newbie R coder here. I have a stacked bar chart in base R that I'd like to reorder numerically by question type (Question 1 Pre, Question 1 Post, Question 2 Pre, Question 2 Post, etc.)
It's probably a fairly simple fix but I can't seem to get the reorder function to work. The other questions on reordering don't quite get to my solution. Maybe reorder isn't the right way to go about it?
Attached my graph and base code. Thank you so much! I appreciate your kind help.
if(!require(psych)){install.packages("psych")}
if(!require(likert)){install.packages("likert")}
library(readxl)
setwd("MSSE 507 Capstone Data Analysis/")
read_xls("ProcessDataMSSE.xls")
Data = read_xls("ProcessDataMSSE.xls")
str(Data) # tbl_df, tbl, and data.frame classes
### Change Likert scores to factor and specify levels; factors because numeric values are ordinal
Data <- Data[, c(3:26)] # Get rid of the other columns! (Drop multiple columns)
Data$`1Pre` <- factor(Data$`1Pre`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`1Post` = factor(Data$`1Post`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`2Pre` <- factor(Data$`2Pre`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`2Post` = factor(Data$`2Post`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`3Pre` <- factor(Data$`3Pre`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`3Post` = factor(Data$`3Post`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`4Pre` <- factor(Data$`4Pre`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`4Post` = factor(Data$`4Post`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`5Pre` <- factor(Data$`5Pre`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`5Post` = factor(Data$`5Post`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`6Pre` <- factor(Data$`6Pre`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`6Post` = factor(Data$`6Post`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`7Pre` <- factor(Data$`7Pre`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`7Post` = factor(Data$`7Post`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`8Pre` <- factor(Data$`8Pre`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`8Post` = factor(Data$`8Post`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`9Pre` <- factor(Data$`9Pre`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`9Post` = factor(Data$`9Post`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`10Pre` <- factor(Data$`10Pre`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`10Post` = factor(Data$`10Post`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`11Pre` <- factor(Data$`11Pre`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`11Post` = factor(Data$`11Post`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`12Pre` <- factor(Data$`12Pre`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data$`12Post` = factor(Data$`12Post`,
levels = c("1", "2", "3", "4"),
ordered = TRUE)
Data <- factor(Data,levels=Data[3:26])
Data
### Double check the data frame
library(psych) # Loads psych package
headTail(Data) # Displays last few and first few data
str(Data) # Shows structure of an object (observations and variables, etc.) - in this case, ordinal factors with 4 levels (1 through 4)
summary(Data) # Summary of the number of times you see a data point
Data$`1Pre` # This allows us to check how many data points are really there
str(Data)
### Remove unnecessary objects, removing the data frame in this case (we've converted that data frame into a table with the read.table function above)
library(likert)
Data <- as.data.frame(Data) # Makes the tibble a data frame
likert(Data) # This will give the percentage responses for each level and group
Result = likert(Data)
summary(Result) # This will give the mean and SD
plot(Result,
main = "Pre and Post Treatment Percentage Responses",
ylab="Questions",
type="bar")
I largely agree with #DzimitryM 's solution. It is unclear to me, however whether this really works. In my solution, I need to use the items variable of the data.frame, not the data.frame as such. There is some comment in the code below (at the bottom) highlighting this.
Anyway this is the reason I made a working example with executable code.
I am aware of the fact, that it may be improved; my focus was on executability.
