仅在使用data.table:=连接两个表时才需要第一个实例

Lam*_*ing 4 merge join r data.table

我有一个policyData,这是我非常庞大的数据集(数百万行),我希望用映射表(数万行)添加一些信息.

样品:

policyData <- data.table(plan=c("c","b","b","d"),v=c(8,7,5,6),foo=c(4,2,8,3))
mapping <- data.table(plan=c("b","b","a","a","c","c"),a=c(1,2,4,5,7,8),b=c(9,8,6,5,3,2))
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的policyData:

   plan v foo
1:    c 8   4
2:    b 7   2
3:    b 5   8
4:    d 6   3
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制图:

   plan a b
1:    b 1 9
2:    b 2 8
3:    a 4 6
4:    a 5 5
5:    c 7 3
6:    c 8 2
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问题是映射有多个实例,我希望只获得第一个匹配.我需要使用内存高效的方式加入这两个:=.

所需的输出是:

   plan v foo  a  b
1:    c 8   4  7  3
2:    b 7   2  1  9
3:    b 5   8  1  9
4:    d 6   3 NA NA
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我试过了:

policyData[mapping, on="plan", `:=`(a=i.a, b=i.b)] 
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它给出了映射表中的最后一个实例:

   plan v foo  a  b
1:    c 8   4  8  2
2:    b 7   2  2  8
3:    b 5   8  2  8
4:    d 6   3 NA NA
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我也尝试过:

policyData[mapping, on="plan", `:=`(a=i.a, b=i.b), mult="first"]
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这给出了奇怪的结果(第二个"b"与映射不匹配):

   plan v foo  a  b
1:    c 8   4  8  2
2:    b 7   2  2  8
3:    b 5   8 NA NA
4:    d 6   3 NA NA
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任何见解都会有所帮助.我已经做了很多搜索.

Jaa*_*aap 5

只是总结mappingmapping[, .SD[1], by = plan]和使用,加盟:

policyData[mapping[, .SD[1], by = plan]
           , on = .(plan)
           , `:=` (a = i.a, b = i.b)]
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它给出了所需的输出:

> policyData
   plan v foo  a  b
1:    c 8   4  7  3
2:    b 7   2  1  9
3:    b 5   8  1  9
4:    d 6   3 NA NA
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  • 或`policyData [unique(mapping,by ="plan"),on ="plan",`:=`(a = ia,b = ib)] []` (2认同)