similar to: Bar Plot Bars "Bleed" off Plotting Area

I've got a list of 5 matrices that are each 5 x 6. I'd like to end up with a 5 x 6 matrix that contains the mean value of the 5 original matrices. I can do this by brute force, but there must be a better way than making each matrix into a vector and then remaking a matrix thanks very much for any help david freedman ll=list(structure(c(9.7, 17.6, 20.8, 24.1, 33.8, 14.5, 25.7, 29.8,

Hello, It's an alternative to use SAS by function in R? I want to plot d histograms by plot.from example bellow: Thank you! plot d 1 1 16.3 2 1 25.0 3 1 57.8 4 1 17.0 5 2 10.8 13 2 96.4 17 3 76.0 18 3 32.0 19 3 11.0 20 3 11.0 24 3 106.0 25 3 12.5 21 4 19.3 22 4 12.0 26 4 15.0 27 5 99.3 32 7 11.0 36

hello, I'd like know how to sort a data frame in R for example how I should do to sort by Catholic with swiss data frame like below thanks Fertility Agriculture Examination Education Catholic Infant.Mortality Courtelary 80.2 17.0 15 12 9.96 22.2 Delemont 83.1 45.1 6 9 84.84 22.2

I have an amd64 7.1 install on a SuperMicro C2SBA+ and I see the following in dmesg.. sio0: configured irq 4 not in bitmap of probed irqs 0 sio0: port may not be enabled sio0: configured irq 4 not in bitmap of probed irqs 0 sio0: port may not be enabled sio0: <16550A-compatible COM port> port 0x3f8-0x3ff irq 4 flags 0x10 on acpi0 sio0: type 16550A sio0: [FILTER] sio1: configured irq 3 not

Dear Community, I have been using two types of survival programs to analyse a data set. The first one is an R function called aftreg. The second one an STATA function called streg. Both of them include the same analyisis with a weibull distribution. Yet, results are very different. Shouldn't the results be the same? Kind regards, J -- View this message in context:

On 2019/7/20 0:13, Michael S. Tsirkin wrote: > On Fri, Jul 19, 2019 at 03:31:29PM +0000, ? jiang wrote: >> On 2019/7/19 22:29, Jiang wrote: >>> On 2019/7/19 10:36, Jason Wang wrote: >>>> On 2019/7/18 ??10:43, Michael S. Tsirkin wrote: >>>>> On Thu, Jul 18, 2019 at 10:42:47AM -0400, Michael S. Tsirkin wrote: >>>>>> On Thu, Jul 18, 2019

On 2019/7/20 0:13, Michael S. Tsirkin wrote: > On Fri, Jul 19, 2019 at 03:31:29PM +0000, ? jiang wrote: >> On 2019/7/19 22:29, Jiang wrote: >>> On 2019/7/19 10:36, Jason Wang wrote: >>>> On 2019/7/18 ??10:43, Michael S. Tsirkin wrote: >>>>> On Thu, Jul 18, 2019 at 10:42:47AM -0400, Michael S. Tsirkin wrote: >>>>>> On Thu, Jul 18, 2019

On 2019/7/19 22:29, Jiang wrote: > > On 2019/7/19 10:36, Jason Wang wrote: >> >> On 2019/7/18 ??10:43, Michael S. Tsirkin wrote: >>> On Thu, Jul 18, 2019 at 10:42:47AM -0400, Michael S. Tsirkin wrote: >>>> On Thu, Jul 18, 2019 at 10:01:05PM +0800, Jason Wang wrote: >>>>> On 2019/7/18 ??9:04, Michael S. Tsirkin wrote: >>>>>> On

Dear all, can somebody please help me how to calculate "McFadden R-square" for a LOGIT model? Corresponding definition can be found here: http://publib.boulder.ibm.com/infocenter/spssstat/v20r0m0/index.jsp?topic=%2Fcom.ibm.spss.statistics.help%2Falg_plum_statistics_rsq_mcfadden.htm Here is my data: Data <- structure(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1,

