如何识别多个列中的重叠 [英] How to identify overlaps in multiple columns
问题描述
我有一个数据集( mydata
),它包含多个列,可以放在另一个数据集中存储的范围内( mycomparison
)。
I have a dataset (mydata
) that contains multiple columns which could fit inside ranges that are stored in another dataset (mycomparison
).
我想加入我的比较
到 mydata
mydata
值在 mycomparison
的范围内。
I would like to join mycomparison
to mydata
where the mydata
values are within the ranges in mycomparison
.
library(data.table)
mydata<-data.table(
id=1:5,
val1=seq(10000, 50000, by=10000),
val2=floor(rnorm(5,mean=400,sd=100)),
val3=rnorm(5,mean=.7,sd=.1)
)
mycomparison<-data.table(
Name=LETTERS[1:3],
minval1=c(0,30000,10000),
maxval1=c(50000,80000,30000),
minval2=c(300,400,300),
maxval2=c(800,800,800),
minval3=c(0,.5,.2),
maxval3=c(1,.9,.8),
correspondingval=c(.1,.2,.3)
)
所需输出
Desired Output
> mydata.withmatches
id val1 val2 val3 Name minval1 maxval1 minval2 maxval2 minval3 maxval3 correspondingval
1: 1 10000 387 0.4844319 A 0 50000 300 800 0 1 0.1
2: 2 20000 425 0.7856313 NA NA NA NA NA NA NA NA
3: 3 30000 324 0.8063969 NA NA NA NA NA NA NA NA
4: 4 40000 263 0.5590113 NA NA NA NA NA NA NA NA
5: 5 50000 187 0.8764396 NA NA NA NA NA NA NA NA
目前的解决方案
这感觉/非常笨重,涉及交叉连接数据(使用 optiRum :: CJ.dt
),做一个大的逻辑检查,然后重新组装
Current solution
This feels/is very clunky and involves cross-joining the data (using optiRum::CJ.dt
), doing a big logical check, and then reassembling the data.
library(optiRum)
workingdt<-CJ.dt(mydata,mycomparison)
matched<-workingdt[val1>=minval1 &
val1<=maxval1 &
val2>=minval2 &
val2<=maxval2 &
val3>=minval3 &
val3<=maxval3][which.min(correspondingval)]
notmatched<-mydata[id!= matched[,id]]
all<-list(matched,notmatched)
mydata.withmatches<- rbindlist(all, fill=TRUE, use.names=TRUE)
寻找更好的解决方案 - UPDATED
我知道 foverlaps
我希望一个更简单,更优雅的解决方案。
I'm hoping for a less clunky and more elegant solution.
推荐答案
我不完全理解您的期望输出,因为多个ID匹配 mycomparison
data.table。使用您的数据(四舍五入到两位小数):
I do not exactly understand your Desired Output, because multiple id's match the mycomparison
data.table. Using your data (rounded to two decimal places):
> mydata
id val1 val2 val3
1: 1 10000 387 0.48
2: 2 20000 425 0.79
3: 3 30000 324 0.81
4: 4 40000 263 0.56
5: 5 50000 187 0.88
和
> mycomparison
Name minval1 maxval1 minval2 maxval2 minval3 maxval3 correspondingval
1: A 0 50000 300 800 0.0 1.0 0.1
2: B 30000 80000 400 800 0.5 0.9 0.2
3: C 10000 30000 300 800 0.2 0.8 0.3
这给出:
> workingdt
id val1 val2 val3 Name minval1 maxval1 minval2 maxval2 minval3 maxval3 correspondingval
1: 1 10000 387 0.48 A 0 50000 300 800 0.0 1.0 0.1
2: 2 20000 425 0.79 A 0 50000 300 800 0.0 1.0 0.1
3: 3 30000 324 0.81 A 0 50000 300 800 0.0 1.0 0.1
4: 4 40000 263 0.56 A 0 50000 300 800 0.0 1.0 0.1
