Is there a way to let biggest value on X (out of X, Y, Z) decide value on V in R?
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I have a dataset (ft.mutate.topics) with five variables (four numeric are ft_technical, ft_performative, ft_procedural and ft_moral). The fifth is "topic_lab" and I would like for it to take on the name (as a character) related to the variable with the highest value of the four other.
The below produces a dataset similar to mine.
set.seed(1) Data <- data.frame( X = sample(1:10), Y = sample(1:10), Z = sample(1:10))
The I would like for a variable - V - to take on either "X", "Y", og "Z" for each observation corresponding to which of these three variables, that takes on the highest value - as an example for X, this is similar again:
if (Data$X > Data$Y & Data$X > Data$Z) Data$label <- "X" Warning message: In if (Data$X > Data$Y & Data$X > Data$Z) Data$label <- "X": the condition has length > 1 and only the first element will be used
In relation to my initial example I have tried the following with a combination of if-commands:
if (ft.mutate.topics$ft_technical > ft.mutate.topics$ft_performative & ft.mutate.topics$ft_technical > ft.mutate.topics$ft_procedural & ft.mutate.topics$ft_technical > ft.mutate.topics$ft_moral) ft.mutate.topics$topic_lab = "technical" if (ft.mutate.topics$ft_performative > ft.mutate.topics$ft_technical & ft.mutate.topics$ft_performative > ft.mutate.topics$ft_procedural & ft.mutate.topics$ft_performative > ft.mutate.topics$ft_moral) ft.mutate.topics$topic_lab = "performative" if (ft.mutate.topics$ft_procedural > ft.mutate.topics$ft_performative & ft.mutate.topics$ft_procedural > ft.mutate.topics$ft_technical & ft.mutate.topics$ft_procedural > ft.mutate.topics$ft_moral) ft.mutate.topics$topic_lab = "procedural" if (ft.mutate.topics$ft_moral > ft.mutate.topics$ft_performative & ft.mutate.topics$ft_moral > ft.mutate.topics$ft_procedural & ft.mutate.topics$ft_moral > ft.mutate.topics$ft_technical) ft.mutate.topics$topic_lab = "moral"
It says: "the condition has length > 1 and only the first element will be used" and substitutes the whole variable with "performative" because it is has the highest value in row 1. Anybody know what is up?
This seems simple. I will use a made up dataset, to adapt to yours should be easy.
nms <- sub("^ft_", "", names(ft)) ft$topic.lab <- apply(ft, 1, function(x) nms[which.max(x)])
This is a simulated dataset.
set.seed(1234) n <- 20 ft <- data.frame(ft_X = rnorm(n, 0, 2), ft_Y = rnorm(n, 0, 3), ft_Z = rnorm(n, 0, 4))
Multivariable Calculus: Early Transcendentals, d In Exercises 28–31, compute d; f(c(t)) at the given value of t. 28. f(x, y) = x + ex, cot) = (31 – 1, t”) at t = 2 29. f(x, y, z) = xz – y”, cot) = (t, to, 1 – 1) 30. at P in the direction V. 32. f(x, y) = x*x”, P = (3,-1), v = 2i +j 33. f(x, y, z) = z:x -xy”, P = (1 Let n + 0 be an integer and r an arbitrary constant. Use the Chain Rule to calculate The p value is calculated for a particular sample mean. Here we assume that we obtained a sample mean, x and want to find its p value. It is the probability that we would obtain a given sample mean that is greater than the absolute value of its Z-score or less than the negative of the absolute value of its Z-score.
You can use
max.col to get the column index for the maximum. You then subset the
names of the dataframe with this.
Data$V <- names(Data)[max.col(Data)]
This defaults to splitting ties at random.
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Here's a possible approach using
# create a fake input with random data set.seed(123) DF <- data.frame(ft_technical=sample(1:10,10), ft_performative=sample(1:10,10), ft_procedural=sample(1:10,10), ft_moral=sample(1:10,10)) # add the columns using apply and which.max mx <- DF[,c('ft_technical','ft_performative','ft_procedural','ft_moral')] DF$topic_lab <- c('technical','performative','procedural','moral')[apply(mx,1,which.max)]
> DF ft_technical ft_performative ft_procedural ft_moral topic_lab 1 3 10 9 10 performative 2 8 5 7 9 moral 3 4 6 6 6 performative 4 7 9 10 8 procedural 5 6 1 4 1 technical 6 1 7 8 3 procedural 7 10 8 3 4 technical 8 9 4 2 7 technical 9 2 3 1 5 moral 10 5 2 5 2 technical
The Encyclopædia Britannica, or, Dictionary of arts, sciences, and , T x = — r; : and these values being substi- x" + y + * tuted in equation (3), there results aV2 + Vy'1 + ctz'2= SECTION V.— ON THE CURVATURE OF SURFACES. Let the normal be taken as the axis of z, and consequently the tangent plane as the plane of xy. To determine the sections of greatest and least curvature. Solve an absolute value equation using the following steps: Get the absolve value expression by itself. Set up two equations and solve them separately.
Nature, PD $ ( x , u , v ) / Dx must vanish , not identically , but in virtue of the relation between x , y A . R . A Treatise on Differential Equations . f ( x , y , z ) = 0 , then this integral is Price 145 . we may take THE value of this useful text - book has been arising out of the conditional vanishlinear partial differential equation of the first The equation $$ y = 2x $$ expresses a relationship in which every y value is double the x value, and $$ y = x + 1 $$ expresses a relationship in which every y value is 1 greater than the x value. So what about a Table Of Values?
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Chapter 2 R basics, You have options of where to live and want to determine the safety of each A more explicit way to ask R to show us the value stored in a is using print like this: You can find out what the function expects and what it does by reviewing the So by not using the names, it assumes the arguments are x followed by base :. It’s often the case that I want to write an R script that loops over multiple datasets, or different subsets of a large dataset, running the same procedure over them: generating plots, or fitting a model, perhaps. I set the script running and turn to another task, only to come back later and find the  Related posts:R annoyances Your Data is Never the Right Shape Survive R
- Any statement will evaluate to many TRUE and FALSEs. Perhaps you need
all(your long statement)?
- It worked! Thank you - will be looking into the apply-family for sure!