similar to: R beginner: matrix algebra

Displaying 20 results from an estimated 2000 matches similar to: "R beginner: matrix algebra"

2013 Jan 25
3
Help with adding 'dates' string as rownames to matrix
Hi, I need help with two related issues: 1. I wish to drop repeating text "BST" from the below 'dates' string: [1] "2005-04-01 BST" "2005-04-04 BST" "2005-04-05 BST" "2005-04-06 BST" "2005-04-07 BST" "2005-04-08 BST" "2005-04-11 BST" "2005-04-12 BST" "2005-04-13 BST" "2005-04-14
2010 Mar 11
4
Forecast
sample report data that i want to forecast quarter quarter_index Revenue 2007 Q1 1 $3,856,799 2007 Q2 2 $4,243,328 2007 Q3 3 $4,930,369 2007 Q4 4 $5,443,579 2008 Q1 5 $5,164,830 2008 Q2 6 $5,104,413 2008 Q3 7
2012 Apr 12
4
Definition of "lag" is opposite in ts and xts objects!
Example: Will ts objects be obsolete or modified? > a [,1] 1983 Q1 2.747365190 1983 Q2 2.791594762 1983 Q3 -0.009953715 1983 Q4 -0.015059485 1984 Q1 -1.190061246 1984 Q2 -0.553031799 1984 Q3 0.686874720 1984 Q4 0.953911035> lag(a,4) [,1] 1983 Q1 NA 1983 Q2 NA 1983 Q3 NA 1983 Q4 NA 1984 Q1 2.747365190 1984 Q2
2012 Feb 03
3
Cannot get "==" operator to return TRUE
I have a data.frame named "df". The dput of df is at the bottom of this e-mail. What I'd like to do is replace the "n/a " values with NA. On Mac OSX, it works to do this: df[df == "n/a"] <- NA However, it does not work on Ubuntu. See below. Thanks in advance, Garrett > x <- df[27, 4] # complete data.frame dput is below > dput(x) "n/a?"
2010 Mar 10
3
see the example and help me
sample report data that i want to forecast quarter quarter_index Revenue 2007 Q1 1 $3,856,799 2007 Q2 2 $4,243,328 2007 Q3 3 $4,930,369 2007 Q4 4 $5,443,579 2008 Q1 5 $5,164,830 2008 Q2 6 $5,104,413 2008 Q3 7
2009 Feb 19
2
table with 3 variables
I have the initial matrice: > *data.frame(Subject=rep(100:101, each=4), Quarter=rep(paste("Q",1:4, sep=""),2), Boolean = rep(c("Y","N"),4))* Subject Quarter Boolean 1 100 Q1 Y 2 100 Q2 N 3 100 Q3 Y 4 100 Q4 N 5 101 Q1 Y 6 101 Q2 N 7 101 Q3 Y 8 101
2008 Mar 02
1
question on lag.zoo
Hi Guys, I'm using zoo package now. I found lag is not doing what I assumed. > x <- zoo(11:21) > z <- zoo(1:10, yearqtr(seq(1959.25, 1961.5, by = 0.25)), frequency = 4) > x 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 19 20 21 > lag(x) 1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20 21 > z 1959 Q2 1959 Q3 1959 Q4 1960 Q1 1960 Q2 1960 Q3 1960 Q4
2005 Nov 12
1
computation on a table
Hello, I have a table (1) of the form q1 q3 q4 q8 q9 A 5 2 0 1 3 B 2 0 2 4 4 I have another table (2): q1 q2 q3 q4 q5 q6 q7 q8 q9 C 10 7 4 2 6 9 3 1 2 I would like to divide the numbers in table (1) by the number of the appropriate column in table (2): q1 q3 q4 q8 q9 A 5/10 2/4 0/2 1/1 3/2 B 2/10 0/4 2/2 4/1 4/2
2012 Feb 17
4
How can I tabulate time series data (in RStudio or any other R editor)?
