[R] Create sequential vector for values in another column

[arun]

Also,

it might be faster to use ?data.table()
library(data.table)
 dt1<- data.table(dat1,key='id.name')
dt1[,x:=seq(.N),by='id.name']
A.K.


On , arun <smartpink111 at yahoo.com> wrote:
Hi,
Try:
dat1<- 

structure(list(id.name = c(123.45, 123.45, 123.45, 123.45, 234.56, 
234.56, 234.56, 234.56, 234.56, 234.56, 234.56, 345.67, 345.67, 
345.67, 456.78, 456.78, 456.78, 456.78, 456.78, 456.78, 456.78, 
456.78, 456.78)), .Names = "id.name", class = "data.frame", row.names = c(NA, 
-23L))
dat1$x <- with(dat1,ave(id.name,id.name,FUN=seq))
A.K.



On Friday, October 11, 2013 9:28 AM, Steven Ranney <steven.ranney at gmail.com> wrote:
Hello all -

I have an example column in a dataFrame

id.name
123.45
123.45
123.45
123.45
234.56
234.56
234.56
234.56
234.56
234.56
234.56
345.67
345.67
345.67
456.78
456.78
456.78
456.78
456.78
456.78
456.78
456.78
456.78
...
[truncated]

And I'd like to create a second vector of sequential values (i.e., 1:N) for
each unique id.name value.  In other words, I need

id.name  x
123.45   1
123.45   2
123.45   3
123.45   4
234.56   1
234.56   2
234.56   3
234.56   4
234.56   5
234.56   6
234.56   7
345.67   1
345.67   2
345.67   3
456.78   1
456.78   2
456.78   3
456.78   4
456.78   5
456.78   6
456.78   7
456.78   8
456.78   9

The number of unique id.name values is different; for some values, nrow()
may be 42 and for others it may be 36, etc.

The only way I could think of to do this is with two nested for loops.  I
tried it but because this data set is so large (nrow = 112,679 with 2,161
unique values of id.name), it took several hours to run.

Is there an easier way to create this vector?  I'd appreciate your thoughts.

Thanks -

SR
Steven H. Ranney

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