17 Apr
2002
17 Apr
'02
1:06 a.m.
Hello IOhannes,
Imz> if you have lists of varying length you can use zexy's Imz> [mean], which gets (1/n)*sum(), and multiply it with the [length] of the Imz> list...
thanks, it's much better this way than using the [unpack ] and [+ ] combo.
thanks !
stm