In my experience package "dplyr" has all functions to deal with this kind
of problems in a simple and compact way
Sergio
Il 03/lug/2015 07:26, "Charles Thuo" ha scritto:
> I have a data frame whose rows are 678013 . I would like to remove rows
> from 30696 to 678013 and then attach a new column
Try
y <- x[ -( 30596:678013 ), ]
Please note that I have replaced 30595 with 30596 which is I think what you
mean.
You can add a new column with
y$new <- new_column # this is your vector of length 30595
Good luck,
Rainer
On Friday 03 July 2015 07:23:28 Charles Thuo wrote:
> I have a data
?precedence
-5:10 is (-5):10
-- Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
-- Clifford Stoll
On Thu, Jul 2, 2015 at 10:23 PM, Charles Thuo wrote:
> I have a data frame whose rows are 678013 . I would like to remove ro
I have a data frame whose rows are 678013 . I would like to remove rows
from 30696 to 678013 and then attach a new column with a length of 30595.
I tried
Y<- X[-30595:678013,] and its not working
In addition how do i add a new column
Kindly assist.
Charles
[[alternative HTML version
Not clear what you did. Is this an example of FAQ 7.16?
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Liv
Thank you Bill and Dennis. grepl worked great.
However, for reason I am not figuring out, the
code worked as I included the procedure
(subroutine) with a source command, viz.,
source("z:\\R\\mylib\\me.R")
Compiling the routine into a library/package, as
I always do, then the command got ig
Try grepl() to do pattern matching in strings. ("%in%" checks for
equality.) E.g., using your original 'out' do
out[ !grepl("sex|rating", rownames(out), ]
to get all but the rows whose names contain the character sequences
"sex" or "rating".
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Su
I like to remove from a data frame rows with labels containing
certain string, e.g., "sex" and "rating". Below is a list of the data
frame and my failed attempt to the rows. Any clues? Thanks.
> out
est se t p disc
p.(Intercept) 26.430 13.605 1.943 0.053
p.sex
ge -
From: "Mossadegh, Ramine N."
To: arun
Cc:
Sent: Thursday, May 23, 2013 10:44 AM
Subject: RE: [R] Removing rows w/ smaller value from data frame
Thank but I get : Error in is.list(by) : 'by' is missing
When I tried ddply(datNew,.(activity),summarize, max_dt=max(max_dt))
r
ddply(datNew,.(activity),summarize, max_dt=tail(sort(max_dt),1))
# activity max_dt
#1 A 2013-03-28
#2 B 2013-03-28
#3 C 2013-03-28
A.K.
- Original Message -
From: ramoss
To: r-help@r-project.org
Cc:
Sent: Thursday, May 23, 2013 10:23 AM
Subject: [R] Removing rows
-bounces@r-
> project.org] On Behalf Of ramoss
> Sent: Thursday, May 23, 2013 4:24 PM
> To: r-help@r-project.org
> Subject: [R] Removing rows w/ smaller value from data frame
>
> Hello,
>
> I have a column called max_date in my data frame and I only want to
> keep the big
Hello,
I have a column called max_date in my data frame and I only want to keep the
bigger values for the same activity. How can I do that?
data frame:
activitymax_dt
A2013-03-05
B 2013-03-28
A 2013-03-28
C 2013-03-28
B 2013-03-01
Thanks very much for your rapid help Arun.
Vince
On Apr 12, 2013, at 4:10 PM, arun kirshna [via R] wrote:
Hi,
>From your example data,
dat1<- read.table(text="
id1 id2 value
a b 10
c d11
b a 10
c e 12
",sep="",header=TRUE,stringsAsFactors=FA
Hi,
>From your example data,
dat1<- read.table(text="
id1 id2 value
a b 10
c d 11
b a 10
c e 12
",sep="",header=TRUE,stringsAsFactors=FALSE)
#it is easier to get the output you wanted
dat1[!duplicated(dat1$value),]
# id1 id2 value
#1 a b
Perhaps I've missed something, but if it's really true that the goal is to
remove rows if the first non-zero element is "D" or "d", then how about
this:
tmp <- gsub('0','',df$ch)
first <- substr(tmp,1,1)
subset(df, tolower(first) != 'd')
and of course it could be rolled up into a single expressio
Got it! Thank you Rui!
cp
On Tue, Jul 3, 2012 at 10:14 AM, Rui Barradas wrote:
> Hello,
>
> I'm glad it helped. See answer inline.
