search for: colordata

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 new column a value based on that old column value? For example, Im doing colordata <- data.frame(id = c(1,2,3,4,5), color = c("blue", "red", "green", "blue", "orange")) for (i in 1:nrow(colordata)){ colordata$response[i] <- ifelse(colordata[i,"color"] == "blue", 1, 0) } which works, but I don't...

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...

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) > > Dav...

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 > > > Citando David Barron <dnbarron at gmail.com>: > > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == &...

== 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 == 'blue', 1, 0) > > Davi...

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: > >> Hi I'm not sure how to ask this, but its a very easy question to answer >> for >> an R person. >&...

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 <dnbarron at gmail.com> wrote: > > ifelse is vectorised, so just use that without the loop. > > > > colordata$response <- ifelse(colordata$color == ...

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) ) 'f(colordata2$color1)' is not a function, character or...

...ction. 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 = c("blue", "red", "green", "blue", "orange"), color2 = c("orange", "green", "blue", "red", "red")) responses <- lapply( colordata2[ -1 ], function(col)...

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 bre...

...end 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 > colordata id color 1 1 blue 2 2 red 3 3 green 4 4 blue 5 5 orange You can get this > colordata$result <- f(colordata$color, "blue") > colordata id color result 1 1 blue TRUE 2 2 red FALSE 3 3 green FALSE 4 4 blue TRUE 5 5 orange FALS...

...line 2), # converts the image to grayscale data (line 3), and calculates the autocorrelation curve # for offset distances of one pixel to 99 pixels (lines 7–10). This code can be used to # perform steps (4) and (6) listed above in the Summary of Steps to Calculate Grain # Size of Natural Sediment. ColorData = imread(‘ImageName.jpg’); x <- read.jpeg("sed2.jpg")# opens the image imshow(ColorData); plot(x) # plot image data = double(rgb2gray(ColorData)); x=rgb2grey(x) # transform in grayscale, the image ; # example of a similar data matrix: data = double(0.54878431372549,0.468,0.553411764705...

...be grateful for anybody who might help to produce the following plot (the code for matlab is below) using the "rgl" package of R? [t,r] = meshgrid(linspace(0,2*pi,361),linspace(-4,4,101)); [x,y] = pol2cart(t,r); P = peaks(x,y); figure('color','white'); polarplot3d(P,'colordata',gradient(P)); view([-18 72]); You may see the code and plot output at: https://plus.google.com/u/0/109789409461372488563/posts/YfcrsMhkjyf Many Thanks A. [[alternative HTML version deleted]]