Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread Michael Bedward
All values in a matrix are the same type, so if you've set up a matrix
with a character column then your numeric values will also be stored
as character. That would explain why they are being converted to
factors.  It would also explain why your query isn't working.

Michael


On 11 November 2010 18:40, Noah Silverman n...@smartmediacorp.com wrote:
 That was a typo.

 It should have read:
 results[results$one  100,]

 It does still fail.

 There is ONE column that is text.  So my guess is that R is seeing that and
 assuming that the entire data.frame should be factors.

 -N

 On 11/10/10 11:16 PM, Michael Bedward wrote:

 Hello Noah,

 If you set these names...

 names(results)- c(one, two, three)

 this won't work...

 results[results$c  100,]

 because you don't have a column called c (unless that's just a typo
 in your post).

 I tried making it a data.frame with
 foo- data.frame(results)

 But that converted all the numeric values to factors!!!

 Not sure what's going on there. If 'results' is a numeric matrix you
 should get a data.frame with numeric cols since under the hood this is
 just calling the as.data.frame function.

 Michael





 On 11 November 2010 16:02, Noah Silvermann...@smartmediacorp.com  wrote:

 Hi,

 I have a process in R that produces a lot of output.  My plan was to
 build
 up a matrix or data.frame row by row, so that I'll have a nice object
 with
 all the resulting data.

 I started with:
 results- matrix(ncol=3)
 names(results)- c(one, two, three)

 Then, when looping through the data:
 results- rbind(results, c(a,b,c))

 This seems to work fine. BUT, my problem arises when I want to filter,
 sort,
 etc.

 I tried (thinking like a data.frame):
 results[results$c  100,]

 But that fails.

 I tried making it a data.frame with
 foo- data.frame(results)

 But that converted all the numeric values to factors!!!  Which causes a
 whole mess of problems.

 Any ideas??

 -N

 __
 R-help@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.




__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread Noah Silverman

That makes perfect sense.

Since I need to build up the results table sequentially as I iterate 
through the data, how would you recommend it??


Thanks,

-N

On 11/11/10 12:03 AM, Michael Bedward wrote:

All values in a matrix are the same type, so if you've set up a matrix
with a character column then your numeric values will also be stored
as character. That would explain why they are being converted to
factors.  It would also explain why your query isn't working.

Michael


On 11 November 2010 18:40, Noah Silvermann...@smartmediacorp.com  wrote:

That was a typo.

It should have read:
results[results$one  100,]

It does still fail.

There is ONE column that is text.  So my guess is that R is seeing that and
assuming that the entire data.frame should be factors.

-N

On 11/10/10 11:16 PM, Michael Bedward wrote:

Hello Noah,

If you set these names...

names(results)- c(one, two, three)

this won't work...

results[results$c100,]

because you don't have a column called c (unless that's just a typo
in your post).


I tried making it a data.frame with
foo- data.frame(results)

But that converted all the numeric values to factors!!!

Not sure what's going on there. If 'results' is a numeric matrix you
should get a data.frame with numeric cols since under the hood this is
just calling the as.data.frame function.

Michael





On 11 November 2010 16:02, Noah Silvermann...@smartmediacorp.comwrote:

Hi,

I have a process in R that produces a lot of output.  My plan was to
build
up a matrix or data.frame row by row, so that I'll have a nice object
with
all the resulting data.

I started with:
results- matrix(ncol=3)
names(results)- c(one, two, three)

Then, when looping through the data:
results- rbind(results, c(a,b,c))

This seems to work fine. BUT, my problem arises when I want to filter,
sort,
etc.

I tried (thinking like a data.frame):
results[results$c100,]

But that fails.

I tried making it a data.frame with
foo- data.frame(results)

But that converted all the numeric values to factors!!!  Which causes a
whole mess of problems.

Any ideas??

-N

__
R-help@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.





__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread Michael Bedward
You can use rbind as in your original post, but if you've got a mix of
character and numeric data start with a data.frame rather than a
matrix.

Michael

On 11 November 2010 20:30, Noah Silverman n...@smartmediacorp.com wrote:
 That makes perfect sense.

 Since I need to build up the results table sequentially as I iterate through
 the data, how would you recommend it??

 Thanks,

 -N

 On 11/11/10 12:03 AM, Michael Bedward wrote:

 All values in a matrix are the same type, so if you've set up a matrix
 with a character column then your numeric values will also be stored
 as character. That would explain why they are being converted to
 factors.  It would also explain why your query isn't working.

 Michael


 On 11 November 2010 18:40, Noah Silvermann...@smartmediacorp.com  wrote:

 That was a typo.

 It should have read:
 results[results$one  100,]

 It does still fail.

 There is ONE column that is text.  So my guess is that R is seeing that
 and
 assuming that the entire data.frame should be factors.

 -N

 On 11/10/10 11:16 PM, Michael Bedward wrote:

 Hello Noah,

 If you set these names...

 names(results)- c(one, two, three)

 this won't work...

 results[results$c    100,]

 because you don't have a column called c (unless that's just a typo
 in your post).

 I tried making it a data.frame with
 foo- data.frame(results)

 But that converted all the numeric values to factors!!!

 Not sure what's going on there. If 'results' is a numeric matrix you
 should get a data.frame with numeric cols since under the hood this is
 just calling the as.data.frame function.

