r-factor

问题 I find manipulating factor variables in R unduly complicated. Things I frequently want to do when cleaning factors include: Resorting levels – not just to set a reference category, but also put all levels in a logical (non-alphabetical order) for summary tables. x <- factor(x, levels = new.order) Recode / rename factor levels – to simplify names and/or collapse multiple categories into one group. For one-to-one recoding levels(x) <- new.levels(x) or plyr::revalue , see here or here for

问题 I have an R dataframe where one of the columns is a factor whose levels have an implicit ordering. How can I convert the factor levels to specific integers in the following manner: "Strongly disagree" --> 1 "Somewhat disagree" --> 2 "Neutral" --> 3 "Somewhat agree" --> 4 "Strongly agree" --> 5 For example, here is my data frame: agree <- c("Strongly agree", "Somewhat disagree", "Somewhat agree", "Neutral", "Strongly agree", "Strongly disagree", "Neutral") age <- c(41, 35, 29, 42, 31, 22, 58)

问题 I would really appreciate your help in this question. I have the following dataset and I would like to create a new variable which would contain the standardized values (z distribution) per level of a given factor variable. x <- data.frame(gender = c("boy","boy","boy","girl","girl","girl"), values=c(1,2,3,6,7,8)) x gender values 1 boy 1 2 boy 2 3 boy 3 4 girl 6 5 girl 7 6 girl 8 My aim is to create one new variable which will contain the z-values calculated separately for each factor level

问题 labs = letters[3:7] vec = rep(1:5,2) How do I get a factor whose levels are "c" "d" "e" "f" "g" ? 回答1: You can do something like this: labs = letters[3:7] vec = rep(1:5,2) factorVec <- factor(x=vec, levels=sort(unique(vec)), labels = c( "c", "d", "e", "f", "g")) I have sorted the unique(vec) , so as to make results consistent. unique() will return unique values based on the first occurrence of the element. By specifying the order, the code becomes more robust. Also by specifying the levels

问题 I have the following setup. df <- data.frame(aa = rnorm(1000), bb = rnorm(1000)) apply(df, 2, typeof) # aa bb #"double" "double" apply(df, 2, class) # aa bb #"numeric" "numeric" Then I try to convert one of the columns to "factor". But as you can see below, I am not getting any "factor" type or classes. Am I doing anything wrong ? df[, 1] <- as.factor(df[, 1]) apply(df, 2, typeof) # aa bb #"character" "character" apply(df, 2, class) # aa bb #"character" "character" 回答1: Sorry I felt my

问题 I have the following setup. df <- data.frame(aa = rnorm(1000), bb = rnorm(1000)) apply(df, 2, typeof) # aa bb #"double" "double" apply(df, 2, class) # aa bb #"numeric" "numeric" Then I try to convert one of the columns to "factor". But as you can see below, I am not getting any "factor" type or classes. Am I doing anything wrong ? df[, 1] <- as.factor(df[, 1]) apply(df, 2, typeof) # aa bb #"character" "character" apply(df, 2, class) # aa bb #"character" "character" 回答1: Sorry I felt my

问题 I have a function that works as expected until I subset it. The function, plotCalendar() is my attempt at a Calendar Heat Map using ggplot2 with facets. The y-axis order is important because it is for the "WeekOfMonth" - when the order is reversed the data viz does not look like a calendar. The code is below, first the calling code, then the function to generate some data - generateData(), then the plot function - plotCalendar() The code works as expected when I used df for the data but when

问题 Not able to fix the below error for the below logistic regression training=(IBM$Serial<625) data=IBM[!training,] dim(data) stock.direction <- data$Direction training_model=glm(stock.direction~data$lag2,data=data,family=binomial) ###Error### ---- Error in eval(family$initialize) : y values must be 0 <= y <= 1 Few rows from the data i am using X Date Open High Low Close Adj.Close Volume Return lag1 lag2 lag3 Direction Serial 1 28-11-2012 190.979996 192.039993 189.270004 191.979996 165.107727

问题 Not able to fix the below error for the below logistic regression training=(IBM$Serial<625) data=IBM[!training,] dim(data) stock.direction <- data$Direction training_model=glm(stock.direction~data$lag2,data=data,family=binomial) ###Error### ---- Error in eval(family$initialize) : y values must be 0 <= y <= 1 Few rows from the data i am using X Date Open High Low Close Adj.Close Volume Return lag1 lag2 lag3 Direction Serial 1 28-11-2012 190.979996 192.039993 189.270004 191.979996 165.107727

问题 Not able to fix the below error for the below logistic regression training=(IBM$Serial<625) data=IBM[!training,] dim(data) stock.direction <- data$Direction training_model=glm(stock.direction~data$lag2,data=data,family=binomial) ###Error### ---- Error in eval(family$initialize) : y values must be 0 <= y <= 1 Few rows from the data i am using X Date Open High Low Close Adj.Close Volume Return lag1 lag2 lag3 Direction Serial 1 28-11-2012 190.979996 192.039993 189.270004 191.979996 165.107727