I've created a plot with geom_line and geom_ribbon (image 1) and the result is okay, but for the sake of aesthetics, I'd like the line and ribbon to be smoother. I know I can use geom_smooth for the line (image 2), but I'm not sure if it's possible to smooth the ribbon.I could create a geom_smooth line for the top and bottom lines of the ribbon (image 3), but is there anyway to fill in the space between those two lines?
A principled way to achieve what you want is to fit a GAM model to your data using the gam() function in mgcv and then apply the predict() function to that model over a finer grid of values for your predictor variable. The grid can cover the span defined by the range of observed values for your predictor variable. The R code below illustrates this process for a concrete example.
# load R packages
library(ggplot2)
library(mgcv)
# simulate some x and y data
# x = predictor; y = response
x <- seq(-10, 10, by = 1)
y <- 1 - 0.5*x - 2*x^2 + rnorm(length(x), mean = 0, sd = 20)
d <- data.frame(x,y)
# plot the simulated data
ggplot(data = d, aes(x,y)) +
geom_point(size=3)
# fit GAM model
m <- gam(y ~ s(x), data = d)
# define finer grid of predictor values
xnew <- seq(-10, 10, by = 0.1)
# apply predict() function to the fitted GAM model
# using the finer grid of x values
p <- predict(m, newdata = data.frame(x = xnew), se = TRUE)
str(p)
# plot the estimated mean values of y (fit) at given x values
# over the finer grid of x values;
# superimpose approximate 95% confidence band for the true
# mean values of y at given x values in the finer grid
g <- data.frame(x = xnew,
fit = p$fit,
lwr = p$fit - 1.96*p$se.fit,
upr = p$fit + 1.96*p$se.fit)
head(g)
theme_set(theme_bw())
ggplot(data = g, aes(x, fit)) +
geom_ribbon(aes(ymin = lwr, ymax = upr), fill = "lightblue") +
geom_line() +
geom_point(data = d, aes(x, y), shape = 1)
This same principle would apply if you were to fit a polynomial regression model to your data using the lm() function.
I am trying to reproduce nice stargazer model (lm) output for model that is not supperted by stargazer.
can linear model stargazer output be produced by hand? Since we can create a dataframe from every model and than insert the created dataframe to stargazer:
library(spdep)
data(afcon, package="spData")
afcon$Y = rnorm(42, 50, 20)
cns <- knearneigh(cbind(afcon$x, afcon$y), k=7, longlat=T)
scnsn <- knn2nb(cns, row.names = NULL, sym = T)
W <- nb2listw(scnsn, zero.policy = TRUE)
ols <- lm(totcon ~ Y, data = afcon)
spatial.lag <- lagsarlm(totcon ~ Y, data = afcon, W)
summary(model)
stargazer(ols, type = "text")
summary(spatial.lag)
data.frame(
spatial.lag$coefficients,
spatial.lag$rest.se
) %>%
rename(coeffs = spatial.lag.coefficients,
se = spatial.lag.rest.se) %>%
stargazer(type = "text", summary = F)
when we do stargazer(ols) output is very nice, I woud like to reproduce same output by hand for spatial.lag is there a way how to do so, how superscript etc...
You mean ^{*}? If so it's not possible in stargazer!! I've already tried it so I recommend you to check the xtable package like I did here.
I will show one approach that can be used: stargazer is really nice and you CAN even create table like above even with the model objects that are not yet supported, e.g. lets say that quantile regression model is not supported by stargazer (even thought is is):
Trick is, you need to be able to obtain coefficients and standart error e.g. as vector. Then supply stargazer with model object that is suppoerted e.g. lm as a template and then mechanically specify which coefficients and standart errors should be used:
library(stargazer)
library(tidyverse)
library(quantreg)
df <- mtcars
model1 <- lm(hp ~ factor(gear) + qsec + disp, data = df)
quantreg <- rq(hp ~ factor(gear) + qsec + disp, data = df)
summary_qr <- summary(quantreg, se = "boot")
# Standart Error for quant reg
se_qr = c(211.78266, 29.17307, 58.61105, 9.70908, 0.12090)
stargazer(model1, model1,
coef = list(NULL, summary_qr$coefficients),
se = list(NULL, se_qr),
type = "text")
I'm trying to set a problem that maximizes the consumption(u) over the time (10 years) subject to the variation of wealth (xt+1 - xt= -ct), where x0=1000. I've tried to solve it, but can't evaluate function at initial parameters. Also, the final conditions are: x10 >= 0 and u10 >= 0. Here's the code I'm using:
library('optimx')
# Parameters
t <- seq(0,10) # time
desc <- 0.1 # tax
beta <- 1/(1+desc) # parameter of function
alpha <- 0.8 # function
xo = 1000 # wealth in t=0
# Variables
x <- rep(xo, length(t)) # wealth in time
# Function
utilidade_total <- function(u, npar=TRUE, print=TRUE){
for (i in lenght(t)){
obj[i]=(beta^i)*(u[i]^alpha)
x[i]=x[i-1]-u[i-1]
}
result <- sum(obj)
return(result)
}
maxuti <- optimx(par=u,fn= utilidade_total,
control = list(maximize=TRUE))
umax <- coef(maxuti)
for (i in 2:lenght(t)){
print(obj[i]=(beta^i)*(u^alpha))
print(stock[i]=x[i-1]-umax[i-1])
}
# Result
for (i in 2:lenght(t)){
obj[i]=(beta^i)*(u^alpha)
x[i]=x[i-1]-u[i-1]
}
plot(t, x, type='b')
plot(t, u, type='b')
I'm trying to create a depth profile graph with the variables depth, distance and temperature. The data collected is from 9 different points with known distances between them (distance 5m apart, 9 stations, 9 different sets of data). The temperature readings are according to these 9 stations where a sonde was dropped directly down, taking readings of temperature every 2 seconds. Max depth at each of the 9 stations were taken from the boat also.