 Michael





 On 11 November 2010 16:02, Noah Silvermann...@smartmediacorp.com
  wrote:

 Hi,

 I have a process in R that produces a lot of output.  My plan was to
 build
 up a matrix or data.frame row by row, so that I'll have a nice object
 with
 all the resulting data.

 I started with:
 results- matrix(ncol=3)
 names(results)- c(one, two, three)

 Then, when looping through the data:
 results- rbind(results, c(a,b,c))

 This seems to work fine. BUT, my problem arises when I want to filter,
 sort,
 etc.

 I tried (thinking like a data.frame):
 results[results$c    100,]

 But that fails.

 I tried making it a data.frame with
 foo- data.frame(results)

 But that converted all the numeric values to factors!!!  Which causes a
 whole mess of problems.

 Any ideas??

 -N

 __
 R-help@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.





__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread Noah Silverman

Still doesn't work.

When using rbind to build the data.frame, it get a structure mostly full 
of NA.
The data is correct, so something about pushing into the data.frame is 
breaking.


Example code:
results - data.frame()

for(i in 1:n){
#do all the work
#a is a test label. b,c,d are numeric.
results - rbind(results, c(a,b,c,d))
}



On 11/11/10 2:00 AM, Michael Bedward wrote:

You can use rbind as in your original post, but if you've got a mix of
character and numeric data start with a data.frame rather than a
matrix.

Michael

On 11 November 2010 20:30, Noah Silvermann...@smartmediacorp.com  wrote:

That makes perfect sense.

Since I need to build up the results table sequentially as I iterate through
the data, how would you recommend it??

Thanks,

-N

On 11/11/10 12:03 AM, Michael Bedward wrote:

All values in a matrix are the same type, so if you've set up a matrix
with a character column then your numeric values will also be stored
as character. That would explain why they are being converted to
factors.  It would also explain why your query isn't working.

Michael


On 11 November 2010 18:40, Noah Silvermann...@smartmediacorp.comwrote:

That was a typo.

It should have read:
results[results$one100,]

It does still fail.

There is ONE column that is text.  So my guess is that R is seeing that
and
assuming that the entire data.frame should be factors.

-N

On 11/10/10 11:16 PM, Michael Bedward wrote:

Hello Noah,

If you set these names...

names(results)- c(one, two, three)

this won't work...

results[results$c  100,]

because you don't have a column called c (unless that's just a typo
in your post).


I tried making it a data.frame with
foo- data.frame(results)

But that converted all the numeric values to factors!!!

Not sure what's going on there. If 'results' is a numeric matrix you
should get a data.frame with numeric cols since under the hood this is
just calling the as.data.frame function.

Michael





On 11 November 2010 16:02, Noah Silvermann...@smartmediacorp.com
  wrote:

Hi,

I have a process in R that produces a lot of output.  My plan was to
build
up a matrix or data.frame row by row, so that I'll have a nice object
with
all the resulting data.

I started with:
results- matrix(ncol=3)
names(results)- c(one, two, three)

Then, when looping through the data:
results- rbind(results, c(a,b,c))

This seems to work fine. BUT, my problem arises when I want to filter,
sort,
etc.

I tried (thinking like a data.frame):
results[results$c  100,]

But that fails.

I tried making it a data.frame with
foo- data.frame(results)

But that converted all the numeric values to factors!!!  Which causes a
whole mess of problems.

Any ideas??

-N

__
R-help@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.





__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread Peter Langfelder
On Thu, Nov 11, 2010 at 11:33 AM, Noah Silverman
n...@smartmediacorp.com wrote:
 Still doesn't work.

 When using rbind to build the data.frame, it get a structure mostly full of
 NA.
 The data is correct, so something about pushing into the data.frame is
 breaking.

 Example code:
 results - data.frame()

 for(i in 1:n){
    #do all the work
    #a is a test label. b,c,d are numeric.
    results - rbind(results, c(a,b,c,d))
 }

Works for me:

results - data.frame()

n = 10
for(i in 1:n){
   a = LETTERS[i];
   b = i;
   c = 3*i + 2
   d = rnorm(1);
   results - rbind(results, c(a,b,c,d))
}


 results

   X.A. X.1. X.5. X.0.142223304589023.
1 A150.142223304589023
2 B280.243612305595176
3 C3   110.476795513990516
4 D4   14  1.0278220664213
5 E5   170.916608672305205
6 F6   20 1.61075985995586
7 G7   230.370423691258896
8 H8   26  -0.0528603547004191
9 I9   29-2.07888666920403
10J   10   32-1.87980721733655

Maybe there's something wrong with the calculation you do?