>
> Em 03-07-2012 17:09, Claudia Penaloza escreveu:
>
> Thank you Rui and Jim, both 'i1' and 'i1new' worked perfectly
>> because there are no instances of 'Dd' or '
Hello,
I'm glad it helped. See answer inline.
Em 03-07-2012 17:09, Claudia Penaloza escreveu:
Thank you Rui and Jim, both 'i1' and 'i1new' worked perfectly
because there are no instances of 'Dd' or 'dD' in the data set (that I
would/not want to include/exclude)... but I understand that 'i1new'
Thank you Rui and Jim, both 'i1' and 'i1new' worked perfectly because there
are no instances of 'Dd' or 'dD' in the data set (that I would/not want to
include/exclude)... but I understand that 'i1new' targets precisely what I
want.
Why isn't a leader of zero's required for either 'i1' or 'i1new',
Hello,
Inline.
Em 03-07-2012 01:15, jim holtman escreveu:
You will have to change the 'i1' expression as follows:
i1 <- grepl("^([0D]|[0d])*$", dd$ch)
i1 # matches strings with d & D in them
[1] TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# second string had 'd' & 'D' in
From: Rui Barradas
To: Claudia Penaloza
Cc: r-help@r-project.org
Sent: Monday, July 2, 2012 7:24 PM
Subject: Re: [R] Removing rows if certain elements are found in character string
Hello,
Try regular expressions instead.
In this data.frame, I've changed row nr.4 to have a row wi
07368;
9 9 T000 0.002456;
A.K.
- Original Message -
From: Claudia Penaloza
To: r-help@r-project.org
Cc:
Sent: Monday, July 2, 2012 6:48 PM
Subject: [R] Removing rows if certain elements are found in character string
I would like to remove rows from the fo
On Jul 2, 2012, at 6:48 PM, Claudia Penaloza wrote:
I would like to remove rows from the following data frame (df) if
there are
only two specific elements found in the df$ch character string (I
want to
remove rows with only "0" & "D" or "0" & "d"). Alternatively, I
would like
to remove ro
You will have to change the 'i1' expression as follows:
> i1 <- grepl("^([0D]|[0d])*$", dd$ch)
> i1 # matches strings with d & D in them
[1] TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
> # second string had 'd' & 'D' in it so it was TRUE above and FALSE below
> i1new <- grepl("^(
Hello,
Try regular expressions instead.
In this data.frame, I've changed row nr.4 to have a row with 'D' as
first non-zero character.
dd <- read.table(text="
ch count
1 00D0 0.007368
2 00d0 0.002456
3
I would like to remove rows from the following data frame (df) if there are
only two specific elements found in the df$ch character string (I want to
remove rows with only "0" & "D" or "0" & "d"). Alternatively, I would like
to remove rows if the first non-zero element is "D" or "d".
On Nov 22, 2011, at 12:43 PM, AC Del Re wrote:
Hi,
Is there an easy way to remove dataframe rows without duplicated
values of
a specified column ('id')? e.g.,
dat <- data.frame(id = c(1,1,1,2,3,3), value = c(5,6,7,4,5,4),
value2 =
c(1,4,3,3,4,3))
dat
id value value2
1 1 5 1
one approach is the following:
dat <- data.frame(id = c(1,1,1,2,3,3), value = c(5,6,7,4,5,4),
value2 = c(1,4,3,3,4,3))
ind <- ave(dat$id, dat$id, FUN = length) > 1
dat[ind, ]
I hope it helps.
