Displaying 20 results from an estimated 30000 matches similar to: "Merging two columns of a data frame"
2011 Apr 28
2
Subsetting Data
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
2012 Aug 02
1
Naive Bayes in R
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
2012 Aug 07
2
Re-grouping data in R
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
2012 Feb 09
1
Grouping together a time variable
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]
2012 Aug 08
1
Calculating percentages across multiple columns
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
2012 Aug 13
3
Using the effects package to plot logit probabilities
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,
2012 Jul 19
3
Removing values from a string
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",
2011 Dec 22
1
Error message with glm
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,
2012 Jul 05
2
Plotting the probability curve from a logit model with 10 predictors
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
2012 Sep 26
1
Specifying a response variable in a Bayesian network
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")
2011 Dec 21
1
Predicting a linear model for all combinations
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,
2011 Dec 15
1
Reordering a numeric variable
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
2011 Nov 17
1
Error When Installing the RODBC Package
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
2012 Sep 11
1
Plotting every probability curve
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
2013 Nov 05
2
Convert date column with two different structures
Let's say I have the following data frame and the date column has two
different ways in which date is presented. How can I use as.Date or the
lubridate package to have one date structure for the entire colum
df = data.frame(Date=c("5/1/13","8/1/13","9/1/13","Apr-10",
"Apr-11","Apr-12","Apr-13"))
It's
2011 Dec 16
1
Zellig Error Message
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
2011 Apr 25
2
Problem installing XML in Ubuntu 10.10
Hello folks,
Here's is info on what system I'm working on.
> sessionInfo()
R version 2.13.0 (2011-04-13)
Platform: i686-pc-linux-gnu (32-bit)
I'm trying to install the XML package. However, I end up with the following
error message.
> install.packages("XML")
....
checking for xml2-config... no
Cannot find xml2-config
ERROR: configuration failed for package ‘XML’
*
2011 Jun 06
1
Merge two columns of a data frame
I have the following data:
prefix <- c("cheap", "budget")
roots <- c("car insurance", "auto insurance")
suffix <- c("quote", "quotes")
prefix2 <- c("cheap", "budget")
roots2 <- c("car insurance", "auto insurance")
roots3 <- c("car insurance", "auto
2012 Feb 09
1
Finding all the coefficients for a logit model
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
2011 Jun 09
1
Using a function inside a function
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 =