Peter

__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread William Dunlap
Peter,

Your example doesn't work for me unless I
set options(stringsAsFactors=TRUE) first.
(If I do set that, then all columns of 'results'
have class character, which I doubt the user
wants.)

 results - data.frame()

 n = 10
 for(i in 1:n){
+a = LETTERS[i];
+b = i;
+c = 3*i + 2
+d = rnorm(1);
+results - rbind(results, c(a,b,c,d))
+ }
There were 36 warnings (use warnings() to see them)
 warnings()[1:5]
$`invalid factor level, NAs generated`
`[-.factor`(`*tmp*`, ri, value = B)

$`invalid factor level, NAs generated`
`[-.factor`(`*tmp*`, ri, value = 2)

$`invalid factor level, NAs generated`
`[-.factor`(`*tmp*`, ri, value = 8)

$`invalid factor level, NAs generated`
`[-.factor`(`*tmp*`, ri, value = -0.305558353507095)

$`invalid factor level, NAs generated`
`[-.factor`(`*tmp*`, ri, value = C)

 results
   X.A. X.1. X.5. X.1.43055780028799.
1 A151.43055780028799
2  NA NA NANA
3  NA NA NANA
4  NA NA NANA
5  NA NA NANA
6  NA NA NANA
7  NA NA NANA
8  NA NA NANA
9  NA NA NANA
10 NA NA NANA

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Peter Langfelder
 Sent: Thursday, November 11, 2010 12:25 PM
 To: Noah Silverman
 Cc: r-help@r-project.org
 Subject: Re: [R] Populating then sorting a matrix and/or data.frame
 
 On Thu, Nov 11, 2010 at 11:33 AM, Noah Silverman
 n...@smartmediacorp.com wrote:
  Still doesn't work.
 
  When using rbind to build the data.frame, it get a 
 structure mostly full of
  NA.
  The data is correct, so something about pushing into the 
 data.frame is
  breaking.
 
  Example code:
  results - data.frame()
 
  for(i in 1:n){
     #do all the work
     #a is a test label. b,c,d are numeric.
     results - rbind(results, c(a,b,c,d))
  }
 
 Works for me:
 
 results - data.frame()
 
 n = 10
 for(i in 1:n){
a = LETTERS[i];
b = i;
c = 3*i + 2
d = rnorm(1);
results - rbind(results, c(a,b,c,d))
 }
 
 
  results
 
X.A. X.1. X.5. X.0.142223304589023.
 1 A150.142223304589023
 2 B280.243612305595176
 3 C3   110.476795513990516
 4 D4   14  1.0278220664213
 5 E5   170.916608672305205
 6 F6   20 1.61075985995586
 7 G7   230.370423691258896
 8 H8   26  -0.0528603547004191
 9 I9   29-2.07888666920403
 10J   10   32-1.87980721733655
 
 Maybe there's something wrong with the calculation you do?
 
 Peter
 
 __
 R-help@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.
 

__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread John Kane


--- On Thu, 11/11/10, William Dunlap wdun...@tibco.com wrote:

 From: William Dunlap wdun...@tibco.com
 Subject: Re: [R] Populating then sorting a matrix and/or data.frame
 To: Peter Langfelder peter.langfel...@gmail.com, r-help@r-project.org
 Received: Thursday, November 11, 2010, 4:19 PM
 Peter,
 
 Your example doesn't work for me unless I
 set options(stringsAsFactors=TRUE) first.


Don't you mean stringsAsFactors=FALSE here?  At least I get the same results 
you do but with stringsAsFactors=FALSE  The TRUE condition is giving multiple 
NAs and error messages





 (If I do set that, then all columns of 'results'
 have class character, which I doubt the user
 wants.)
 
  results - data.frame()
 
  n = 10
  for(i in 1:n){
 +    a = LETTERS[i];
 +    b = i;
 +    c = 3*i + 2
 +    d = rnorm(1);
 +    results - rbind(results, c(a,b,c,d))
 + }
 There were 36 warnings (use warnings() to see them)
  warnings()[1:5]
 $`invalid factor level, NAs generated`
 `[-.factor`(`*tmp*`, ri, value = B)
 
 $`invalid factor level, NAs generated`
 `[-.factor`(`*tmp*`, ri, value = 2)
 
 $`invalid factor level, NAs generated`
 `[-.factor`(`*tmp*`, ri, value = 8)
 
 $`invalid factor level, NAs generated`
 `[-.factor`(`*tmp*`, ri, value = -0.305558353507095)
 
 $`invalid factor level, NAs generated`
 `[-.factor`(`*tmp*`, ri, value = C)
 
  results
    X.A. X.1. X.5. X.1.43055780028799.
 1     A    1   
 5    1.43055780028799
 2  NA NA NA   
             NA
 3  NA NA NA   
             NA
 4  NA NA NA   
             NA
 5  NA NA NA   
             NA
 6  NA NA NA   
             NA
 7  NA NA NA   
             NA
 8  NA NA NA   
             NA
 9  NA NA NA   
             NA
 10 NA NA NA     
           NA
 
 Bill Dunlap
 Spotfire, TIBCO Software
 wdunlap tibco.com  
 
  -Original Message-
  From: r-help-boun...@r-project.org
 
  [mailto:r-help-boun...@r-project.org]
 On Behalf Of Peter Langfelder
  Sent: Thursday, November 11, 2010 12:25 PM
  To: Noah Silverman
  Cc: r-help@r-project.org
  Subject: Re: [R] Populating then sorting a matrix
 and/or data.frame
  
  On Thu, Nov 11, 2010 at 11:33 AM, Noah Silverman
  n...@smartmediacorp.com
 wrote:
   Still doesn't work.
  
   When using rbind to build the data.frame, it get
 a 
  structure mostly full of
   NA.
   The data is correct, so something about pushing
 into the 
  data.frame is
   breaking.
  