Best,
Dimitris
On 11/22/2011 6:43 PM, AC Del Re wrote:
Hi,
Is there an easy way to remove dat
Sorry, you need this first:
L <- split(dat, dat$id)
do.call(rbind, lapply(L, function(d) if(nrow(d) > 1) return(d)))
D.
On Tue, Nov 22, 2011 at 10:38 AM, Dennis Murphy wrote:
> Hi:
>
> Here's one way:
>
> do.call(rbind, lapply(L, function(d) if(nrow(d) > 1) return(d)))
> id value value2
> 1.
Hi:
Here's one way:
do.call(rbind, lapply(L, function(d) if(nrow(d) > 1) return(d)))
id value value2
1.1 1 5 1
1.2 1 6 4
1.3 1 7 3
3.5 3 5 4
3.6 3 4 3
HTH,
Dennis
On Tue, Nov 22, 2011 at 9:43 AM, AC Del Re wrote:
> Hi,
>
> Is there an easy
This is ugly, but it gets what you want.
dat[which(dat[,1] %in% unique((dat[duplicated(dat[,1], fromLast = T),
1]))),]
AC Del Re wrote
>
> Hi,
>
> Is there an easy way to remove dataframe rows without duplicated values of
> a specified column ('id')? e.g.,
>
> dat <- data.frame(id = c(1
Hi,
Is there an easy way to remove dataframe rows without duplicated values of
a specified column ('id')? e.g.,
dat <- data.frame(id = c(1,1,1,2,3,3), value = c(5,6,7,4,5,4), value2 =
c(1,4,3,3,4,3))
dat
id value value2
1 1 5 1
2 1 6 4
3 1 7 3
4 2 4 3
5
On Jun 2, 2011, at 11:35 AM, Petr Savicky wrote:
On Thu, Jun 02, 2011 at 11:23:28AM -0400, Jim Silverton wrote:
Hi,
Can someone tell me how to remove rows of zeros from a matrix?
For example if I have the following matrix,
0 0
0 1
2 8
0 0
4 56
I should end up with
0 1
2 8
4 56
Hi.
Try the
On Thu, Jun 02, 2011 at 11:23:28AM -0400, Jim Silverton wrote:
> Hi,
> Can someone tell me how to remove rows of zeros from a matrix?
> For example if I have the following matrix,
>
> 0 0
> 0 1
> 2 8
> 0 0
> 4 56
>
> I should end up with
> 0 1
> 2 8
> 4 56
Hi.
Try the following
a <- matrix(c
Assuming the matrix is named X:
X[which(rowSums(X) > 0),]
should work.
Also, this list is a text-only list. As you are using gmail, sending
text only messages is very easy, and may clear confusion in future
posts.
HTH,
Jon
On Thu, Jun 2, 2011 at 11:23 AM, Jim Silverton wrote:
> Hi,
> Can som
Hi,
Can someone tell me how to remove rows of zeros from a matrix?
For example if I have the following matrix,
0 0
0 1
2 8
0 0
4 56
I should end up with
0 1
2 8
4 56
--
Thanks,
Jim.
[[alternative HTML version deleted]]
__
R-help@r-project.or
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ali Salekfard
> Sent: Wednesday, December 29, 2010 6:25 AM
> To: r-help@r-project.org
> Subject: Re: [R] Removing rows with earlier dates
>
> Than
David,
Thanks alot. Your code is worked fine on the whole dataset (no memory error
as I had with the other ideas). I do like the style - especialy the fact
that it is all in one line - , but for large datasets it takes longer than
what I wrote. I ran it on the same machine with the same set of rul
On Dec 29, 2010, at 11:03 AM, Ali Salekfard wrote:
> David,
>
> Thanks alot. Your code is worked fine on the whole dataset (no
> memory error as I had with the other ideas). I do like the style -
> especialy the fact that it is all in one line - , but for large
> datasets it takes longer th
On Dec 29, 2010, at 9:24 AM, Ali Salekfard wrote:
Thanks to everyone. Joshua's response seemed the most concise one,
but it
used up so much memory that my R just gave error. I checked the other
replies and all in all I came up with this, and thought to share it
with
others and get comments
> David Winsemius
> on Fri, 24 Dec 2010 11:47:05 -0500 writes:
> On Dec 24, 2010, at 11:04 AM, David Winsemius wrote:
>>
>> On Dec 24, 2010, at 8:45 AM, Ali Salekfard wrote:
>>
>>> Hi all,
>>>
>>> I'm new to the list but have benfited from it quite exte
Thanks to everyone. Joshua's response seemed the most concise one, but it
used up so much memory that my R just gave error. I checked the other
replies and all in all I came up with this, and thought to share it with
others and get comments.
