm stretching, but, what the hell

Anyway, the big problem is that without a very complete dictionary and correctly handling the grammar, there is no way to decide between the two. For simplicity, you have to decide that a hyphen either a word delimiter or part of the word.

I agree, what we need is an intelligent algorithm. Find my former editor and her half-dozen lackeys and do some of that so-called expert-systems stuff - maybe it works now

If I was being paid, or I was paying, by the number of words in the text, I would care about the algorithm, but differently

BR