   Example code:
   results - data.frame()
  
   for(i in 1:n){
      #do all the work
      #a is a test label. b,c,d are numeric.
      results - rbind(results, c(a,b,c,d))
   }
  
  Works for me:
  
  results - data.frame()
  
  n = 10
  for(i in 1:n){
     a = LETTERS[i];
     b = i;
     c = 3*i + 2
     d = rnorm(1);
     results - rbind(results, c(a,b,c,d))
  }
  
  
   results
  
     X.A. X.1. X.5. X.0.142223304589023.
  1     A    1 
   5    0.142223304589023
  2     B    2 
   8    0.243612305595176
  3     C   
 3   11    0.476795513990516
  4     D   
 4   14      1.0278220664213
  5     E   
 5   17    0.916608672305205
  6     F   
 6   20 
    1.61075985995586
  7     G   
 7   23    0.370423691258896
  8     H   
 8   26  -0.0528603547004191
  9     I   
 9   29    -2.07888666920403
  10   
 J   10   32   
 -1.87980721733655
  
  Maybe there's something wrong with the calculation you
 do?
  
  Peter
  
  __
  R-help@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.
  
 
 __
 R-help@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.
 



__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread Peter Langfelder
On Thu, Nov 11, 2010 at 1:19 PM, William Dunlap wdun...@tibco.com wrote:
 Peter,

 Your example doesn't work for me unless I
 set options(stringsAsFactors=TRUE) first.
 (If I do set that, then all columns of 'results'
 have class character, which I doubt the user
 wants.)

You probably mean stringsAsFactors=FALSE.

What you say makes sense, because the c() function produces a vector
in which all components have the same type, wnd it will be character.
If you don't want to have characters, my solution would be

n = 10
results - data.frame(a = rep(, n), b = rep(0, n), c = rep(0, n), d
= rep(0, n))
for(i in 1:n){
   a = LETTERS[i];
   b = i;
   c = 3*i + 2
   d = rnorm(1);
   results$a[i] = a
   results$b[i] = b
   results$c[i] = c
   results$d[i] = d
}

 results
   a  b  c   d
1  A  1  5 -1.31553805
2  B  2  8  0.09198054
3  C  3 11 -0.05860804
4  D  4 14  0.77796136
5  E  5 17  1.28924697
6  F  6 20  0.47631483
7  G  7 23 -1.23727076
8  H  8 26  0.83595295
9  I  9 29  0.69435349
10 J 10 32 -0.30922930

 mode(results[, 1])
[1] character
 mode(results[, 2])
[1] numeric
 mode(results[, 3])
[1] numeric
 mode(results[, 4])
[1] numeric



or alternatively

n = 10
num - data.frame(b = rep(0, n), c = rep(0, n), d = rep(0, n))
labels = rep(, n);
for(i in 1:n){
   a = LETTERS[i];
   b = i;
   c = 3*i + 2
   d = rnorm(1);
   labels[i] = a
   num[i, ] = c(b, c, d)
}
results = data.frame(a = labels, num)

 results
   a  b  c   d
1  A  1  5 -0.47150097
2  B  2  8 -1.30507313
3  C  3 11 -1.09860425
4  D  4 14  0.91326330
5  E  5 17 -0.09732841
6  F  6 20 -0.75134162
7  G  7 23  0.31360908
8  H  8 26 -1.54406716
9  I  9 29 -0.36075743
10 J 10 32 -0.23758269
 mode(results[, 1])
[1] character
 mode(results[, 2])
[1] numeric
 mode(results[, 3])
[1] numeric
 mode(results[, 4])
[1] numeric


Peter

__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread William Dunlap
You are right, I mistyped it.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

 -Original Message-
 From: John Kane [mailto:jrkrid...@yahoo.ca] 
 Sent: Thursday, November 11, 2010 1:58 PM
 To: Peter Langfelder; r-help@r-project.org; William Dunlap
 Subject: Re: [R] Populating then sorting a matrix and/or data.frame
 
 
 
 --- On Thu, 11/11/10, William Dunlap wdun...@tibco.com wrote:
 
  From: William Dunlap wdun...@tibco.com
  Subject: Re: [R] Populating then sorting a matrix and/or data.frame
  To: Peter Langfelder peter.langfel...@gmail.com, 
 r-help@r-project.org
  Received: Thursday, November 11, 2010, 4:19 PM
  Peter,
  
  Your example doesn't work for me unless I
  set options(stringsAsFactors=TRUE) first.
 
 
 Don't you mean stringsAsFactors=FALSE here?  At least I get 
 the same results you do but with stringsAsFactors=FALSE  The 
 TRUE condition is giving multiple NAs and error messages
 
 
 
 
 
  (If I do set that, then all columns of 'results'
  have class character, which I doubt the user
  wants.)
  
   results - data.frame()
  
   n = 10
   for(i in 1:n){
  +    a = LETTERS[i];
  +    b = i;
  +    c = 3*i + 2
  +    d = rnorm(1);
  +    results - rbind(results, c(a,b,c,d))
  + }
  There were 36 warnings (use warnings() to see them)
   warnings()[1:5]
  $`invalid factor level, NAs generated`
  `[-.factor`(`*tmp*`, ri, value = B)
  
  $`invalid factor level, NAs generated`
  `[-.factor`(`*tmp*`, ri, value = 2)
  
  $`invalid factor level, NAs generated`
  `[-.factor`(`*tmp*`, ri, value = 8)
  
  $`invalid factor level, NAs generated`
  `[-.factor`(`*tmp*`, ri, value = -0.305558353507095)
  
