similar to: simple question on data frames assignment

ifelse is vectorised, so just use that without the loop. colordata$response <- ifelse(colordata$color == 'blue', 1, 0) David On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com> wrote: > Hi I'm not sure how to ask this, but its a very easy question to answer for > an R person. > > What is an easy way to check for a column value and then assigne a

Hello, Or even simpler, without ifelse, colordata$response <- colordata$color == 'blue' + 0 Hope this helps, Rui Barradas ? Citando David Barron <dnbarron at gmail.com>: > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0) > > David > > On 7 April 2016 at 12:41, Michael

Thaks so much! And how would you incorporate lapply() here? On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> wrote: > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0) > > David > > On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com> wrote: >

If you are not using an anonymous function and say you had written the function out The below gives me the error > 'f(colordata2$color1)' is not a function, character or symbol' But then how is the anonymous function working? f <- function(col){ ifelse(col == 'blue', 1, 0) } responses <- lapply(colordata2[ -1 ], f(colordata2$color1) )

== is also vectorised, and you're better off with TRUE and FALSE rather than 1 and 0, so I'd recommend: colordata$response <- colordata$color == 'blue' Hadley On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> wrote: > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color ==

Fyi, This statement returned the following error 'Error in "Yes" + 0 : non-numeric argument to binary operator' On Thu, Apr 7, 2016 at 8:43 AM, <ruipbarradas at sapo.pt> wrote: > Hello, > > Or even simpler, without ifelse, > > colordata$response <- colordata$color == 'blue' + 0 > > Hope this helps, > > Rui Barradas > > >

Lapply is not a vectorized function. It is compact to read, but it would not be worth using for this calculation. However, if your data frame had multiple color columns in your data frame that you wanted to make responses for then you might want to use lapply as a more compact version of a for loop to repeat this operation. colordata2 <- data.frame(id = c(1,2,3,4,5), color1 =

lapply(colordata2[ -1 ], f ) When you put the parentheses on, you are calling the function yourself before lapply gets a chance. The error pops up because you are giving a vector of numbers (the answer f gave you) to the second argument of lapply instead of a function. -- Sent from my phone. Please excuse my brevity. On April 7, 2016 7:31:18 AM PDT, Michael Artz <michaeleartz at

Why am I better off with true and false? On Thu, Apr 7, 2016 at 8:41 AM, Hadley Wickham <h.wickham at gmail.com> wrote: > == is also vectorised, and you're better off with TRUE and FALSE > rather than 1 and 0, so I'd recommend: > > colordata$response <- colordata$color == 'blue' > > Hadley > > On Thu, Apr 7, 2016 at 6:52 AM, David Barron

Hi The question does not make much sense so as your code. Maybe you shall spend some time with R tutorials. 1. lapply or sapply is basically hidden cycle 2. function shall return something, yours does not So if you want some binary outcome from a vector you can use e.g. f <- function(vector, token) { response <- vector==token response } So with your colordata >

That didn't work Jim! Thanks anyway On Mon, Apr 18, 2016 at 9:02 PM, Jim Lemon <drjimlemon at gmail.com> wrote: > Hi Michael, > At a guess, try this: > > iqr<-function(x) { > return(paste(round(quantile(x,0.25),0),round(quantile(x,0.75),0),sep="-") > } > > .col3_Range=iqr(datat$tenure) > > Jim > > > > On Tue, Apr 19, 2016 at

I need the output to have groups and the probability any given record in that group then has of being in the response class. Just like my email in the beginning i need the output that looks like if A and if B and if C then %77 it will be D. The examples you provided are just simply not similar. They are different and would take interpretation to get what i need. On Apr 14, 2016 1:26 AM,

I still need the output to match my requiremnt in my original post. With decision rules "clusters" and probability attached to them. The examples are sort of similar. You just provided links to general info about trees. Sent from my Verizon, Samsung Galaxy smartphone<div> </div><div> </div><!-- originalMessage --><div>-------- Original message

Ok is there a way to do it with decision tree? I just need to make the decision rules. Perhaps I can pick one of the trees used with Random Forrest. I am somewhat familiar already with Random Forrest with respective to bagging and feature sampling and getting the mode from the leaf nodes and it being an ensemble technique of many trees. I am just working from the perspective that I need

Since you only have 3 predictors, each categorical with a small number of categories, you can use expand.grid to make a data.frame containing all possible combinations and give that the predict method for your model to get all possible predictions. Something like the following untested code. newdata <- expand.grid( Humidity = levels(Humidity), #(High, Medium,Low)

If you show us, not just tell us about, a self-contained example someone might show you a non-hacky way of getting the job done. (I don't see an argument to plyr::ddply called 'transform'.) Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Apr 19, 2016 at 12:18 PM, Michael Artz <michaeleartz at gmail.com> wrote: > Oh thanks for that clarification Bert! Hope you enjoyed

Tjats great that you are familiar and thanks for responding. Have you ever done what I am referring to? I have alteady spent time going through links and tutorials about decision trees and random forrests and have even used them both before. Mike On Apr 13, 2016 5:32 PM, "Sarah Goslee" <sarah.goslee at gmail.com> wrote: It sounds like you want classification or regression trees.

Again, IQR returns two both a .25 and a .75 value and it failed, which is why I didn't use it before. Also, the first function just returns tha same value repeating. Since they are the same, before the second call, using the mode function is just a way to grab one value. I could have used average, min, max, they all would have returned the same thing. Mike On Tue, Apr 19, 2016 at 7:24 PM,

Well, instead of your functions try: Mode <- function(x) { tabx <- table(x) tabx[which.max(tabx)] } and use R's IQR function instead of yours. ... so I still don't get why you want to return a character string instead of a value for the IQR; and the mode of a sample defined as above is generally a bad estimator of the mode of the distribution. To say more than that would

> That didn't work Jim! It always helps to say how the suggestion did not work. Jim's function had a typo in it - was that the problem? Or did you not change the call to ddply to use that function. Here is something that might "work" for you: library(plyr) data <- data.frame(groupColumn=rep(1:5,1:5), col1=2^(0:14)) myIqr <- function(x) {