Jek*_*ina 14 join r data.table
我有2个data.tables:
library(data.table)
dt1 <- data.table(id = 1:5, value1 = 11:15, value2 = 21:25, value3 = 36:40)
dt2 <- data.table(name = c("value1", "value1", "value1", "value1",
"value2", "value2", "value2", "value3", "value3"),
valueMin = c(10, 13, 14, 18, 21, 24, 25, 36, 38),
valueMax = c(13, 14, 18, 20, 24, 25, 27, 38, 42),
label = c(101:104, 201:203, 301:302))
> dt1
id value1 value2 value3
1: 1 11 21 36
2: 2 12 22 37
3: 3 13 23 38
4: 4 14 24 39
5: 5 15 25 40
> dt2
name valueMin valueMax label
1: value1 10 13 101
2: value1 13 14 102
3: value1 14 18 103
4: value1 18 20 104
5: value2 21 24 201
6: value2 24 25 202
7: value2 25 27 203
8: value3 36 38 301
9: value3 38 42 302
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我希望得到的结果是这样的:从加入标签dt2
,以dt1
通过这样的事实value1
在dt1
是valueMin和valueMax之间dt2
和dt2$name
匹配value1
).这是我的解决方案(给出正确的结果):
varName <- "value1"
dt2_temp <- dt2[name == varName,]
dt1[dt2_temp, on = .(value1 > valueMin, value1 <= valueMax), nomatch = 0] %>%
select(id, label)
id label
1: 1 101
2: 2 101
3: 3 101
4: 4 102
5: 5 103
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我想对(使用循环)中的label
所有其余列(value2
,value3
)执行相同的(获取列)dt1
,因此需要将连接中的列名替换为其value1
存储的名称varName
,例如:
dt1[dt2_temp, on = .(varName > valueMin, varName <= valueMax), nomatch = 0]
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不幸的是,我没有成功使用:简单varName
,eval(varName)
,as.name(varName)
.你知道如何解决这个问题吗?
错误消息类似于:
Run Code Online (Sandbox Code Playgroud)Error in `[.data.table`(dt1, dt2_temp, on = .(varName > valueMin, varName <= valueMax), : Column(s) [varName,varName] not found in x
Jaa*_*aap 10
为什么不在没有循环的情况下一次性完成所有操作?
可能的解决方案:
melt(dt1, id = 1)[dt2, on = .(variable = name, value > valueMin, value <= valueMax), lbl := i.label
][, dcast(.SD, id ~ variable, value.var = c("value","lbl"))]
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这使:
Run Code Online (Sandbox Code Playgroud)id value_value1 value_value2 value_value3 lbl_value1 lbl_value2 lbl_value3 1: 1 11 21 36 101 NA NA 2: 2 12 22 37 101 201 301 3: 3 13 23 38 101 201 301 4: 4 14 24 39 102 201 302 5: 5 15 25 40 103 202 302