Hi! I have 2 data.frames: "fish" and "popn": >fish xloc yloc id birth size weight energy gonad 20 15 15 54 -60 107.9 63.0 15952.9 8.0 21 15 15 32 -60 105.1 61.4 15538.8 7.8 91 4 43 96 -60 118.9 69.4 17573.2 8.8 71 32 4 64 -60 121.6 71.0 17976.0 9.0 34 2 64 20 -60 116.2 67.9 17173.0 8.6 95 6 20 58 -60 106.5

Dear r-users,   Originally my data is in notepad.  I use data.frame(...) to convert the data into columns but it has no header.  The data below is what I got in R.  I would like to extract these data:   19710629        39.3  19701126  19720629        33.8  19720517 ... when I use data2_30min[1, cbind(1,3,5)] command the result is NULL.   Then I tried data2_30min[1], it gives me the whole set of

hello to all. I am working on macroeconomic data series of India, which in a yearly basis. I am unable to convert my data frame into time series. kindly help me. also using zoo and xts packages. but they take only monthly observations. 'data.frame': 30 obs. of 4 variables: $ year: int 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 ... $ cnsm: num 174 175 175 172 173 ... $ incm:

Dear anyone who knows more about R than me (so everyone). I have been bashing my head on the keyboard all day trying to do something with my table. I have some data, like so: yyyy-mm Rainfall(mm) 1 1977-02 17.4 2 1977-03 34.0 3 1977-04 26.2 4 1977-05 42.6 5 1977-06 58.6 6 1977-07 23.2 7 1977-08 26.8 8 1977-09 48.4 9

Hi, Suppose I have the following vector (data points): > x [1] 36.0 57.3 73.3 92.0 300.4 80.9 19.8 31.4 85.8 44.9 24.6 48.0 [13] 28.0 38.3 85.2 103.6 154.4 128.5 38.3 72.4 122.7 123.1 41.8 21.7 [25] 143.6 120.2 46.6 29.2 44.8 25.0 57.3 96.4 29.4 62.9 66.4 30.0 [37] 24.1 14.8 56.6 102.4 117.5 90.4 37.2 79.6 27.8 17.1 26.6 16.3 [49] 41.4 48.9 24.1

Dear r-users,   I have saved data in word pad.  I would like to extract certain part of data only, for example   19710629 080000(PARTIAL) 39.3 at interval beginning 19701126 010326 19720629 080000(PARTIAL) 33.8 at interval beginning 19720517 144507 19730629 080000(PARTIAL) 32.2 at interval beginning 19720910 135747   and so on...   The original data set are given below:      

R 2.10 Windows 7 I am trying to plot three graphs on top of each other. I need to have the axises perfectly aligned. For some reason the ticks on the y axes are slightly off so they do not perfectly align. Can someone tell me how I can get the to overlay each other perfectly? I thought the yaxp parameter would solve my problem, but it does not. My data and code follows: >

Hi R Users, I am going to run a multiple linear regression with around 57 independent variables. Each time I run the model with just 11 variables, the results are reasonable. With increasing the number of independent variables more than 11, the coefficients will get ?NA? in the output. Is there any limitation for the number of independent variables in multiple linear regressions in R? I attached

Assuming the input data.frame, DF, is of the form shown reproducibly in the Note below, to convert the series to zoo or ts: library(zoo) # convert to zoo z <- read.zoo(DF) # convert to ts as.ts(z) # Note: DF <- structure(list(year = c(1980, 1981, 1982, 1983, 1984), cnsm = c(174, 175, 175, 172, 173), incm = c(53.4, 53.7, 53.5, 53.2, 53.3), with = c(60.3, 60.5, 60.2, 60.1, 60.7)),

Dear R-help list, I've got a data set in long format - each subject can have several (varying in number) measurements, with each record representing one measurement. I want to assign a sequence number to each measurement, starting at 1 for a person's first measurement. I can do this with the by function, but there must be an easier way. Here's my code - id is id number, age is

You can just use the same code that I provided before but now use your dataset. Like this df <- read.csv(file="data2.csv",header=TRUE) dates <- as.Date(paste(df$year,"-01-01",sep="")) myXts <- xts(df,order.by=dates) head(myXts) #The last command "head(myXts)" shows you the first few rows of the xts object year cnsm incm wlth