  $`invalid factor level, NAs generated`
  `[-.factor`(`*tmp*`, ri, value = C)
  
   results
     X.A. X.1. X.5. X.1.43055780028799.
  1     A    1   
  5    1.43055780028799
  2  NA NA NA   
              NA
  3  NA NA NA   
              NA
  4  NA NA NA   
              NA
  5  NA NA NA   
              NA
  6  NA NA NA   
              NA
  7  NA NA NA   
              NA
  8  NA NA NA   
              NA
  9  NA NA NA   
              NA
  10 NA NA NA     
            NA
  
  Bill Dunlap
  Spotfire, TIBCO Software
  wdunlap tibco.com  
  
   -Original Message-
   From: r-help-boun...@r-project.org
  
   [mailto:r-help-boun...@r-project.org]
  On Behalf Of Peter Langfelder
   Sent: Thursday, November 11, 2010 12:25 PM
   To: Noah Silverman
   Cc: r-help@r-project.org
   Subject: Re: [R] Populating then sorting a matrix
  and/or data.frame
   
   On Thu, Nov 11, 2010 at 11:33 AM, Noah Silverman
   n...@smartmediacorp.com
  wrote:
Still doesn't work.
   
When using rbind to build the data.frame, it get
  a 
   structure mostly full of
NA.
The data is correct, so something about pushing
  into the 
   data.frame is
breaking.
   
Example code:
results - data.frame()
   
for(i in 1:n){
   #do all the work
   #a is a test label. b,c,d are numeric.
   results - rbind(results, c(a,b,c,d))
}
   
   Works for me:
   
   results - data.frame()
   
   n = 10
   for(i in 1:n){
      a = LETTERS[i];
      b = i;
      c = 3*i + 2
      d = rnorm(1);
      results - rbind(results, c(a,b,c,d))
   }
   
   
results
   
      X.A. X.1. X.5. X.0.142223304589023.
   1     A    1 
    5    0.142223304589023
   2     B    2 
    8    0.243612305595176
   3     C   
  3   11    0.476795513990516
   4     D   
  4   14      1.0278220664213
   5     E   
  5   17    0.916608672305205
   6     F   
  6   20 
     1.61075985995586
   7     G   
  7   23    0.370423691258896
   8     H   
  8   26  -0.0528603547004191
   9     I   
  9   29    -2.07888666920403
   10   
  J   10   32   
  -1.87980721733655
   
   Maybe there's something wrong with the calculation you
  do?
   
   Peter
   
   __
   R-help@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.
   
  
  __
  R-help@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.
  
 
 
 

__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread Peter Langfelder
On Thu, Nov 11, 2010 at 1:19 PM, William Dunlap wdun...@tibco.com wrote:
 Peter,

 Your example doesn't work for me unless I
 set options(stringsAsFactors=TRUE) first.

Yes, you need to set options(stringsAsFactors=FALSE) (note the FALSE).
I do it always so I forgot about that, sorry.

__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread Noah Silverman
Your errors look exactly like mine.

Changing the option flag does allow me to create the data.frame without
any errors.  A quick look confirms that all the values are there and
correct.

However, R has coerced all of my numeric values to strings. 

Using your sample code also turns all the numeric values to strings. 
Perhaps it has something to do with the way R in interpreting the first
column?

Thanks!

-N



On 11/11/10 1:59 PM, William Dunlap wrote:
 You are right, I mistyped it.

 Bill Dunlap
 Spotfire, TIBCO Software
 wdunlap tibco.com  

 -Original Message-
 From: John Kane [mailto:jrkrid...@yahoo.ca] 
 Sent: Thursday, November 11, 2010 1:58 PM
 To: Peter Langfelder; r-help@r-project.org; William Dunlap
 Subject: Re: [R] Populating then sorting a matrix and/or data.frame



 --- On Thu, 11/11/10, William Dunlap wdun...@tibco.com wrote:

 From: William Dunlap wdun...@tibco.com
 Subject: Re: [R] Populating then sorting a matrix and/or data.frame
 To: Peter Langfelder peter.langfel...@gmail.com, 
 r-help@r-project.org
 Received: Thursday, November 11, 2010, 4:19 PM
 Peter,

 Your example doesn't work for me unless I
 set options(stringsAsFactors=TRUE) first.

 Don't you mean stringsAsFactors=FALSE here?  At least I get 
 the same results you do but with stringsAsFactors=FALSE  The 
 TRUE condition is giving multiple NAs and error messages





 (If I do set that, then all columns of 'results'
 have class character, which I doubt the user
 wants.)

 results - data.frame()

 n = 10
 for(i in 1:n){
 +a = LETTERS[i];
 +b = i;
 +c = 3*i + 2
 +d = rnorm(1);
 +results - rbind(results, c(a,b,c,d))
 + }
 There were 36 warnings (use warnings() to see them)
 warnings()[1:5]
 $`invalid factor level, NAs generated`
 `[-.factor`(`*tmp*`, ri, value = B)

 $`invalid factor level, NAs generated`
 `[-.factor`(`*tmp*`, ri, value = 2)

 $`invalid factor level, NAs generated`
 `[-.factor`(`*tmp*`, ri, value = 8)

 $`invalid factor level, NAs generated`
 `[-.factor`(`*tmp*`, ri, value = -0.305558353507095)

 $`invalid factor level, NAs generated`
 `[-.factor`(`*tmp*`, ri, value = C)

 results
X.A. X.1. X.5. X.1.43055780028799.
 1 A1   
 51.43055780028799
 2  NA NA NA   
 NA
 3  NA NA NA   
 NA
 4  NA NA NA   
 NA
 5  NA NA NA   
 NA
 6  NA NA NA   
 NA
 7  NA NA NA   
 NA
 8  NA NA NA   
 NA
 9  NA NA NA   
 NA
 10 NA NA NA 
   NA

 Bill Dunlap
 Spotfire, TIBCO Software
 wdunlap tibco.com  

 -Original Message-
 From: r-help-boun...@r-project.org
 [mailto:r-help-boun...@r-project.org]
 On Behalf Of Peter Langfelder
 Sent: Thursday, November 11, 2010 12:25 PM
 To: Noah Silverman
 Cc: r-help@r-project.org
 Subject: Re: [R] Populating then sorting a matrix
 and/or data.frame
 On Thu, Nov 11, 2010 at 11:33 AM, Noah Silverman
 n...@smartmediacorp.com
 wrote:
 Still doesn't work.

 When using rbind to build the data.frame, it get
 a 
 structure mostly full of
 NA.
 The data is correct, so something about pushing
 into the 
 data.frame is
 breaking.

 Example code:
 results - data.frame()

 for(i in 1:n){
#do all the work
#a is a test label. b,c,d are numeric.
results - rbind(results, c(a,b,c,d))
 }
 Works for me:

 results - data.frame()

 n = 10
 for(i in 1:n){
 a = LETTERS[i];
 b = i;
 c = 3*i + 2
 d = rnorm(1);
 results - rbind(results, c(a,b,c,d))
 }


 results
 X.A. X.1. X.5. X.0.142223304589023.
 1 A1 
   50.142223304589023
 2 B2 
   80.243612305595176
 3 C   
 3   110.476795513990516
 4 D   
 4   14  1.0278220664213
 5 E   
 5   170.916608672305205
 6 F   
 6   20 
1.61075985995586
 7 G   
 7   230.370423691258896
 8 H   
 8   26  -0.0528603547004191
 9 I   
 9   29-2.07888666920403
 10   
 J   10   32   
 -1.87980721733655
 Maybe there's something wrong with the calculation you
 do?
 Peter

 __
 R-help@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.
 __
 R-help@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.



 __
 R-help@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.

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the 

Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread Noah Silverman
That makes perfect sense.  All of my numbers are being coerced into
strings by the c() function.  Subsequently, my data.frame contains all
strings.

I can't know the length of the data.frame ahead of time, so can't
predefine it like your example.
One thought would be to make it arbitrarily long filled with 0 and
delete off the unused rows.  But this seems rather wasteful.

-N

On 11/11/10 2:02 PM, Peter Langfelder wrote:
 On Thu, Nov 11, 2010 at 1:19 PM, William Dunlap wdun...@tibco.com wrote:
 Peter,

 Your example doesn't work for me unless I
 set options(stringsAsFactors=TRUE) first.
 (If I do set that, then all columns of 'results'
 have class character, which I doubt the user
 wants.)
 You probably mean stringsAsFactors=FALSE.

 What you say makes sense, because the c() function produces a vector
 in which all components have the same type, wnd it will be character.
 If you don't want to have characters, my solution would be

 n = 10
 results - data.frame(a = rep(, n), b = rep(0, n), c = rep(0, n), d
 = rep(0, n))
 for(i in 1:n){
a = LETTERS[i];
b = i;
c = 3*i + 2
d = rnorm(1);
results$a[i] = a
results$b[i] = b
results$c[i] = c
results$d[i] = d
 }

 results
a  b  c   d
 1  A  1  5 -1.31553805
 2  B  2  8  0.09198054
 3  C  3 11 -0.05860804
 4  D  4 14  0.77796136
 5  E  5 17  1.28924697
 6  F  6 20  0.47631483
 7  G  7 23 -1.23727076
 8  H  8 26  0.83595295
 9  I  9 29  0.69435349
 10 J 10 32 -0.30922930

 mode(results[, 1])
 [1] character
 mode(results[, 2])
 [1] numeric
 mode(results[, 3])
 [1] numeric
 mode(results[, 4])
 [1] numeric



 or alternatively

 n = 10
 num - data.frame(b = rep(0, n), c = rep(0, n), d = rep(0, n))
 labels = rep(, n);
 for(i in 1:n){
a = LETTERS[i];
b = i;
c = 3*i + 2
d = rnorm(1);
labels[i] = a
num[i, ] = c(b, c, d)
 }
 results = data.frame(a = labels, num)

 results
a  b  c   d
 1  A  1  5 -0.47150097
 2  B  2  8 -1.30507313
 3  C  3 11 -1.09860425
 4  D  4 14  0.91326330
 5  E  5 17 -0.09732841
 6  F  6 20 -0.75134162
 7  G  7 23  0.31360908
 8  H  8 26 -1.54406716
 9  I  9 29 -0.36075743
 10 J 10 32 -0.23758269
 mode(results[, 1])
 [1] character
 mode(results[, 2])
 [1] numeric
 mode(results[, 3])
 [1] numeric
 mode(results[, 4])
 [1] numeric


 Peter

 __
 R-help@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.

__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread Peter Langfelder
I see 4 ways to write the code:

1. make the frame very long at the start and use my code - this is
practical if you know that your data frame will not be longer than a
certain number of rows, be it a million;

2a. use something like

result1 = data.frame(a=a, b=b, c=c, d=d)

within the loop to create a 1x4 data frame that you can rbind to
results within the loop;

2b. make the code a bit more intelligent, for example by allocating
blocks of say n=1000 at a time as needed and rbind-ing them to result;

3. fill up results with characters using your rbind(results,
c(a,b,c,d)), then use something like

results[, c(2:4)] = apply(apply(results[, c(2:4), 2, as.character), 2,
as.numeric)

to convert the characters in columns 2:4 to numbers (this construct
also works with factors)

The difference between 2a and 2b is that 2b may be faster if n is
large, because 2a grows 4 objects by 1 unit n times, which is quite
slow. The same holds for solution 3. In that sense solution 1 may be
less wasteful than solutions 2a or 3 although it may not look like
that.

Peter



On Thu, Nov 11, 2010 at 3:38 PM, Noah Silverman n...@smartmediacorp.com wrote:
 That makes perfect sense.  All of my numbers are being coerced into
 strings by the c() function.  Subsequently, my data.frame contains all
 strings.

 I can't know the length of the data.frame ahead of time, so can't
 predefine it like your example.
 One thought would be to make it arbitrarily long filled with 0 and
 delete off the unused rows.  But this seems rather wasteful.

 -N

__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread David Winsemius


On Nov 11, 2010, at 6:38 PM, Noah Silverman wrote:


That makes perfect sense.  All of my numbers are being coerced into
strings by the c() function.  Subsequently, my data.frame contains all
strings.

I can't know the length of the data.frame ahead of time, so can't
predefine it like your example.
One thought would be to make it arbitrarily long filled with 0 and
delete off the unused rows.  But this seems rather wasteful.


Although it might be faster, though. Here is a non-c() method using  
instead the list function (with options(stringsAsFactors=FALSE). List  
does not coerce to same mode and rbind.dta.frame will accept a list as  
a row argument:


results - data.frame(a=vector(mode=character, length=0) ,
  b=vector(mode=numeric, length=0),
 cc=vector(mode=numeric, length=0), # note:   
avoid c as name

  d=vector(mode=numeric, length=0))
 n = 10
 for(i in 1:n){
 a = LETTERS[i];
 b = i;
 cc = 3*i + 2
 d = rnorm(1);
 results - rbind(results, list(a=a,b=b,cc=cc,d=c))
  }
 results
   a  b cc  d
2  A  1  5  5
21 B  2  8  8
3  C  3 11 11
4  D  4 14 14
5  E  5 17 17
6  F  6 20 20
7  G  7 23 23
8  H  8 26 26
9  I  9 29 29
10 J 10 32 32

OOOPs used d=c and there was a c vector hanging around to be picked  
up.


--
David.



-N

On 11/11/10 2:02 PM, Peter Langfelder wrote:
On Thu, Nov 11, 2010 at 1:19 PM, William Dunlap wdun...@tibco.com  
wrote:

Peter,

Your example doesn't work for me unless I
set options(stringsAsFactors=TRUE) first.
(If I do set that, then all columns of 'results'
have class character, which I doubt the user
wants.)

You probably mean stringsAsFactors=FALSE.

What you say makes sense, because the c() function produces a vector
in which all components have the same type, wnd it will be character.
If you don't want to have characters, my solution would be

n = 10
results - data.frame(a = rep(, n), b = rep(0, n), c = rep(0, n), d
= rep(0, n))
for(i in 1:n){
  a = LETTERS[i];
  b = i;
  c = 3*i + 2
  d = rnorm(1);
  results$a[i] = a
  results$b[i] = b
  results$c[i] = c
  results$d[i] = d
}


results

  a  b  c   d
1  A  1  5 -1.31553805
2  B  2  8  0.09198054
3  C  3 11 -0.05860804
4  D  4 14  0.77796136
5  E  5 17  1.28924697
6  F  6 20  0.47631483
7  G  7 23 -1.23727076
8  H  8 26  0.83595295
9  I  9 29  0.69435349
10 J 10 32 -0.30922930


mode(results[, 1])

[1] character

mode(results[, 2])

[1] numeric

mode(results[, 3])

[1] numeric

mode(results[, 4])

[1] numeric



or alternatively

n = 10
num - data.frame(b = rep(0, n), c = rep(0, n), d = rep(0, n))
labels = rep(, n);
for(i in 1:n){
  a = LETTERS[i];
  b = i;
  c = 3*i + 2
  d = rnorm(1);
  labels[i] = a
  num[i, ] = c(b, c, d)
}
results = data.frame(a = labels, num)


results

  a  b  c   d
1  A  1  5 -0.47150097
2  B  2  8 -1.30507313
3  C  3 11 -1.09860425
4  D  4 14  0.91326330
5  E  5 17 -0.09732841
6  F  6 20 -0.75134162
7  G  7 23  0.31360908
8  H  8 26 -1.54406716
9  I  9 29 -0.36075743
10 J 10 32 -0.23758269

mode(results[, 1])

[1] character

mode(results[, 2])

[1] numeric

mode(results[, 3])

[1] numeric

mode(results[, 4])

[1] numeric


Peter

__
R-help@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.


__
R-help@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.


David Winsemius, MD
West Hartford, CT

__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-11 Thread Noah Silverman
David,

Great solution.  While a bit longer to enter, it lets me explicitly
define a type for each column. 