My structure was as follows:
ACCOUNT RULE DATE
A1
Whenever a task calls for breaking a data object into pieces, operate on the
pieces, then put it back together, then think about using the plyr package.
Sent from my iPod
On Dec 24, 2010, at 6:58 AM, "Ali Salekfard" wrote:
> Hi all,
>
> I'm new to the list but have benfited from it quite ext
> with(YourDataFrame, tapply(`Effective Date`, `RULE COLUMNS`,
> function(x) x[which.max(x)]))
David pointed out that this will just return a table of dates. One
work around is:
do.call("rbind", by(DataFrame, DataFrame[, "RULE COLUMNS"],
function(x) x[which.max(x[, "Effective Date"]), ]))
bu
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ali Salekfard
> Sent: Friday, December 24, 2010 5:46 AM
> To: r-help@r-project.org
> Subject: [R] Removing rows with earlier dates
>
> Hi all,
>
>
Hi,
On Fri, Dec 24, 2010 at 5:45 AM, Ali Salekfard wrote:
[snip]
> I have a data frame that contains mapping rules in this way:
>
> ACCOUNT, RULE COLUMNS, Effective Date
>
>
> The dataframe comes from a database that stores all dates. What I would like
> to do is to create a data frame with only
On Dec 24, 2010, at 11:04 AM, David Winsemius wrote:
On Dec 24, 2010, at 8:45 AM, Ali Salekfard wrote:
Hi all,
I'm new to the list but have benfited from it quite extensively.
Straight to
my rather strange question:
I have a data frame that contains mapping rules in this way:
ACCOUNT,
On Dec 24, 2010, at 8:45 AM, Ali Salekfard wrote:
Hi all,
I'm new to the list but have benfited from it quite extensively.
Straight to
my rather strange question:
I have a data frame that contains mapping rules in this way:
ACCOUNT, RULE COLUMNS, Effective Date
The dataframe comes from
Hi all,
I'm new to the list but have benfited from it quite extensively. Straight to
my rather strange question:
I have a data frame that contains mapping rules in this way:
ACCOUNT, RULE COLUMNS, Effective Date
The dataframe comes from a database that stores all dates. What I would like
to do
thanks Henrique , it did work with a slight modif
subset(merge(X, Y, by.x = 'groups', by.y = 1, all = TRUE), var2http://r.789695.n4.nabble.com/removing-rows-from-a-matrix-using-condition-within-groups-tp3004132p3005899.html
Sent from the R help mailing list archive at Nabble.com.
___
I tired this and seems to capture only a few
--
View this message in context:
http://r.789695.n4.nabble.com/removing-rows-from-a-matrix-using-condition-within-groups-tp3004132p3005894.html
Sent from the R help mailing list archive at Nabble.com.
__
R-
Try this:
subset(merge(X, Y, by.x = 'groups', by.y = 1, all = TRUE), values < V2, -V2)
On Wed, Oct 20, 2010 at 1:37 PM, swam wrote:
>
> I am needing some help in removing certain rows in a data.matrix and then
> do
> some calculation. So Iam need to removing certain values above a threshold
>
I am needing some help in removing certain rows in a data.matrix and then do
some calculation. So Iam need to removing certain values above a threshold
or value from another vector.
