Question

我想使用另一列作为参考，从较大的表中为一个列分配一个值。

E.g。数据：

require(data.table)
dt <- data.table(N=c(1:5),GPa1=c(sample(0:5,5)),GPa2=c(sample(5:15,5)),
GPb1=c(sample(0:20,5)),GPb2=c(sample(0:10,5)),id=c("b","a","b","b","a"))

   N GPa1 GPa2 GPb1 GPb2 id
1: 1    4   10    7    0  b
2: 2    5   15   19    7  a
3: 3    1    5   20    5  b
4: 4    0   13    3    4  b
5: 5    3    7    8    1  a

我们的想法是获得新列Val1和Val2。以1结尾的任何GP列都有资格获得Val1，任何2结尾都有资格获得Val2。要插入到列中的值由每行id列确定。

因此，您可以看到Val1，您会在GPb1列，然后GPa1，GPb1，GPb1上再次绘制，最后GPa1 1}}。

最终结果将是;

   N GPa1 GPa2 GPb1 GPb2 id Val1 Val2
1: 1    4   10    7    0  b   7    0
2: 2    5   15   19    7  a   5   15
3: 3    1    5   20    5  b  20    5
4: 4    0   13    3    4  b   3    4
5: 5    3    7    8    1  a   3    7

我确实得到了答案，但在熔化之后有很多行等等，但我确信在data.table中必须有一种优雅的方式来做到这一点。我最初感到沮丧的是paste0无法在data.table;

中使用

dt[1,paste0("GP",id,"1")]

# The following gives a vector that is correct for Val1 (and works for 2)
diag(as.matrix(dt[,.SD,.SDcols=dt[,paste0("GP",id,"1")]]))

# I think the answer lies in `set`, but i've not had any luck.
for (i in 1:nrow(dt)) set(dt, i=dt[i,.SD,.SDcols=dt[,paste0("GP",id,"2")]], j=i, value=0)

但;

$('#myForm').validate({
rules: {
aiterms: {
required: true
    }
},
submitHandler: function(form) {
$$.ajaxSetup({ cache: false });
var myterms;
var mybookingid;
$("#submit").unbind().click(function() {
myterms = $("#myterms").val();
mybookingid = $("#my_booking_id").val();
if ($('#myterms').is(":checked")==false)
{
myApp.alert('You need to accept terms to proceed further!', 'Terms Not Accepted!'); 
    }
else
{
$$.ajax({
type: "POST",
url: "submit1.php",
dataType : 'html',
data: {myterms1: myterms,mybookingid1: mybookingid },
cache: false,
success: function(result){
$("#mycontent" ).empty();
$("#mycontent").html(result);
},//success
error: function(result) 
     {   
myApp.alert('We are unable to process data due to technical reasons!', 'Error');
      }  
   });
}//else
return false;
     });
   }//submit handler
}); //validate

这样的数据非常难看，所以也许最好只使用融化方法。

Answer 1

dt[id == "a", c("Val1", "Val2") := .(GPa1, GPa2)]
dt[id == "b", c("Val1", "Val2") := .(GPb1, GPb2)]
#   N GPa1 GPa2 GPb1 GPb2 id Val1 Val2
#1: 1    2   13    5    8  b    5    8
#2: 2    3    8    7    2  a    3    8
#3: 3    5   11   19    1  b   19    1
#4: 4    4    5    6    9  b    6    9
#5: 5    1   15    1   10  a    1   15

R数据表：根据参考列为列分配值

1 个答案: