I've just started learning R, and was wondering, say I have the dataset quake, and I want to generate the probability histogram of quakes near Fiji, would the code simply be hist(quakes$lat,freq=F)?
A histogram shows the frequency or proportion of a given value out of all the values in a data set. You need a numeric vector as the x argument for hist(). There is no flat variable in quakes, but there is a lat variable. hist(quakes$lat, freq = F) would show the following:
This shows the north/south geographical distribution of earthquakes, centering around -20, and, since it is approximately normal (with a left skew) suggests that there is a mechanism for earthquake generation that centers around a specific latitude.
The best way to learn is to try. If you wonder if that would be the way to do it, try it.
You might also want to look at this tutorial on creating kernel density plots with ggplot.
Related
I am having issues trying to generate a code that will cleanly produce a mean (specifically a weighted average) based on a simple plot of points using interpolation.
For Example;
ex=c(1,2,3,4,5)
why=c(2,5,9,15,24)
This shows the kind of information I am working with.
plot(ex, why, type="o")
At this point, I want to actually have each point "binned" so the lines between them are straight. To do this, I have been adding points to the x values manually in excel as (x+0.01).
This is the new output:
why=c(2,2,5,5,9,9,15,15,24,24)
ex=c(1,2,2.01,3,3.01,4,4.01,5,5.01,6)
plot(ex, why, type="o")
So this is where my question comes in to play. I have to do this many times and do not want to generate a ton of new vectors and objects. To get a weighted average, I have been interpolating y values for increments of x at 0.01 using interpolation into a new object. I am then able to go into this new object and get a mean when a point falls between the actual ex values, i.e.
mean(newy[1:245])
Because I made new y values for 100 increments of x that (basically) follow a straight line, I am getting a weighted average here for x= 1 to 2.45.
Is there an easier and more elegant way to embed the interpolate code into the mean code so I could just say "average of interpolated y for nonreal x to nonreal x?"
It doesn't do exactly what you want, but you should consider the stepfun function -- this creates a step function out of two series.
plot(stepfun(ex[-1], why))
stepfun is handy because it gives you a function defined over that interval, so you can easily interpolate just by evaluating anywhere. The downside to it is that it is not strictly defined on the range given (which is why we have to cut off the first value in ex).
Based on your second plotting example, I think you are probably looking for this:
library(ggplot2)
qplot(ex, why, geom="step")
this gives:
Or if you want the line to go vertical first, you can use:
qplot(ex, why, geom="step", direction = "vh")
which gives:
what are some good kriging/interpolation idea/options that will allow heavily-weighted points to bleed over lightly-weighted points on a plotted R map?
the state of connecticut has eight counties. i found the centroid and want to plot poverty rates of each of these eight counties. three of the counties are very populated (about 1 million people) and the other five counties are sparsely populated (about 100,000 people). since the three densely-populated counties have more than 90% of the total state population, i would like those the three densely-populated counties to completely "overwhelm" the map and impact other points across the county borders.
the Krig function in the R fields package has a lot of parameters and also covariance functions that can be called, but i'm not sure where to start?
here is reproducible code to quickly produce a hard-bordered map and then three differently-weighted maps. hopefully i can just make changes to this code, but perhaps it requires something more complex like the geoRglm package? two of the three weighted maps look almost identical, despite one being 10x as weighted as the other..
https://raw.githubusercontent.com/davidbrae/swmap/master/20141001%20how%20to%20modify%20the%20Krig%20function%20so%20a%20huge%20weight%20overwhelms%20nearby%20points.R
thanks!!
edit: here's a picture example of the behavior i want-
disclaimer - I am not an expert on Krigging. Krigging is complex and takes a good understanding of the underlying data, the method and the purpose to achieve the correct result. You may wish to try to get input from #whuber [on the GIS Stack Exchange or contact him through his website (http://www.quantdec.com/quals/quals.htm)] or another expert you know.
That said, if you just want to achieve the visual effect you requested and are not using this for some sort of statistical analysis, I think there are some relatively simple solutions.
EDIT:
As you commented, though the suggestions below to use theta and smoothness arguments do even out the prediction surface, they apply equally to all measurements and thus do not extend the "sphere of influence" of more densely populated counties relative to less-densely populated. After further consideration, I think there are two ways to achieve this: by altering the covariance function to depend on population density or by using weights, as you have. Your weighting approach, as I wrote below, alters the error term of the krigging function. That is, it inversely scales the nugget variance.
As you can see in the semivariogram image, the nugget is essentially the y-intercept, or the error between measurements at the same location. Weights affect the nugget variance (sigma2) as sigma2/weight. Thus, greater weights mean less error at small-scale distances. This does not, however, change the shape of the semivariance function or have much effect on the range or sill.
I think that the best solution would be to have your covariance function depend on population. however, I'm not sure how to accomplish that and I don't see any arguments to Krig to do so. I tried playing with defining my own covariance function as in the Krig example, but only got errors.
Sorry I couldn't help more!
Another great resource to help understand Krigging is: http://www.epa.gov/airtrends/specialstudies/dsisurfaces.pdf
As I said in my comment, the sill and nugget values as well as the range of the semivariogram are things you can alter to affect the smoothing. By specifying weights in the call to Krig, you are altering the variance of the measurement errors. That is, in a normal use, weights are expected to be proportional to the accuracy of the measurement value so that higher weights represent more accurate measurements, essentially. This isn't actually true with your data, but it may be giving you the effect you desire.
To alter the way your data is interpolated, you can adjust two (and many more) parameters in the simple Krig call you are using: theta and smoothness. theta adjusts the semivariance range, meaning that measured points farther away contribute more to the estimates as you increase theta. Your data range is
range <- data.frame(lon=range(ct.data$lon),lat=range(ct.data$lat))
range[2,]-range[1,]
lon lat
2 1.383717 0.6300484
so, your measurement points vary by ~1.4 degrees lon and ~0.6 degrees lat. Thus, you can play with specifying your theta value in that range to see how that affects your result. In general, a larger theta leads to more smoothing since you are drawing from more values for each prediction.
Krig.output.wt <- Krig( cbind(ct.data$lon,ct.data$lat) , ct.data$county.poverty.rate ,
weights=c( size , 1 , 1 , 1 , 1 , size , size , 1 ),Covariance="Matern", theta=.8)
r <- interpolate(ras, Krig.output.wt)
r <- mask(r, ct.map)
plot(r, col=colRamp(100) ,axes=FALSE,legend=FALSE)
title(main="Theta = 0.8", outer = FALSE)
points(cbind(ct.data$lon,ct.data$lat))
text(ct.data$lon, ct.data$lat-0.05, ct.data$NAME, cex=0.5)
Gives:
Krig.output.wt <- Krig( cbind(ct.data$lon,ct.data$lat) , ct.data$county.poverty.rate ,
weights=c( size , 1 , 1 , 1 , 1 , size , size , 1 ),Covariance="Matern", theta=1.6)
r <- interpolate(ras, Krig.output.wt)
r <- mask(r, ct.map)
plot(r, col=colRamp(100) ,axes=FALSE,legend=FALSE)
title(main="Theta = 1.6", outer = FALSE)
points(cbind(ct.data$lon,ct.data$lat))
text(ct.data$lon, ct.data$lat-0.05, ct.data$NAME, cex=0.5)
Gives:
Adding the smoothness argument, will change the order of the function used to smooth your predictions. The default is 0.5 leading to a second-order polynomial.
