search for: labourer's

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

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

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

Hi all, I am struggling with a strange issue in R that I have not encountered before and I am not sure how to resolve this. The model looks like this, with all irrelevant variables left out: LABOUR - a dummy variable NONLABOUR = 1 - LABOUR AGE - a categorical variable / factor VOTE - a dummy variable glm(VOTE ~ 0 + LABOUR + NONLABOUR + LABOUR : AGE + NONLABOUR : AGE,

Hi everyone, This is not so much of an R question as a statistics question. I currently work for the largest pre employment screening company in Canada. Upper management has noticed that noticed that usually a month or so before any big kind of economic shock happens, that our incoming files (requests for a background check) jump up or down. As the company statistician, they've asked me

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",

...out the mean score of z for NDP managers, Conservative managers and Liberal managers and then for a few other configurations. Ive played around with aggregate, tapply and by, but I can't get it to work. Cordially, Simon Kiss mydata=data.frame(cbind(x,y,z)) mydata$x=as.factor(sample(c("labourers", "salaried", "managers"), size=300, replace=TRUE)) mydata$y=as.factor(sample(c("NDP", "Green", "Liberal", "Conservative"), size=300, replace=TRUE)) mydata$z=as.numeric(sample(1:4, size=300, replace=TRUE)) **********************...

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:

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

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

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

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)

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 --

> 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

Hola queridos usuarios de R Estoy intentando hacer un gráfico de una variable en el tiempo para un conjunto de países usando R base, con el comando plot. Tengo 40 países y son 15 años. Lo quiero salvar como formato png, cada uno de ellos. Tengo el siguiente código, for (i in 1:length(countrylist)){ currcty <- countrylist[i] filename <-

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

> > > 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

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

Pero eso es para crear variables binarias tipo 0-1 si el pais pertence a un determinado grupo. Lo que quiero es crear una variable de tipo factor con esos 5 niveles, sabiendo que tengo en el dataframe una variable llamada Country, con el nombre del pais. Gracias On Wed, 6 Mar 2019 at 15:27, Xavier-Andoni Tibau Alberdi < xavitibau en gmail.com> wrote: > Buenas, > > Para ello yo uso

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