5: 5 50000 187 0.88 A 0 50000 300 800 0.0 1.0 0.1
6: 1 10000 387 0.48 B 30000 80000 400 800 0.5 0.9 0.2
7: 2 20000 425 0.79 B 30000 80000 400 800 0.5 0.9 0.2
8: 3 30000 324 0.81 B 30000 80000 400 800 0.5 0.9 0.2
9: 4 40000 263 0.56 B 30000 80000 400 800 0.5 0.9 0.2
10: 5 50000 187 0.88 B 30000 80000 400 800 0.5 0.9 0.2
11: 1 10000 387 0.48 C 10000 30000 300 800 0.2 0.8 0.3
12: 2 20000 425 0.79 C 10000 30000 300 800 0.2 0.8 0.3
13: 3 30000 324 0.81 C 10000 30000 300 800 0.2 0.8 0.3
14: 4 40000 263 0.56 C 10000 30000 300 800 0.2 0.8 0.3
15: 5 50000 187 0.88 C 10000 30000 300 800 0.2 0.8 0.3
并暂停您的 which.min()
:
> workingdt[val1>=minval1 & val1<= maxval1 & val2>=minval2 &
val2<=maxval2 & val3>=minval3 & val3<=maxval3]
id val1 val2 val3 Name minval1 maxval1 minval2 maxval2 minval3 maxval3 correspondingval
1: 1 10000 387 0.48 A 0 50000 300 800 0.0 1.0 0.1
2: 2 20000 425 0.79 A 0 50000 300 800 0.0 1.0 0.1
3: 3 30000 324 0.81 A 0 50000 300 800 0.0 1.0 0.1
4: 1 10000 387 0.48 C 10000 30000 300 800 0.2 0.8 0.3
5: 2 20000 425 0.79 C 10000 30000 300 800 0.2 0.8 0.3
如果使用data.table group-by功能,你可以为每个 id
选择 min(correspondingval)
(我暂时离开不匹配的数据):
If you use the data.table group-by functionality, you can pick the min(correspondingval)
for each id
(I am leaving off the unmatched data for the moment):
> workingdt[val1>=minval1 & val1<= maxval1 & val2>=minval2 &
val2<=maxval2 & val3>=minval3 & val3<=maxval3]
[,.SD[which.min(correspondingval)], by=id]
id val1 val2 val3 Name minval1 maxval1 minval2 maxval2 minval3 maxval3 correspondingval
1: 1 10000 387 0.48 A 0 50000 300 800 0 1 0.1
2: 2 20000 425 0.79 A 0 50000 300 800 0 1 0.1
3: 3 30000 324 0.81 A 0 50000 300 800 0 1 0.1
或者,如果您愿意, max(correspondingval)
/ p>
Or, the max(correspondingval)
if you prefer:
> workingdt[val1>=minval1 & val1<= maxval1 & val2>=minval2 &
val2<=maxval2 & val3>=minval3 & val3<=maxval3]
[,.SD[which.max(correspondingval)], by=id]
id val1 val2 val3 Name minval1 maxval1 minval2 maxval2 minval3 maxval3 correspondingval
1: 1 10000 387 0.48 C 10000 30000 300 800 0.2 0.8 0.3
2: 2 20000 425 0.79 C 10000 30000 300 800 0.2 0.8 0.3
3: 3 30000 324 0.81 A 0 50000 300 800 0.0 1.0 0.1
如果您需要 - 如所需输出中所示 - 是第一行最小 correspondingval
和其他所有 NA
有更简单的方法来做到这一点。如果你想知道每个 id
匹配一个范围 - 如我在输出中显示的那样 - 那么一个更清洁,更优雅的解决方案是不同的。
If all you want--as shown in your Desired Output--is the first row with the minimum correspondingval
and everything else with NA
there are easier ways to do this. If you want to know where each id
matches a range--as I have shown in my output--then a cleaner, more elegant solution is different.
让我知道。
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