So the data I have is:
Depth at each of the 9 stations (y axis)
Temperature readings at each of the 9 stations, at around .2m intervals vertical until the bottom was reached (fill area)
distance between the stations, (x axis)
Is it possible to create a depth profile similar to this? (obviously without the greater resolution in this graph)
I've already tried messing around with ggplot2 and raster but I just can't seem to figure out how to do this.
One of the problems I've come across is how to make ggplot2 distinguish between say 5m depth temperature reading at station 1 and 5m temperature reading at station 5 since they have the same depth value.
Even if you can guide me towards another program that would allow me to create a graph like this, that would be great
[ REVISION ]
(Please comment me if you know more suitable interpolation methods, especially not needing to cut under bottoms data.)
ggplot() needs long data form.
library(ggplot2)
# example data
max.depths <- c(1.1, 4, 4.7, 7.7, 8.2, 7.8, 10.7, 12.1, 14.3)
depth.list <- sapply(max.depths, function(x) seq(0, x, 0.2))
temp.list <- list()
set.seed(1); for(i in 1:9) temp.list[[i]] <- sapply(depth.list[[i]], function(x) rnorm(1, 20 - x*0.5, 0.2))
set.seed(1); dist <- c(0, sapply(seq(5, 40, 5), function(x) rnorm(1, x, 1)))
dist.list <- sapply(1:9, function(x) rep(dist[x], length(depth.list[[x]])))
main.df <- data.frame(dist = unlist(dist.list), depth = unlist(depth.list) * -1, temp = unlist(temp.list))
# a raw graph
ggplot(main.df, aes(x = dist, y = depth, z = temp)) +
geom_point(aes(colour = temp), size = 1) +
scale_colour_gradientn(colours = topo.colors(10))
# a relatively raw graph (don't run with this example data)
ggplot(main.df, aes(x = dist, y = depth, z = temp)) +
geom_raster(aes(fill = temp)) + # geom_contour() +
scale_fill_gradientn(colours = topo.colors(10))
If you want a graph such like you showed, you have to do interpolation. Some packages give you spatial interpolation methods. In this example, I used akima package but you should think seriously that which interpolation methods to use.
I used nx = 300 and ny = 300 in below code but I think it would be better to decide those values carefully. Large nx and ny gives a high resolution graph, but don't foreget real nx and ny (in this example, real nx is only 9 and ny is 101).
library(akima); library(dplyr)
interp.data <- interp(main.df$dist, main.df$depth, main.df$temp, nx = 300, ny = 300)
interp.df <- interp.data %>% interp2xyz() %>% as.data.frame()
names(interp.df) <- c("dist", "depth", "temp")
# draw interp.df
ggplot(interp.df, aes(x = dist, y = depth, z = temp)) +
geom_raster(aes(fill = temp)) + # geom_contour() +
scale_fill_gradientn(colours = topo.colors(10))
# to think appropriateness of interpolation (raw and interpolation data)
ggplot(interp.df, aes(x = dist, y = depth, z = temp)) +
geom_raster(aes(fill = temp), alpha = 0.3) + # interpolation
scale_fill_gradientn(colours = topo.colors(10)) +
geom_point(data = main.df, aes(colour = temp), size = 1) + # raw
scale_colour_gradientn(colours = topo.colors(10))
Bottoms don't match !!I found ?interp says "interpolation only within convex hull!", oops... I'm worrid about the interpolation around the problem-area, is it OK ? If no problem, you need only cut the data under the bottoms. If not, ... I can't answer immediately (below is an example code to cut).