Hello, I have a question on how to tabulate the time series data. I use RStudio, but if can be done in any other R editor, it should work in RStudio as well. > a1<-11:22 > a1ts<-ts(a1, frequency=4, start=c(1978,1)) > a1ts Qtr1 Qtr2 Qtr3 Qtr4 1978 11 12 13 14 1979 15 16 17 18 1980 19 20 21 22 If I click the variable "a1ts" on the
2008 Feb 01
2
the "union" of several data frame rows
Hi, I have a question about how to obtain the union of several data frame rows. I'm trying to create a common key for several tests composed of different items. Here is a small scale version of the problem. These are keys for 4 different tests, not all mutually exclusive: id q1 q2 q3 q4 q5 q6 1 A C 2 B D 3 A D B 4 C D B D I would like
2009 Feb 19
2
table with 3 varialbes
I have the initial matrice: > *data.frame(Subject=rep(100:101, each=4), Quarter=rep(paste("Q",1:4, sep=""),2), Boolean = rep(c("Y","N"),4))* Subject Quarter Boolean 1 100 Q1 Y 2 100 Q2 N 3 100 Q3 Y 4 100 Q4 N 5 101 Q1 Y 6 101 Q2 N 7 101 Q3 Y 8 101
2011 Mar 14
2
data.frame transformation
Hi R users, I have following data frame df<-data.frame(q1=c(0,0,33.33,"check"),q2=c(0,33.33,"check",9.156), q3=c("check","check",25,100),q4=c(7.123,35,100,"check")) and i would like to replace every element that is less then 10 with . (dot) in order to obtain this: q1 q2 q3 q4 1 . . check . 2 . 33.33 check 35
2009 Jul 11
3
Reading data entered within an R program
Dear R-helpers, I know of two ways to reading data within an R program, using textConnection and stdin (demo program below). I've Googled about and looked in several books for comparisons of the two approaches but haven't found anything. Are there any particular advantages or disadvantages to these two approaches? If you were teaching R beginners, which would you present? Thanks, Bob
2011 Jul 06
3
Tables and merge
----- Original Message ----- From: "Silvano" <silvano at uel.br> To: <r-help at r-project.org> Sent: Thursday, June 30, 2011 9:07 AM Subject: Tables and merge > Hi, > > I have 21 files which is common variable CODE. > Each file refers to a question. > > I would like to join the 21 files into one, to construct > tables for each question by CODE. >
2018 Jan 28
2
Plotting quarterly time series
I have a data set with quarterly time series for several variables. The time index is recorded in column 1 of the dataframe as a character vector "Q1 1961", "Q2 1961","Q3 1961", "Q4 1961", "Q1 1962", etc. I want to produce line plots with ggplot2, but it seems I need to convert the time index from character to date class. Is that right? If so, how
2004 Aug 06
2
@Christian Buchner: speex acm & netmeeting
Hi, > I can't tell why. Try to register mode 8kHz Mono mode Q3 or Q4 for Ok, I will try it. I tried 16kHz Q3 mono at 9,8kbit. > and upwards use 2 frames. All this is defined in codec.c in the > QualityInfo[] table. Oh OK. I looked a bit in the sources, but it was too confusing for me. ;-) > I would be interested in seeing the code for registering the Speex codec > with
2012 Jan 25
2
having a bit of regression trouble
I got the code for how to do regression without an intercept out of the back of my book and the next part of the homework asks me to do it with an intercept. The problem is, Q1 disappears whenever I try. Here is my code: Without the intercept: load("tsa3.rda") > > Q=factor(rep(1:4,21)) > reg=lm(log(jj)~0+trend+Q,na.action=NULL) > model.matrix(reg) trend Q1 Q2 Q3 Q4 1
2009 Jul 01
2
sorting question
I've asked about custom sorting before and it appears that -- in terms of a user-defined order -- it can only be done either by defining a custom class or using various tricks with "order" Just wondering if anyone has a clever way to order "vintages" of the form 2002, 2003H1, 2003H2, 2004, 2005Q1, 2005Q2, etc some have H1 or H2, some have Q1,Q2,Q3,Q4, some are just plain
2006 Dec 03
1
passing an argument to a function which is also to be a dataframe column name
any suggestions on the following gratefully welcome, I have a dataframe, which I am subsetting via labels atpi[, creativity] where (for example) atpi = as.data.frame(matrix(1:50, ncol = 5, nrow = 10)) names(atpi) = c("Q1", "Q2", "Q3", "Q4", "Q5") and creativity = c("Q1", "Q3", "Q4") I want to add an extra column
2011 Mar 13
1
replace with quantile value for a large data frame...
Dear R-Experts I am sure this might look simple question for experts, at least is problem for me. I have a large data frame with over 1000 variables and each have different distribution( i.e. have different quantile). I want to create a new grouped data frame, where the new variables where the value falling in first (<25%), second (25% to <50%), third (50% to <75%) and fourth quantiles