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Steven Ranney

2013-Oct-11 13:26 UTC

[R] Create sequential vector for values in another column

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

	[[alternative HTML version deleted]]

arun

2013-Oct-11 13:43 UTC

head link

[R] Create sequential vector for values in another column

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

??? [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

PIKAL Petr

2013-Oct-11 13:49 UTC

head link

[R] Create sequential vector for values in another column

Hi

I named your data test

test$x<-1
test$x<-unlist(lapply(split(test$x, test$id.name),
cumsum))> test   id.name x
1   123.45 1
2   123.45 2
3   123.45 3
4   123.45 4
5   234.56 1
6   234.56 2
7   234.56 3
8   234.56 4
9   234.56 5
10  234.56 6
11  234.56 7
12  345.67 1
13  345.67 2
14  345.67 3
15  456.78 1
16  456.78 2
17  456.78 3
18  456.78 4
19  456.78 5
20  456.78 6
21  456.78 7
22  456.78 8
23  456.78 9>Two comments:
This works only when your data are sorted
Beware of FAQ 7.31

Regards
Petr

> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of Steven Ranney
> Sent: Friday, October 11, 2013 3:26 PM
> To: r-help at r-project.org
> Subject: [R] Create sequential vector for values in another column
> 
> 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
> 
> 	[[alternative HTML version deleted]]
> 
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.

Berend Hasselman

2013-Oct-11 13:58 UTC

head link

[R] Create sequential vector for values in another column

On 11-10-2013, at 15:26, 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.
I named your dataframe dat1.
You can also do this

unlist(sapply(rle(dat1$id.name)$lengths, function(k) 1:k ))

And as Petr told you: beware of FAQ 7.31

Berend

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