similar to: reference category for factor in regression

Displaying 20 results from an estimated 900 matches similar to: "reference category for factor in regression"

2010 May 11
1
(svy)glm and weights question
Hi all, I am running a set of logistic regressions, where we want to use some weights, and I am not sure whether what I am doing is reasonable or not. The dependent variable is turnout in an election - i.e. survey respondents were asked whether or not they voted. The percentage of those who say they voted is much higher than the actual turnout, probably due both to non-response bias and social
2013 Apr 09
1
sorting the VAR model output according to variable names??
I was wondering if one can have the coefficients of VAR model sorted according to variable names rather than lags. If you notice below, the output is sorted according to lags. >VAR(cbind(fossil,labour),p=2,type="const") VAR Estimation Results: ======================= Estimated coefficients for equation fossil: =========================================== Call: fossil = fossil.l1
2011 Nov 03
2
query about counting rows of a dataframe
Dear R users, I have got the following data frame, called my_df: gender day_birth month_birth year_birth labour 1 F 22 10 2001 1 2 M 29 10 2001 2 3 M 1 11 2001 1 4 F 3 11
2005 Dec 24
8
Prototype OOP example
Hi, Here is what I want to do: Labour = Class.create(); Labour.prototype = { initialize:function(name){ this.name = name; } } What I want to do is create a class called "Worker" which will inherit from "Labour", and the signature of "initialize" is "function(name, position)". May I ask what should do? Thank you all very much for the
2010 Jul 19
2
Grouping and stacking bar plot for categorical variables
Hi all, I have a series of cateogiral variables that look just like this: welfare=sample(c("less", "same", "more"), 1000, replace=TRUE) education=sample(c("less", "same", "more"), 1000, replace=TRUE) defence=sample(c("less", "same", "more"), 1000, replace=TRUE) egp=sample(c("salariat",
2011 Mar 31
1
Cluster analysis, factor variables, large data set
Dear R helpers, I have a large data set with 36 variables and about 50.000 cases. The variabels represent labour market status during 36 months, there are 8 different variable values (e.g. Full-time Employment, Student,...) Only cases with at least one change in labour market status is included in the data set. To analyse sub sets of the data, I have used daisy in the cluster-package to create
2019 Feb 19
2
Cambiar el formato de datos
Después del "gather()" puedes hacer un "arrange()" que es una ordenación. Y dentro de "arrange()" le indicas la variable por la que ordenas (no hacen falta comillas)... Lo ordenará alfabéticamente. Saludos, Carlos Ortega www.qualityexcellence.es El mar., 19 feb. 2019 a las 13:47, Antonio Rodriguez Andres (< antoniorodriguezandres70 en gmail.com>) escribió:
2019 Mar 06
2
Crear una variable tipo factor a partir de un vector de caracteres
Si lo que quiero es crear una variable llamada por ejemplo region (del tipo factor) con esos 5 valores On Wed, 6 Mar 2019 at 15:41, Xavier-Andoni Tibau Alberdi < xavitibau en gmail.com> wrote: > No, No. Fíjate en el Ifelse(condición, valor si positivo, valor si > negativo). > > Si, x %in% ca entonces el valor devuelto es "ca", un factor. En caso > negativo, vamos
2019 Sep 04
2
Plot. window error- Usando R base. Gráfico
No me sale error pero no me imprime la linea, ni ningun gráfico por pais. On Wed, 4 Sep 2019 at 18:42, neo <ericconchamunoz en gmail.com> wrote: > podría ser el problema el tipo de dato en X y el tipo de gráfico que > intentas hacer ? > > si Year es entero y estas pidiendo que el gráfico sea tipo "I", podría > ser que quizá eso te produce el error ? > > si
2019 Sep 04
2
Plot. window error- Usando R base. Gráfico
Lo que obtengo es dim(currcty) = NULL lo que hice es crear una lista de paises countrylist <- unique(length(eco_freedom2$Countries) Los datos son de esta forma head(eco_freedom2, 5) Year ISO_Code Countries SUMMARY.INDEX X1..Size.of.Government 641 2000 AGO Angola NA NA 601 2001 AGO Angola NA NA 561 2002
2019 Oct 10
