search for: cuttings

Displaying 20 results from an estimated 10706 matches for "cuttings".

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2001 Oct 09
2
Assert in jbd-kernel.c
Hello. I have installed the ext3 file system on a test system, and sometimes I have a problem: I get an assert from within jbd-kernel.c, and whatever prgram was writing to the disk when this happens is unable to continue. The system is a server I built, which I named "dax". It is running Debian unstable, and I updated it to all the latest packages in Debian unstable as of today.
2002 Dec 18
6
Can I build an array of regrssion model?
Hi, I am trying to use piecewise linear regression to approximate a nonlinear function. Actually, I don't know how many linear functions I need, therefore, I want build an array of regression models to automate the approximation job. Could you please give me any clue? Attached is ongoing code: rawData = scan("c:/zyang/mass/data/A01/1.PRN", what=list(numeric(),numeric())); len =
2008 Oct 11
1
problem with cut.Date/date plotting in ggplot2
I've hit a problem in ggplot2 which I can trace back to cut.Date , which is either a bug or (??) ggplot2 trying to do something it shouldn't with cut.Date (although its use of cut.Date seems OK). Apparently any (?) call of the form cut(as.Date("2008-07-07"),"weeks") where the date *begins the week*, gives the error Error in 1:(1 + max(which(breaks < maxx))) :
2008 Mar 18
4
cut.Date and cut.POSIXt problem
cut.Date and cut.POSIXt indicate that the breaks argument can be an integer followed by a space followed by "year", etc. but it seems the integer is ignored. For example, I assume that breaks = "3 months" is supposed to cut it into quarters but, in fact, it cuts it into months as if 3 had not been there. > d <- seq(Sys.Date(), length = 12, by = "month") >
2011 Jan 31
2
identify subsets based on two grouping factors
Hi, I have a data.frame that has a categorical variable, for which I would like to look at the distribution of levels of this variable, based on a grouping of two other variables. As an example: x <- data.frame(obs=sample(c('low', 'high'),100, replace=TRUE), grp1=sample(1:10, 100, replace=TRUE), grp2=runif(100)) cut.grp1 <- cut(x$grp1, 3) cut.grp2 <- cut(x$grp2, 3)
2007 Dec 13
3
what does cut(data, breaks=n) actually do?
Hello, I'm trying to bin a quantity into 2-3 bins for calculating entropy and mutual information. One of the approaches I'm exploring is the cut() function, which is what the mutualInfo function in binDist uses. When it's called in the format cut(data, breaks=n), it somehow splits the data into n distinct bins. Can anyone tell me how cut() decides where to cut? Thanks, Melissa
2009 Dec 04
2
Dividing a pixel image into factors - (cut.im(), cut.default())
Hi, I have a numeric pixel image which I would like to divide into factors for analysis in Spatstat. I have found that I can use cut.im() function to divide the range of pixel values into a series of equal length intervals (e.g. if my pixels values range from 0 to 60, cut.im(X.im,breaks=2) will produce two factors one containing pixel values 0-30 and one containing pixel values of 30 - 60, or
2008 Oct 13
1
cut.Date problem when starting on first day of week (PR#13159)
Apparently any (?) call of the form cut(date,"weeks") where the date *begins the week*, gives the error Error in 1:(1 + max(which(breaks < maxx))) : result would be too long a vector In addition: Warning message: In max(which(breaks < maxx)) : no non-missing arguments to max; returning -Inf To my surprise, this was first reported as a problem in 2006 (!) (version 2.3.1
2005 Sep 02
6
question on cut
I am trying to use cut to tell me the suffix of a file. for example: echo /home/silentm/log/file.machine.log | cut -d . -f 1- I was expecting to get .log or log but I get the entire string echoed back. doing the opposite gave me what I expected: echo /home/silentm/log/file.machine.log | cut -d . -f 1 gives me /home/silentm/log/file I am trying to find a way to test if the file ends in .log?
