similar to: Convert date column with two different structures

I'm trying to run a logit model and plot the probability curve for a number of the important predictors. I'm trying to do this with the Effects package. df=data.frame(income=c(5,5,3,3,6,5), won=c(0,0,1,1,1,0), age=c(18,18,23,50,19,39), home=c(0,0,1,0,0,1)) str(df) md1 = glm(factor(won) ~ income + age + home, data=df,

So I have the following data frame and I want to know how I can remove all "NA" values from each string, and also remove all "|" values from the START of the string. So they should something like "auto|insurance" or "auto|insurance|quote" one = data.frame(keyword=c("|auto", "NA|auto|insurance|quote", "NA|auto|insurance",

I have a logit model with about 10 predictors and I am trying to plot the probability curve for the model. Y=1 = 1 / 1+e^-z where z=B0 + B1X1 + ... + BnXi If the model had only one predictor, I know to do something like below. mod1 = glm(factor(won) ~ as.numeric(bid), data=mydat, family=binomial(link="logit")) all.x <- expand.grid(won=unique(won), bid=unique(bid)) y.hat.new

I'm working on a logistic regression in R with the car package but keep getting the following error message. It's only and warning and not an error, but I'm just not sure how to resolve the issues. glm.fit: algorithm did not converge glm.fit: fitted probabilities numerically 0 or 1 occurred d1 = data.frame(mwin=c(mwin), mbid=c(mbid)) m1 = zelig(mwin ~ mbid, data=d1,

I'm developing a naive bayes in R. I have the following data and am trying to predict on returned (class). dat = data.frame(home=c(0,1,1,0,0), gender=c("M","M","F","M","F"), returned=c(0,0,1,1,0)) str(dat) dat$home <- as.factor(dat$home) dat$returned <- as.factor(dat$returned) library(e1071) m <- naiveBayes(returned ~ ., dat) m

Lets say I have a linear model and I want to find the average expented value of the dependent variable. So let's assume that I'm studying the price I pay for coffee. Price = B0 + B1(weather) + B2(gender) + ... What I'm trying to find is the predicted price for every possible combination of values in the independent variables. So Expected price when: weather=1, gender=male weather=1,

I have the following variable, time, which is a character variable and it's structured as follows. > head(as.character(dat$time), 30) [1] "00:00:01" "00:00:16" "00:00:24" "00:00:25" "00:00:25" "00:00:40" "00:01:50" "00:01:54" "00:02:33" "00:02:43" "00:03:22" [12]

I have the following data and am trying to find the percentage of bid values purchased for that price. So let's say I have a bid of 5 and it's sold 2 times for $3 and $5. Since the original bid was $5, the percentage of times that that bid value results in a sold purchase AT that specific bid level was 1/3 because of the three time where the bid was three, it ended up being sold for $5

I'm running a linear model in R using the car package. I have a variable education, which i have recoded and regrouped to my wishes. However, R seems to place each element of that variable in alphabetical order. When I am running the model, don't I need the model order from lowest to highest to make an inference that a one unit change in one variable produced a one unit change in

I don't have a logistic regression model and am trying to generate probability curves for all possible combinations of the variables. My logit model has 5+ variables, and I want to draw curves for every scenario. See code below. When home_owner is 0 and 1, I want curves. The same goes for all other variables categories, so that I have permutations for all possible combinations. I've

I have a data frame with a column of values that I want to bucket (group) into specific levels. > str(dat)'data.frame': 3678 obs. of 39 variables: $ id : int 23 76 129 156 166 180 200 214 296 344 ... $ final_purchase_amount : Factor w/ 32 levels "\\N","1082","1109",..: 1 1 1 1 1 1 1 1 1 1 ... So I ran the following to

I'm running R in Ubuntu 10.10 and am trying to install the RODBC package. However, I get the following error message: ERROR: configuration failed for package ‘RODBC’ * removing ‘/home/amathew/R/i686-pc-linux-gnu-library/2.13/RODBC’ The downloaded packages are in ‘/tmp/RtmpekzPOQ/downloaded_packages’ Warning message: In install.packages() : installation of package 'RODBC' had

I'm using the subset() function in R. dat <- data.frame(one=c(6,7,8,9,10), Number=c(5,15,13,1,13)) subset(dat, Number >= 10) However, I want to find the number of all rows who meet the Number>=10 condition. I've done this in the past with something like colSums or rowSums or another similar function. But I don't remember how to get the number of elements which meet that

Hi,? I have a data set with various date formats in one column and not sure how to unify it.Here is a few formats: 02091702/22/170221201703/17/160015-08-239/2/1500170806May-2-201522-March-2014 I tried parse_date_time from lubridate library but it failed.Thanks so much.?Best,Farnoosh [[alternative HTML version deleted]]

I'm trying to teach myself about Bayesian Networks and am working with the following data and the bnlearn package. I understand the conceptual aspects of BNs, but I'm not sure how to specify the response variables in R when constructing a dag plot. I've cecked ?hc and done numerous google searches without luck. Can anyone help? library("bnlearn")

I'm trying to calculate predicted probabilities in R with Zelig and keep getting the following error. Can anyone help? > x.low <- setx(mod, type=1)Error in dta[complete.cases(mf), names(dta) %in% vars, drop = FALSE] : incorrect number of dimensions When I ran the model, I ran everything but the explanatory variable as a numeric variable. Now, I'm trying everything and no

Hi folks, I have a simple question that I just can't solve. I'm trying to merge two columns in my data frame. > sessionInfo() R version 2.13.0 (2011-04-13) Platform: i686-pc-linux-gnu (32-bit) > head(dat) Year Month Number 2002 Jan 0 2002 Feb 0 2002 March 0 2002 April 1 2002 May 0 2002 June 0 I tried to do the following, but it

Let's say I have a variable, day, which is saved as a factor with 7 levels, and I use it in a logistic regression model. I ran the model using the car package in R and printed out the results. mod1 = glm(factor(status1) ~ factor(day), data=mydat, family=binomial(link="logit")) print(summary(mod1)) The result I get is: Coefficients: Estimate Std. Error z value

I'm trying to run a function inside a function but get an error message. lst <- list(roots = c("car insurance", "auto insurance"), roots2 = c("insurance"), prefix = c("cheap", "budget"), prefix2 = c("low cost"), suffix = c("quote", "quotes"), suffix2 = c("rate", "rates"), suffix3 =

Hey, Are all the dates connected? So no comma or space btw? Regards, Christoph > On 29 Jun 2017, at 2:02 pm, Farnoosh Sheikhi via R-help <r-help at r-project.org> wrote: > > Hi, > I have a data set with various date formats in one column and not sure how to unify it.Here is a few formats: > 02091702/22/170221201703/17/160015-08-239/2/1500170806May-2-201522-March-2014