Thanks!!!

-N

On 11/11/10 4:02 PM, David Winsemius wrote:

 On Nov 11, 2010, at 6:38 PM, Noah Silverman wrote:

 That makes perfect sense.  All of my numbers are being coerced into
 strings by the c() function.  Subsequently, my data.frame contains all
 strings.

 I can't know the length of the data.frame ahead of time, so can't
 predefine it like your example.
 One thought would be to make it arbitrarily long filled with 0 and
 delete off the unused rows.  But this seems rather wasteful.

 Although it might be faster, though. Here is a non-c() method using
 instead the list function (with options(stringsAsFactors=FALSE). List
 does not coerce to same mode and rbind.dta.frame will accept a list as
 a row argument:

 results - data.frame(a=vector(mode=character, length=0) ,
   b=vector(mode=numeric, length=0),
  cc=vector(mode=numeric, length=0), # note: 
 avoid c as name
   d=vector(mode=numeric, length=0))
  n = 10
  for(i in 1:n){
  a = LETTERS[i];
  b = i;
  cc = 3*i + 2
  d = rnorm(1);
  results - rbind(results, list(a=a,b=b,cc=cc,d=c))
   }
  results
a  b cc  d
 2  A  1  5  5
 21 B  2  8  8
 3  C  3 11 11
 4  D  4 14 14
 5  E  5 17 17
 6  F  6 20 20
 7  G  7 23 23
 8  H  8 26 26
 9  I  9 29 29
 10 J 10 32 32

 OOOPs used d=c and there was a c vector hanging around to be picked up.


__
R-help@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.


[R] Populating then sorting a matrix and/or data.frame

2010-11-10 Thread Noah Silverman

Hi,

I have a process in R that produces a lot of output.  My plan was to 
build up a matrix or data.frame row by row, so that I'll have a nice 
object with all the resulting data.


I started with:
results - matrix(ncol=3)
names(results) - c(one, two, three)

Then, when looping through the data:
results - rbind(results, c(a,b,c))

This seems to work fine. BUT, my problem arises when I want to filter, 
sort, etc.


I tried (thinking like a data.frame):
results[results$c  100,]

But that fails.

I tried making it a data.frame with
foo - data.frame(results)

But that converted all the numeric values to factors!!!  Which causes a 
whole mess of problems.


Any ideas??

-N

__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-10 Thread Michael Bedward
Hello Noah,

If you set these names...
 names(results) - c(one, two, three)

this won't work...
 results[results$c  100,]

because you don't have a column called c (unless that's just a typo
in your post).

 I tried making it a data.frame with
 foo - data.frame(results)

 But that converted all the numeric values to factors!!!

Not sure what's going on there. If 'results' is a numeric matrix you
should get a data.frame with numeric cols since under the hood this is
just calling the as.data.frame function.

Michael





On 11 November 2010 16:02, Noah Silverman n...@smartmediacorp.com wrote:
 Hi,

 I have a process in R that produces a lot of output.  My plan was to build
 up a matrix or data.frame row by row, so that I'll have a nice object with
 all the resulting data.

 I started with:
 results - matrix(ncol=3)
 names(results) - c(one, two, three)

 Then, when looping through the data:
 results - rbind(results, c(a,b,c))

 This seems to work fine. BUT, my problem arises when I want to filter, sort,
 etc.

 I tried (thinking like a data.frame):
 results[results$c  100,]

 But that fails.

 I tried making it a data.frame with
 foo - data.frame(results)

 But that converted all the numeric values to factors!!!  Which causes a
 whole mess of problems.

 Any ideas??

 -N

 __
 R-help@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.


__
R-help@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.


Re: [R] Populating then sorting a matrix and/or data.frame

2010-11-10 Thread Noah Silverman

That was a typo.

It should have read:
results[results$one  100,]

It does still fail.

There is ONE column that is text.  So my guess is that R is seeing that 
and assuming that the entire data.frame should be factors.


-N

On 11/10/10 11:16 PM, Michael Bedward wrote:

Hello Noah,

If you set these names...

names(results)- c(one, two, three)

this won't work...

results[results$c  100,]

because you don't have a column called c (unless that's just a typo
in your post).


I tried making it a data.frame with
foo- data.frame(results)

But that converted all the numeric values to factors!!!

Not sure what's going on there. If 'results' is a numeric matrix you
should get a data.frame with numeric cols since under the hood this is
just calling the as.data.frame function.

Michael





On 11 November 2010 16:02, Noah Silvermann...@smartmediacorp.com  wrote:

Hi,

I have a process in R that produces a lot of output.  My plan was to build
up a matrix or data.frame row by row, so that I'll have a nice object with
all the resulting data.

I started with:
results- matrix(ncol=3)
names(results)- c(one, two, three)

Then, when looping through the data:
results- rbind(results, c(a,b,c))

This seems to work fine. BUT, my problem arises when I want to filter, sort,
etc.

I tried (thinking like a data.frame):
results[results$c  100,]

But that fails.

I tried making it a data.frame with
foo- data.frame(results)

But that converted all the numeric values to factors!!!  Which causes a
whole mess of problems.

Any ideas??

-N

__
R-help@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.



__
R-help@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.