library(likert)
### mimic some of your data, with 'accepted' naming
Data <- data.frame(
C01Pre = as.character(c( rep(1, 10), rep(2, 60), rep(3, 25), rep(4, 5) )),
C01Post = as.character(c( rep(1, 25), rep(2, 52), rep(3, 21), rep(4, 2) )),
C02Pre = as.character(c( rep(1, 25), rep(2, 68), rep(3, 5), rep(4, 2) )),
C02Post = as.character(c( rep(1, 30), rep(2, 53), rep(3, 13), rep(4, 4) )),
C03Pre = as.character(c( rep(1, 20), rep(2, 52), rep(3, 25), rep(4, 3) )),
C03Post = as.character(c( rep(1, 20), rep(2, 39), rep(3, 35), rep(4, 6) ))
)
### coerce to ordered factor
Data$C01Pre <- factor(Data$C01Pre, levels = c("1", "2", "3", "4"), ordered = TRUE)
Data$C01Post <- factor(Data$C01Post, levels = c("1", "2", "3", "4"), ordered = TRUE)
Data$C02Pre <- factor(Data$C02Pre, levels = c("1", "2", "3", "4"), ordered = TRUE)
Data$C02Post <- factor(Data$C02Post, levels = c("1", "2", "3", "4"), ordered = TRUE)
Data$C03Pre <- factor(Data$C03Pre, levels = c("1", "2", "3", "4"), ordered = TRUE)
Data$C03Post <- factor(Data$C03Post, levels = c("1", "2", "3", "4"), ordered = TRUE)
Result = likert(Data)
### show the "natural" order when processed by likert()
summary(Result)
# Item low neutral high mean sd
# 6 C03Post 59 0 41 2.27 0.8510837
# 1 C01Pre 70 0 30 2.25 0.7017295
# 5 C03Pre 72 0 28 2.11 0.7506899
# 2 C01Post 77 0 23 2.00 0.7385489
# 4 C02Post 83 0 17 1.91 0.7666667
# 3 C02Pre 93 0 7 1.84 0.5983141
plot(Result,
group.order = names(Result$items)) ## this is the key!
## difference with other answer is:
## names of the "items" variable of the df
## not the data.frame itself
This results in the following graph:
Grouping option can be added to plot() in order to get the plot, that is ordered by the column names of the initial dataset:
plot(Result,
group.order = names(Data),
type="bar")
Related
I am trying to make a model using the lapply function where lapply indexes through each column of response variables and creates a linear model using the predictor variables. I am then each individual linear model to the stepAIC funciton and then to the stepVIF function after that. I can make this work in a dataset with no na values, however my dataset is full of na values which is giving me a variable length differ error when I pass the linear models to the stepAIC function.
This is what I have tried so far. I made the miultiple.func variable in an attempt to remove na values from column at a time buit it does not work and I think that it would end up removing all of the rows of data except for the fourth column due to how the complete.cases() works. This is why I only want to remove the Na values from one column at a time of the response variables (the column being called in the model), and all of the Na values from the predictor variables (col d, e and f).
data_dummy <- data.frame(first_column = c("A", "B", "c", "d", "e", "f"),
second_column = c("1", "Na", "3", "4", "5", "6"),
third_column = c("Na", "7", "3", "Na", "2", "6"),
fourth_column = c("5", "8", "3", "4", "5", "1"),
fith_column = c("2", "Na", "3", "na", "2", "6"),
sixth_column = c("5", "9", "3", "4", "na", "1")
)
g <- 3
multiple.func <- function(g) {
c(data34[complete.cases(data_dummy[[,c(g)]]),], lm(reformulate(names(data34)[4:7], response=names(data_dummy)[g]), data_dummy)) #trying to remove NA
}
full.model <- lapply(data_dummy, multiple.func)
step.model <- lapply(full.model, function(x)MASS::stepAICIC(x, direction = "both", trace = FALSE)) #Fit stepwise regression model
stepmod3 <- lapply(step.model, function(x)pedometrics::stepVIF(model = x, threshold = 10, verbose = TRUE))
I'm looking for a way to generate multiple 3-level variables from an older 5-level variable, while keeping the old variables.
