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发表于 2010-5-11 13:19:52
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用list吧( C1 n2 b0 O) z1 u
patches-own [wealth rank]
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ca
3 f5 ]& \. y: u ask patches [set wealth random 100 set pcolor green ]
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2 N5 p3 f. C s- Wto go
9 ?7 S3 h! J3 `& \2 A5 U let mylist1 []- L, P) n! o2 ?& O- Y3 |; K
set mylist1 sort-by [[wealth] of ?1 < [wealth] of ?2] patches0 ^5 v8 X! l: U% A W" Y
let mylist2 []
1 ?( W8 S, f& V- P* {1 D set mylist2 n-values count patches [? + 1]
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( foreach mylist1 mylist2
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z; Z: Y7 }# ~% G: q" _# k ask ?1[ set rank ?2]
; b7 a) J& d: X4 y' s show [rank] of ?1; [/ k2 ]0 s, U" Y5 X" l. S0 a
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4 X+ O6 i7 O+ q: W6 R; w2 A let mylist3 []) X9 a# N# b" j$ `
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set mylist3 sort-by [[rank] of ?1 < [rank] of ?2] patches
# b/ |) O U- F" C" Z show count patches8 U, F/ S7 C+ B! Z, |
show mylist1 @- h p M1 @7 D- q, t. u7 _
show mylist3" {* Z; T3 M' d5 z0 N' }
end
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/ ]7 W+ z+ k9 y- o6 R这时的mylist1 和mylist3 应该是一样的 |
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