search for: labouring

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",

hello, can someone tell me how to generate the means for a data frame that looks like this? My data frame has many more variables, but I won't bother you with those; these are the one's that I'm interested in. Needless to say, z is the variable in which I'm interested. I'd like to find out the mean score of z for NDP managers, Conservative managers and Liberal managers

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

Hello all, After a prolonged labour with a few re-issues of the tag and many pushes during the day, NUT v2.8.2 was found in a strange egg by a bunny hopping away from the April fools. All that - blessed under the shine of Polar lights in the middle of the icy North Ocean. A python module was released but slipped away with an earlier commit. Other artifacts should follow shortly. Jim Klimov

Hello all, After a prolonged labour with a few re-issues of the tag and many pushes during the day, NUT v2.8.2 was found in a strange egg by a bunny hopping away from the April fools. All that - blessed under the shine of Polar lights in the middle of the icy North Ocean. A python module was released but slipped away with an earlier commit. Other artifacts should follow shortly. Jim Klimov

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