bottoms <- max.depths * -1
# calculate bottom values using linear interpolation
approx.bottoms <- approx(dist, bottoms, n = 300) # n must be the same value as interp()'s nx
# change temp values under bottom into NA
library(dplyr)
interp.cut.df <- interp.df %>% cbind(bottoms = approx.bottoms$y) %>%
mutate(temp = ifelse(depth >= bottoms, temp, NA)) %>% select(-bottoms)
ggplot(interp.cut.df, aes(x = dist, y = depth, z = temp)) +
geom_raster(aes(fill = temp)) +
scale_fill_gradientn(colours = topo.colors(10)) +
geom_point(data = main.df, size = 1)
If you want to use stat_contour
It is harder to use stat_contour than geom_raster because it needs a regular grid form. As far as I see your graph, your data (depth and distance) don't form a regular grid, it means it is much difficult to use stat_contour with your raw data. So I used interp.cut.df to draw a contour plot. And stat_contour have a endemic problem (see How to fill in the contour fully using stat_contour), so you need to expand your data.
library(dplyr)
# 1st: change NA into a temp's out range value (I used 0)
interp.contour.df <- interp.cut.df
interp.contour.df[is.na(interp.contour.df)] <- 0
# 2nd: expand the df (It's a little complex, so please use this function)
contour.support.func <- function(df) {
colname <- names(df)
names(df) <- c("x", "y", "z")
Range <- as.data.frame(sapply(df, range))
Dim <- as.data.frame(t(sapply(df, function(x) length(unique(x)))))
arb_z = Range$z[1] - diff(Range$z)/20
df2 <- rbind(df,
expand.grid(x = c(Range$x[1] - diff(Range$x)/20, Range$x[2] + diff(Range$x)/20),
y = seq(Range$y[1], Range$y[2], length = Dim$y), z = arb_z),
expand.grid(x = seq(Range$x[1], Range$x[2], length = Dim$x),
y = c(Range$y[1] - diff(Range$y)/20, Range$y[2] + diff(Range$y)/20), z = arb_z))
names(df2) <- colname
return(df2)
}
interp.contour.df2 <- contour.support.func(interp.contour.df)
# 3rd: check the temp range (these values are used to define contour's border (breaks))
range(interp.cut.df$temp, na.rm=T) # 12.51622 20.18904
# 4th: draw ... the bottom border is dirty !!
ggplot(interp.contour.df2, aes(x = dist, y = depth, z = temp)) +
stat_contour(geom="polygon", breaks = seq(12.51622, 20.18904, length = 11), aes(fill = ..level..)) +
coord_cartesian(xlim = range(dist), ylim = range(bottoms), expand = F) + # cut expanded area
scale_fill_gradientn(colours = topo.colors(10)) # breaks's length is 11, so 10 colors are needed
# [Note]
# You can define the contour's border values (breaks) and colors.
contour.breaks <- c(12.5, 13.5, 14.5, 15.5, 16.5, 17.5, 18.5, 19.5, 20.5)
# = seq(12.5, 20.5, 1) or seq(12.5, 20.5, length = 9)
contour.colors <- c("darkblue", "cyan3", "cyan1", "green3", "green", "yellow2","pink", "darkred")
# breaks's length is 9, so 8 colors are needed.
# 5th: vanish the bottom border by bottom line
approx.df <- data.frame(dist = approx.bottoms$x, depth = approx.bottoms$y, temp = 0) # 0 is dummy value
ggplot(interp.contour.df2, aes(x = dist, y = depth, z = temp)) +
stat_contour(geom="polygon", breaks = contour.breaks, aes(fill = ..level..)) +
coord_cartesian(xlim=range(dist), ylim=range(bottoms), expand = F) +
scale_fill_gradientn(colours = contour.colors) +
geom_line(data = approx.df, lwd=1.5, color="gray50")
bonus: legend technic
library(dplyr)
interp.contour.df3 <- interp.contour.df2 %>% mutate(temp2 = cut(temp, breaks = contour.breaks))
interp.contour.df3$temp2 <- factor(interp.contour.df3$temp2, levels = rev(levels(interp.contour.df3$temp2)))
ggplot(interp.contour.df3, aes(x = dist, y = depth, z = temp)) +
stat_contour(geom="polygon", breaks = contour.breaks, aes(fill = ..level..)) +
coord_cartesian(xlim=range(dist), ylim=range(bottoms), expand = F) +
scale_fill_gradientn(colours = contour.colors, guide = F) + # add guide = F
geom_line(data = approx.df, lwd=1.5, color="gray50") +
geom_point(aes(colour = temp2), pch = 15, alpha = 0) + # add
guides(colour = guide_legend(override.aes = list(colour = rev(contour.colors), alpha = 1, cex = 5))) + # add
labs(colour = "temp") # add
You want to treat this as a 3-D surface with temperature as the z dimension. The given plot is a contour plot and it looks like ggplot2 can do that with stat_contour.