2009 Sep 11
1
bar chart with means - using ggplot
Like this? # example using qplot library(ggplot2) meanprice <- tapply(diamonds$price, diamonds$cut, mean);meanprice cut <- factor(levels(diamonds$cut), levels = levels(diamonds$cut)) qplot(cut, meanprice, geom="bar", stat="identity", fill = I("grey50")) dev.new() # create a new graph to compare with qplot # Example using ggplot ggdata <-
2003 Jul 12
2
using cut on matrices
Dear list, I'd like to use the function cut() on matrices, ie that when I apply it to a matrix, it would return a matrix of the same dimensions instead of a vector. I wonder if there is a better (more elegant) solution than matrix(cut(a, ...), ncol=ncol(a), nrow=nrow(a)) because I would like to use cut on both vectors and matrices and avoid testing whether a is a matrix. Thanks, Tamas
2002 May 14
0
RE: cut.dendrogram (PR#1552)
I'm resending this bug report with a new example. As seen below, cut.dendrogram gives an error message for some heights, but not for others and with some datasets adn not others. I can't see why. Last time I unwittingly sent my message with HTML formatting. This time I'm travelling and using an e-mail system that I am unfamiliar with. As far as I can see, I am not using HTML.
2008 Aug 09
2
levels values of cut()
Dear list, I have the following example, from which I am hoping to retrieve numeric values of the factor levels (that is, without the brackets): > > x <- seq(1, 15, length=100) > y <- sin(x) > > my.cuts <- cut(which(abs(y) < 1e-1), 3) > levels(my.cuts) hist() does not suit me for this, as it does not necessarily respect the number of breaks. getAnywhere
2005 Aug 02
1
cut.Date functionality for chron date/time objects
Hello, I've encountered the need to cut some chron objects of the form: R> mychron <- chron(sort(runif(10, 0, 10))) R> mychron [1] (01/01/70 16:36:20) (01/02/70 00:08:46) (01/03/70 16:54:49) [4] (01/04/70 06:45:00) (01/07/70 06:21:24) (01/07/70 18:28:44) [7] (01/08/70 00:47:05) (01/08/70 05:11:44) (01/10/70 01:07:53) [10] (01/10/70 17:46:53) into arbitrary (e.g. a given number
2017 Nov 04
2
ntfs user mappings?
. DOMAIN_ADMIN_PASSWD.sh echo ${PASSWD} | kinit ${ADMIN}@${DOMAIN} echo -n > /etc/ntfs-3g.usermap for DOMAIN_USER in $(wbinfo -u);do RPCLOOKUPID=$(rpcclient -P -c "lookupnames ${DOMAIN_USER}" ${DOMAIN}) if [ "${RPCLOOKUPID:0:7}" != "ERROR: " ] && [ "${RPCLOOKUPID:0:7}" != "Failed " ];then SID=$(echo ${RPCLOOKUPID}|awk '{print
2010 Dec 14
3
Question about cut()
Dear all, I would like to use cut() to make numerics to factors, the sample codes are as follows. However, the result is not what I want, since r[3] =?9 should be in the interval of "8-10%" rather than "2-4%". Maybe cut() is not the right function to use for my situation. Please help. > r <- c(1,1,9,1,1,1) > col_no <-
2012 Jan 26
2
How do I use the cut function to assign specific cut points?
I am new to R, and I am trying to cut a continuous variable BMI into different categories and can't figure out how to use it. I would like to cut it into four groups: <20, 20-25, 25-30 and >= 30. I am having difficulty figuring the code for <20 and >=30? Please help. Thank you. -- View this message in context:
2016 Oct 30
7
Power Cut
Dear All I am using a centos server for cdr billing and mediation device on a remote network. I am experiencing problem that I am suspicious it comes from main supply power cut at the remote site. The power supply to the remote site comes from battery charger that will be automatically switched in circuit under main supply power cut but cannot provide adequate power for more than 2 hours . I am
2011 Jun 14
1
Expand DF with all levels of a variable
Dear list, I would like to expand a DF with all the missing levels of a variable. a <- c(2,2,3,4,5,6,7,8,9) a.cut <- cut(a, breaks=c(0,2,6,9,12), right=FALSE ) (x <- data.frame(a, a.cut)) # In 'x' the level "[0,2)" is "missing". AddMissingLevel <- function(xdf) { xfac <- factor( c("[0,2)", "[2,6)", "[6,9)",
2006 Nov 29
4
Why the factor levels returned by cut() are not ordered?
What is the reason, that the levels of the factor returned by cut() are not marked as ordered levels? > is.ordered( cut( breaks=3, sample(10 ) ) ) FALSE > help(factor) ... If 'ordered' is 'TRUE', the factor levels are assumed to be ordered. ... Wolfram