For eg.. in the below data matrix X there are 6 groups (A, B, C, D, E ,F)
>X
rowsgroups values
1 A
try this:
myDF[complete.cases(myDF), ]
Best,
Dimitris
milton ruser wrote:
Dear all,
I have a data.frame with some NAs that can
happens anywere, and I would like to keep
only rows without NAs. Thanks in advance
for your help, and Merry Xmas.
myvect<-runif(100)
myvect[sample(1:100)[1:5]]<-NA
Dear all,
I have a data.frame with some NAs that can
happens anywere, and I would like to keep
only rows without NAs. Thanks in advance
for your help, and Merry Xmas.
myvect<-runif(100)
myvect[sample(1:100)[1:5]]<-NA
myDF<-data.frame(matrix(myvect,ncol=5))
myDF
miltinho
[[alternative HTM
Hi All,
> act_2
DateDtime Hour Min Second Rep
51 2006-02-22 14:52:18 14 52 18 useractivity_act
52 2006-02-22 14:52:18 14 52 18 4
55 2006-02-22 14:52:49 14 52 49 4
57 2006-02-22 14:52:51 14 52 51
On Thu, Nov 20, 2008 at 12:28 PM, Gavin Simpson <[EMAIL PROTECTED]> wrote:
> But Prof. Ripley has pointed out (off list) that
>
> ffg[rowSums(ffg) > 0, ]
I suggested much the same solution off-list (using apply rather than
rowSums, as I'm
apparently incapable of remembering the existence of the l
On Thu, 2008-11-20 at 17:08 +, Gavin Simpson wrote:
> On Thu, 2008-11-20 at 12:01 -0500, stephen sefick wrote:
> > ##I want to remove the rows where the row sums are zero and this is as
> > far as I have gotten
>
> Given your ffg,
>
> ## the which() call returns row indices for rows with rowS
On Thu, 2008-11-20 at 12:01 -0500, stephen sefick wrote:
> ##I want to remove the rows where the row sums are zero and this is as
> far as I have gotten
Given your ffg,
## the which() call returns row indices for rows with rowSum > 0
ffg[which(rowSums(ffg) > 0, ]
does the trick
HTH
G
>
> ff
##I want to remove the rows where the row sums are zero and this is as
far as I have gotten
ffg <- (structure(list(CD = c(0, 0, 0, 0, 3.125, 0, 0, 0, 0, 1.6, 3.125,
0, 0, 6.25, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3.125, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.6, 0, 0, 0, 0, 0,
0, 0, 0, 0
R is doing what you are telling it to do. You aren't assigning the
result of your indexing to a new data.frame, and so it is simply
print()ing the result. Example,
x <- 1:10
x[-1]
x
vs.
x <- 1:10
x <- x[-1]
x
Joe Trubisz wrote:
Hi...
I have a rather large dataframe that I'm trying to rem
Hi...
I have a rather large dataframe that I'm trying to remove rows from.
I'm issuing the command:
dtx[-which(dtx$rdate > "2008-06-16"),]
and it tries to print out over 170,000 lines of output. So...I did:
options(max.print=1e6)
and ran it again. It worked, but when I did a:
which(dtx$rdate
> Hi, I have a problem regarding matrix handeling. I am working with
> a serie of matrixes containing several columns. Now I would like to
> delete those rows of the matrixes,that in one of the columns contain
> values less than 50 or greater than 1000.
Try this:
m <- matrix(runif(150, 0, 1050
Try:
x <- matrix(sample(10:1100, 100), 10)
x[!apply(x < 50 | x > 1000, 1, any),]
On Fri, May 2, 2008 at 6:48 AM, Monna Nygård <[EMAIL PROTECTED]> wrote:
>
> Hi, I have a problem regarding matrix handeling. I am working with a
> serie of matrixes containing several columns. Now I would like to d
Hi, I have a problem regarding matrix handeling. I am working with a serie of
matrixes containing several columns. Now I would like to delete those rows of
the matrixes,that in one of the columns contain values less than 50 or greater
than 1000. How would this be possible, I have tried to crea
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