2
how to automatically create the home directory
> > > If you're on a RedHat system with selinux (RHEL, CentOS, fedora), then > > it looks like > > <https://danwalsh.livejournal.com/69837.html> pam_oddjob_mkhomedir > > will create the home directories for you and also ensure that the > > correct selinux labels are applied. I have this on my todo list, as > > I'm currently using the ADUC
2019 Mar 25
2
Uso de merge
Jose Luis Column `Country` joining factors with different levels, coercing to character vector common_col_names <- intersect(names(sub_kei), names(knowledge)) > common_col_names [1] "Country" "Year" nrow(sub_kei) <- 132 nrow(knowledge) <- 3864 Tiene distinto numero de pais como de año, en el sub_kei aparecen 5 años y en el otro dataset (knowledge) datos anuales
2019 Mar 25
2
Uso de merge
Hola usuarios de R Estoy tratando de usar merge, para dos data frame, sin embargo al usarlo me da resultado correcto, en términos de emparejamiento de pais y año, pero lo que me hace es que el dataframe *y* me hace como un append por filas. Las variables comunes son país y año. Alguna sugerencia? combine = merge(sub_kei, knowledge, by = common_col_names, all.x = TRUE, all.y = TRUE) Saludos --
2019 Oct 10
2
how to automatically create the home directory
If you're on a RedHat system with selinux (RHEL, CentOS, fedora), then it looks like <https://danwalsh.livejournal.com/69837.html> pam_oddjob_mkhomedir will create the home directories for you and also ensure that the correct selinux labels are applied. I have this on my todo list, as I'm currently using the ADUC method, which is labour intensive. -- Mason On Thu, 10 Oct 2019 at
2005 Jun 29
3
IMP 4
Dear All, anyone installed IMP4 on Centos 4.1 yet, http://www.horde.org/imp/4.0/ if so what problems did you have, I have a webmail server I am going to upgrade to Centos 4.1 and Imp 4 from its current centos 3.3 implementation and imp3 This is going to be a labour intensive manual task, and I wondered if anyone else out there was mad enough to have tried it already.... Failing that just
2019 Sep 04
3
Plot. window error- Usando R base. Gráfico
Carlos Al especificar los limites, no me sale ningun error, aunque no consigue graficar, nada, # Look at the time series for each country for the time period, for instance GDPPC for (i in 1:length(countrylist)){ currcty <- countrylist[i] filename <- paste("index",currcty,".png",sep="") png(filename,width=800,height=600)
2010 May 27
2
help calculating variable based on factor level of another
Dear colleagues, I want to calculate the value of x2 based on the value of x1. x1 is a factor with three separate levels. I want to make sure that missing values remain as NA in X2, but non-missing values take on a value of either 0 or 1 dependending on the value in x1. This is the code I'm working with...Can any one help? I've seen some other requests on a topic like this, but not
2019 Feb 19
2
Cambiar el formato de datos
> gather(pobla, key = year, value = totpop, year60:year63) Country year totpop 1 Afghanistan year60 8996351 2 Albania year60 1608800 3 Algeria year60 11124888 4 Andorra year60 13411 Gracias Carlos Antonio On Tue, 19 Feb 2019 at 12:54, Carlos Ortega <cof en qualityexcellence.es> wrote: > Sí, tienes varias formas. > > Mira la función
2005 Jan 26
2
reshape (a better way)
Hi, I am using the NLSY79 data (longitudinal data from the Bureau of labour stats in the US). The extractor exctracts this data in a "wide" format and I need to reshape it into a long format. What I am doing right now is to do it in chuncks for each and evry variable that is varying and then I merge the data together. This is taking a long time. my question is: How do I specify that
2017 Jun 19
2
Problema con Histograma con porcentajes usando ggplot
Creo que esto me da para DK, y luego veré como aplicar el barplot ess %>% filter(cntry %in% c("DK")) %>% count (stflife) %>% mutate (freq = (n /sum(n)*100))%>% print 2017-06-18 19:01 GMT-05:00 Antonio Rodriguez Andres < antoniorodriguezandres70 en gmail.com>: > He conseguido el total para un país, pero no me deja usar percent = > count() /sum(count),