I'm not sure how the contour lines are computed (often it's linear interpolation along a Delaunay triangulation). If you want more control over how to interpolate between your x/y grid points, you can calculate a surface model first and feed those z coordinates into ggplot2.
I use the following function to convert the topicmodels output to JSON output to use in ldavis.
topicmodels_json_ldavis <- function(fitted, corpus, doc_term){
## Required packages
library(topicmodels)
library(dplyr)
library(stringi)
library(tm)
library(LDAvis)
## Find required quantities
phi <- posterior(fitted)$terms %>% as.matrix
theta <- posterior(fitted)$topics %>% as.matrix
vocab <- colnames(phi)
doc_length <- vector()
for (i in 1:length(corpus)) {
temp <- paste(corpus[[i]]$content, collapse = ' ')
doc_length <- c(doc_length, stri_count(temp, regex = '\\S+'))
}
temp_frequency <- inspect(doc_term)
freq_matrix <- data.frame(ST = colnames(temp_frequency),
Freq = colSums(temp_frequency))
rm(temp_frequency)
## Convert to json
json_lda <- LDAvis::createJSON(phi = phi, theta = theta,
vocab = vocab,
doc.length = doc_length,
term.frequency = freq_matrix$Freq)
return(json_lda)
}
but I receive the following error
Error in LDAvis::createJSON(phi = phi, theta = theta, vocab = vocab, doc.length = doc_length, : Length of doc.length not equal
to the number of rows in theta; both should be equal to the number of
documents in the data.
Here is my complete code:
data <- read.csv("textmining.csv")
corpus <- Corpus(DataframeSource(data.frame(data$reasonforleaving)))
# Remove punctuations and numbers because they are generally uninformative.
corpus <- tm_map(corpus, removePunctuation)
corpus <- tm_map(corpus, removeNumbers)
# Convert all words to lowercase.
corpus <- tm_map(corpus, content_transformer(tolower))
# Remove stopwords such as "a", "the", etc.
corpus <- tm_map(corpus, removeWords, stopwords("english"))
# Use the SnowballC package to do stemming.
library(SnowballC)
corpus <- tm_map(corpus, stemDocument)
# remove extra words
toSpace <- content_transformer(function (x , pattern ) gsub(pattern, " ", x))
corpus <- tm_map(corpus, toSpace, "still")
corpus <- tm_map(corpus, toSpace, "also")
# Remove excess white spaces between words.
corpus <- tm_map(corpus, stripWhitespace)
# Inspect the first document to see what it looks like.
corpus[[1]]$content
dtm <- DocumentTermMatrix(corpus)
# remove empty documents
library(slam)
dtm = dtm[row_sums(dtm)>0,]
# Use topicmodels package to conduct LDA analysis.
burnin <- 500
iter <- 1000
keep <- 30
k <- 5
result55 <- LDA(dtm, 5)
ldaoutput = topicmodels_json_ldavis(result55,corpus, dtm)
Do you know why I receive the error?
Thanks
I had the same issue with same code, and found this function here :
topicmodels2LDAvis <- function(x, ...){
post <- topicmodels::posterior(x)
if (ncol(post[["topics"]]) < 3) stop("The model must contain > 2 topics")
mat <- x#wordassignments
LDAvis::createJSON(
phi = post[["terms"]],
theta = post[["topics"]],
vocab = colnames(post[["terms"]]),
doc.length = slam::row_sums(mat, na.rm = TRUE),
term.frequency = slam::col_sums(mat, na.rm = TRUE)
)
}
Much much simpler to use, just put as argument your LDA result :
result55 <- LDA(dtm, 5)
serVis(topicmodels2LDAvis(result55))
Problem
Your problem is in for (i in 1:length(corpus)) in
doc_length <- vector()
for (i in 1:length(corpus)) {
temp <- paste(corpus[[i]]$content, collapse = ' ')
doc_length <- c(doc_length, stri_count(temp, regex = '\\S+'))
}
Remember, you have removed some "empty" documents from your DocumentTermMatrix in dtm = dtm[row_sums(dtm)>0,],
so your vector length here is going to be too big.
Suggestion
You may want to keep a vector of the empty docs around as it will help you not only to generate the JSON but also to go back and forth between your empty and full document sets.
doc.length = colSums( as.matrix(tdm) > 0 )[!empty.docs]
My suggestion assumes you have the full tdm with empty docs in place