This is how it is now:
structure(list(Quesiton1 = c("I", "5", "4", "4"), Question2 = c("I",
"5", "4", "4"), Question3 = c("I", "3", "2", "4")), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -4L))
I would like this:
structure(list(Quesiton1 = c("I", "5", "4", "4"), Question2 = c("I",
"5", "4", "4"), Question3 = c("I", "3", "2", "4"), Question1_3l = c("NA",
"3", "3", "3"), Question2_3l = c("NA", "3", "3", "3"), Question3_3l = c("NA",
"2", "1", "3")), row.names = c(NA, -4L), class = c("tbl_df",
"tbl", "data.frame"))
I have this code to recode the 5-level variable
df2 %>%
mutate_at(vars(Question1, Question2, Question3), recode,'1'=1, '2'=1, '3'=3, '4'=5, '5'=5, 'l' = NA)
But what I want to do is to keep the old variable and generate the 3 level variable into something like Question1_3l, Question2_3l, Question3_3l.
It shouldn't be too difficult. In Stata it looks something like this:
foreach i of varlist ovsat-not_type_number {
local lbl : variable label `i'
recode `i' (1/2=1)(3=2)(4/5=3), gen(`i'_3l)
}
Thank you.
Not the most elegant, not the fastest (but still pretty fast), not the most idiomatic, but this does what you want (I think) and should be easy to read and customize.
dt <- structure(list(Quesiton1 = c("I", "5", "4", "4"),
Question2 = c("I", "5", "4", "4"),
Question3 = c("I", "3", "2", "4")),
class = c("tbl_df", "tbl", "data.frame"),
row.names = c(NA, -4L))
#transfor your data into a data.table
setDT(dt)
#define the names of the columns that you want to recode
vartoconv <- names(dt)
#define the names of the recoded columns
newnames <- paste0(vartoconv, "_3l")
#define an index along the vector of the names of the columns to recode
for(varname_loopid in seq_along(vartoconv)){
#identify the name of the column to recode for each iteration
varname_loop <- vartoconv[varname_loopid]
#identify the name of the recoded column for each iteration
newname_loop <- newnames[varname_loopid]
#create the recoded variable by using conditionals on the variable to recode
dt[get(varname_loop) %in% c(1, 2), (newname_loop) := 1]
dt[get(varname_loop) == 3, (newname_loop) := 2]
dt[get(varname_loop) %in% c(4, 5), (newname_loop) := 3]
}
Try:
library(tidyverse)
library(stringr)
df2 <- replicate(6, sample(as.character(1:5), 50, replace = TRUE), simplify = "matrix") %>%
as_tibble(.name_repair = ~str_c("Question", 1:6))
df2 %>%
mutate_at(vars(Question1:Question3),
~case_when(.x %in% c('1', '2') ~ 1L, # 1L means integer 1
.x %in% c('3') ~ 3L,
.x %in% c('4', '5') ~ 5L,
TRUE ~ as.integer(NA)))
I am working on large data sets, for which i have written a code to perform row by row operation on a data frame, which is sequential. The process is slow.
I am trying to perform the operation using parallel processing to make it fast.
Here is code
library(geometry)
# Data set - a
data_a = structure(c(10.4515034409741, 15.6780890052356, 12.5581992918563,
9.19067944250871, 14.4459166666667, 11.414, 17.65325, 12.468,
11.273, 15.5945), .Dim = c(5L, 2L), .Dimnames = list(c("1", "2",
"3", "4", "5"), c("a", "b")))
# Data set - b
data_b = structure(c(10.4515034409741, 15.6780890052356, 12.5581992918563,
9.19067944250871, 14.4459166666667, 11.3318076923077, 13.132273830156,
6.16003995082975, 11.59114820435, 10.9573192090395, 11.414, 17.65325,
12.468, 11.273, 15.5945, 11.5245, 12.0249, 6.3186, 13.744, 11.0921), .Dim = c(10L,
2L), .Dimnames = list(c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10"), c("a",
"b")))
conv_hull_1 <- convhulln( data_a, options = "FA") # Draw Convex Hull
test = c()
for (i in 1:nrow(data_b)){
df = c()
con_hull_all <- inhulln(conv_hull_1, matrix(data_b[i,], ncol = 2))
df$flag <- ifelse(con_hull_all[1] == TRUE , 0 , ifelse(con_hull_all[1] == FALSE , 1, 2))
test <- as.data.frame(rbind(test, df))
print(i)
}
test
Is there any way to parallelize row wise computation?
