Regular Member
Joined: 01/08/2023
Location: AustraliaPosts: 49
| Posted: 03:43am 09 Aug 2023 |
Hi. For an one dimensional array with numbers for elements, how would I obtain the "lowest positive' element value, excluding zero(0)? Thanks.
Guru
Joined: 04/06/2022
Location: GermanyPosts: 1479
| Posted: 06:21am 09 Aug 2023 |
Hi Mark,
If i understand your question right.
Therange of floating point numbers is
1.797693134862316e+308 to 2.225073858507201e-308.
The range of 64-bit integers (whole numbers) that can be manipulated is ± 9223372036854775807.
Cheers
Mart!n
Edited 2023-08-09 16:38 by Martin H.
Guru
Joined: 05/03/2018
Location: NetherlandsPosts: 5991
| Posted: 07:38am 09 Aug 2023 |
Find the max of the array.
Add 10xmax to all numbers that are equal zero or negative.
Find the min of the array
That is your answer.
Volhout
Regular Member
Joined: 01/08/2023
Location: AustraliaPosts: 49
| Posted: 07:47am 09 Aug 2023 |
Thanks Martin and Volhout.
I should have given an example.
One dim array e.g. (12,0,14.5,3,62)
The lowest element in the array is 0, but I wanted to get 3 instead (the lowest min above zero).
MMBasic function ====> MATH(MIN arrayname()) is a solution.
But it returns 0 if I use it on the above example array. Whereas I would like to get 3.
MATH(MIN arrayname() [,index%])
Perhaps the index can be used, but I don't know how to use it.
-thanks
Edited 2023-08-09 18:38 by MarkF
Guru
Joined: 11/12/2012
Location: United KingdomPosts: 11614
| Posted: 08:51am 09 Aug 2023 |
Can't think of any way to do this with the MATH commands or functions so you will need to just write a loop
General solution
max = 1.797693134862316e+308 ' or 9223372036854775807 if integer array
for i=bound(array(),0) to bound(array())
if(array(i)>0 and array(i)<max then max = array(i)
next i
print max
for i=bound(array(),0) to bound(array())
if(array(i)>0 and array(i)<max then max = array(i)
next i
print max
Guru
Joined: 07/08/2011
Location: AustraliaPosts: 6543
| Posted: 10:23am 09 Aug 2023 |
Not the fastest:
'
DIM a(4) = (12,0,14.5,3,62)
DO
m = MATH(MIN a(), index%)
IF m = 0 THEN
a(index%) = 100000
ENDIF
LOOP UNTIL m > 0
PRINT index%, m
DIM a(4) = (12,0,14.5,3,62)
DO
m = MATH(MIN a(), index%)
IF m = 0 THEN
a(index%) = 100000
ENDIF
LOOP UNTIL m > 0
PRINT index%, m
Make a copy of the array first if you need to retain all elements.
Another way is 'sort' then step through to skip over the zero values.
Jim
Guru
Joined: 05/10/2019
Location: United KingdomPosts: 8955
| Posted: 10:39am 09 Aug 2023 |
Don't reject zero values. Add 0.001 to everything and don't have them to start with. ;)
Regular Member
Joined: 01/08/2023
Location: AustraliaPosts: 49
| Posted: 01:33pm 09 Aug 2023 |
Thanks Matt, Jim and Mick
. Jim's example of the MATHS MIN function is very interesting. - Mark