similar to: Converting an ftable (contingency table) to a dataframe in R

Hi all, When I try the following with pkg Survey it returns the error below: ftable(svyby(~INCOME, ~AGECL+RACECL, svymean, design=q50), rownames=list(AGECL=c("<35", "35-44", "45-54", "55-64", "65-74", ">=75"), RACECL=c("white non hispanic", "non white or

Dear all, an hopefully quick table question. I have the following data: Two objects that are 2*9 matrix with nine column names (Dis1, ..., Dis9) and the row names (2010,2020). The content are frequencies (numeric). In want to create a table that is along the lines of ftable(UCBAdmissions) and should looks like this: Dis1 | ...| Dis9 2010|2020|....|2010|2020 (first row,first column is the value

try resetting your factor levels and re-run? q50 <- update( q50 , INCOME = factor( INCOME ) , AGECL = factor( AGECL ) , RACECL = factor( RACECL ) ) On Sun, Jul 9, 2017 at 2:59 PM, Orsola Costantini via R-help < r-help at r-project.org> wrote: > Hi all, > > When I try the following with pkg Survey it returns the error below: > > ftable(svyby(~INCOME, ~AGECL+RACECL,

Hi all: I have 3 questions about the poisson regression of contingency table. Q1¡¢How to understand the "independent poisson process"as many books or paper mentioned? For instance: Table1 ------------------------------------------- treat caner non-cancer sum ------------------------------------------- treat1 52(57.18) 19(13.82) 71 treat2

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

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

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?"

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

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

I am trying to run a confirmatory factor analysis using the SEM package. My data are ordinal. I have read http://socserv.mcmaster.ca/jfox/Misc/sem/SEM-paper.pdf. When I apply the hetcor function, I receive the following error: Error in checkmvArgs(lower = lower, upper = upper, mean = mean, corr = corr, : at least one element of 'lower' is larger than 'upper' Example:

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

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

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

How can I export an ftable object in the same format that appears in R command window? For testing that i was using this example that is in help of this function. ## Start with a contingency table. ftable(Titanic, row.vars = 1:3) ftable(Titanic, row.vars = 1:2, col.vars = "Survived") ftable(Titanic, row.vars = 2:1, col.vars = "Survived") ## Start with a data frame. x <-

----- 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. >

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

Dear R-developers, I would like to suggest a 'method' slot for format.ftable() (see an adjusted 'format.ftable()' below, taken from the source of R-2.15.2). At the moment, format.ftable() contains several empty cells due to the way the row and column labels are printed. This creates problems (= unwanted empty columns/rows) when converting an ftable to a LaTeX table; see an

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

Hi All, There are a variety of functions that can be applied to a variable (column) in a data frame: mean, min, max, sd, range, IQR, etc. I am aware of only two that work on the rows, using q1-q3 as example variables: rowMeans(cbind(q1,q2,q3),na.rm=T) #mean of multiple variables rowSums (cbind(q1,q2,q3),na.rm=T) #sum of multiple variables Can the standard column functions (listed in the

Hi, Why the qr() produces a negative Q compared with Gram-Schmidt? (note example below, except Q[2,3]) Here is an example, I calculate the Q by Gram-Schmidt process and compare the output with qr.Q() a <- c(1,0,1) b <- c(1,0,0) c <- c(2,1,0) x <- matrix(c(a,b,c),3,3) ########################## # Gram-Schmidt ########################## A <- matrix(a,3,1) q1 <-