As you can observe, for small datasets the computational time is really low, but as soon as i increase the data size, the computation time increases drastically.
Can you provide solution with the code.
Thanks in advance.
You could take advantage of the parameter to the inhulln function. This allows more than one row of points to be tested to be passed in.
I've tried the code below on a 320,000 row matrix that I made from the original data and it's quick.
library(geometry)
library(dplyr)
# Data set - a
data_a = structure(
c(
10.4515034409741,
15.6780890052356,
12.5581992918563,
9.19067944250871,
14.4459166666667,
11.414,
17.65325,
12.468,
11.273,
15.5945
),
.Dim = c(5L, 2L),
.Dimnames = list(c("1", "2",
"3", "4", "5"), c("a", "b"))
)
# Data set - b
data_b = structure(
c(
10.4515034409741,
15.6780890052356,
12.5581992918563,
9.19067944250871,
14.4459166666667,
11.3318076923077,
13.132273830156,
6.16003995082975,
11.59114820435,
10.9573192090395,
11.414,
17.65325,
12.468,
11.273,
15.5945,
11.5245,
12.0249,
6.3186,
13.744,
11.0921
),
.Dim = c(10L,
2L),
.Dimnames = list(c(
"1", "2", "3", "4", "5", "6", "7", "8", "9", "10"
), c("a",
"b"))
)
conv_hull_1 <- convhulln( data_a, options = "FA") # Draw Convex Hull
#Make a big data_b
for (i in 1:15) {
data_b = rbind(data_b, data_b)
}
In_Or_Out <- inhulln(conv_hull_1, data_b)
result <- data.frame(data_b) %>% bind_cols(InOrOut=In_Or_Out)
I use dplyr::bind_cols to bind the in or out result to a data frame version of the original data so you might need some changes for your specific environment.
I'm trying to use msSurv for a multi-state modelling problem that looks at an individuals transition to different stages. Part of that is creating a tree object which is where I think I'm making a mistake but I can't understand what it is. I'll include the minimum workable example here.
Nodes <- c("1", "2", "3", "4", "5", "6")
Edges <- list("1" = list(edges = c("2", "3", "4", "5", "6")),
"2" = list(edges = c("1", "3", "4", "5", "6")),
"3" = list(edges = c("1", "2", "4", "5", "6")),
"4" = list(edges = c("1", "2", "3", "5", "6")),
"5" = list(edges = c("3", "4", "6")),
"6" = list(edges = NULL))
treeobj <- new("graphNEL", nodes = Nodes, edgeL = Edges, edgemode = "directed")
fit3 <- msSurv(df, treeobj, bs = TRUE, LT = TRUE)
The error I'm getting is as follows.
No states eligible for exit distribution calculation.
Entry distributions calculated for states 6 .
Error in bs.IA[, , j, b] : subscript out of bounds
The dataset in question can be found here.
Any help is sincerely appreciated.
I may be misunderstanding, but your 6 group doesn't have 1-6 as an edge, thus the program returns an error because in essence you're saying 6 isn't connected to the calculation. In relation to the solution, I believe 6 should have edges, as in this line may need to have edges: "6" = list(edges = NULL))
I intend to change the order of levels of some factors.
The intention is to apply this command
Df$X1 <- ordered(Df$X1, levels = c("5", "4", "3", "2", "1"))
to a list of variables (X1 to X2)
Df <- data.frame(
X1 = ordered(sample(1:5,30,r=T)),
X2 = ordered(sample(1:5,30,r=T)),
X3 = as.factor(sample(1:5,30,r=T)),
Y = as.factor(sample(1:5,30,r=T))
)
tmplistporadove <- as.list(paste("Df$",names(Df)[1:2],sep=""))
zmena <- lapply(tmplistporadove, function(x) substitute(x <- ordered(x, levels = c("5", "4", "3", "2", "1"))) )
eval(zmena)
But R just prints this:
X[[i]] <- ordered(X[[i]], levels = c("5", "4", "3", "2", "1"))