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关注：1
2013-05-23 12:21
求翻译：扫描事物数据库,计算 中各个项集的支持度。将 中不满足最小支持度的项剔除,形成频繁k 项集构成的集合。是什么意思？ 待解决
 悬赏分：1
- 离问题结束还有
扫描事物数据库,计算 中各个项集的支持度。将 中不满足最小支持度的项剔除,形成频繁k 项集构成的集合。
问题补充： |
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2013-05-23 12:21:38
Scanning the database of things, the calculation of each item set support. Will does not meet the minimum support of items removed to form a frequent itemset consisting of k sets.
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2013-05-23 12:23:18
Scanning things in the database, the calculation of the level of support. will be in the smallest degree of support does not meet the defunct K, formed the Group constitute the collection frequency.
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2013-05-23 12:24:58
Scanning thing database, in computation each item of collection support.Will not satisfy a smallest support rejection, will form the frequent k item of collection constitution the set.
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2013-05-23 12:26:38
Scanning databases, each set in the calculation in the polls. Items that do not meet the minimal support removed, forming frequent collection of k-sets.
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2013-05-23 12:28:18
Scanning databases, each set in the calculation in the polls. Items that do not meet the minimal support removed, forming frequent collection of k-sets.
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