[REBOL] Re: count characters. (Was: Simple things should be simple to do ?)

From: tooki:widebay:au at: 16-Nov-2000 15:27

Hi All,

I've found the answer to my own question:

> One thing though: isn't there a more elegant way of doing the
> append(append(append count char) 1) "^/") stuff?

Turns out you can do:

append count reduce [char 1 "^/"]

with the same result.

The whole code now reads:

count-char: func [
	str [string!]
	/local count look
][
	count: make block! 0
	foreach char str [
		either look: find count char [
			change (next look) ((second look) + 1)
		][
			append count reduce [char 1 "^/"]
		]
	]
	print count
]

It does do the trick:

>> count-char "aaabbbccddddddd"
a 3
b 3
c 2
d 7

So, let's see if anyone else comes up with even more elegant solutions. I
love to see other people's code. The ideas some people come up with never
stop to astound and delight me.

Bard

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