Krig.output.wt <- Krig( cbind(ct.data$lon,ct.data$lat) , ct.data$county.poverty.rate ,
weights=c( size , 1 , 1 , 1 , 1 , size , size , 1 ),
Covariance="Matern", smoothness = 0.6)
r <- interpolate(ras, Krig.output.wt)
r <- mask(r, ct.map)
plot(r, col=colRamp(100) ,axes=FALSE,legend=FALSE)
title(main="Theta unspecified; Smoothness = 0.6", outer = FALSE)
points(cbind(ct.data$lon,ct.data$lat))
text(ct.data$lon, ct.data$lat-0.05, ct.data$NAME, cex=0.5)
Gives:
This should give you a start and some options, but you should look at the manual for fields. It is pretty well-written and explains the arguments well.
Also, if this is in any way quantitative, I would highly recommend talking to someone with significant spatial statistics know how!
Kriging is not what you want. (It is a statistical method for accurate--not distorted!--interpolation of data. It requires preliminary analysis of the data--of which you do not have anywhere near enough for this purpose--and cannot accomplish the desired map distortion.)
The example and the references to "bleed over" suggest considering an anamorph or area cartogram. This is a map which will expand and shrink the areas of the county polygons so that they reflect their relative population while retaining their shapes. The link (to the SE GIS site) explains and illustrates this idea. Although its answers are less than satisfying, a search of that site will reveal some effective solutions.
lot's of interesting comments and leads above.
I took a look at the Harvard dialect survey to get a sense for what you are trying to do first. I must say really cool maps. And before I start in on what I came up with...I've looked at your work on survey analysis before and have learned quite a few tricks. Thanks.
So my first take pretty quickly was that if you wanted to do spatial smoothing by way of kernel density estimation then you need to be thinking in terms of point process models. I'm sure there are other ways, but that's where I went.
So what I do below is grab a very generic US map and convert it into something I can use as a sampling window. Then I create random samples of points within that region, just pretend those are your centroids. After I attach random values to those points and plot it up.
I just wanted to test this conceptually, which is why I didn't go through the extra steps to grab cbsa's and also sorry for not projecting, but I think these are the fundamentals. Oh and the smoothing in the dialect study is being done over the whole country. I think. That is the author is not stratifying his smoothing procedure within polygons....so I just added states at the end.
code:
library(sp)
library(spatstat)
library(RColorBrewer)
library(maps)
library(maptools)
# grab us map from R maps package
usMap <- map("usa")
usIds <- usMap$names
# convert to spatial polygons so this can be used as a windo below
usMapPoly <- map2SpatialPolygons(usMap,IDs=usIds)
# just select us with no islands
usMapPoly <- usMapPoly[names(usMapPoly)=="main",]
# create a random sample of points on which to smooth over within the map
pts <- spsample(usMapPoly, n=250, type='random')
# just for a quick check of the map and sampling locations
plot(usMapPoly)
points(pts)
# create values associated with points, be sure to play aroud with
# these after you get the map it's fun
vals <-rnorm(250,100,25)
valWeights <- vals/sum(vals)
ptsCords <- data.frame(pts#coords)
# create window for the point pattern object (ppp) created below
usWindow <- as.owin(usMapPoly)
# create spatial point pattern object
usPPP <- ppp(ptsCords$x,ptsCords$y,marks=vals,window=usWindow)
# create colour ramp
col <- colorRampPalette(brewer.pal(9,"Reds"))(20)
# the plots, here is where the gausian kernal density estimation magic happens
# if you want a continuous legend on one of the sides get rid of ribbon=FALSE
# and be sure to play around with sigma
plot(Smooth(usPPP,sigma=3,weights=valWeights),col=col,main=NA,ribbon=FALSE)
map("state",add=TRUE,fill=FALSE)
example no weights:
example with my trivial weights
There is obviously a lot of work in between this and your goal of making this type of map reproducible at various levels of spatial aggregation and sample data, but good luck it seems like a cool project.
p.s. initially I did not use any weighting, but I suppose you could provide weights directly to the Smooth function. Two example maps above.
Is there a simple way to plot the difference between two probability density functions?
I can plot the pdfs of my data sets (both are one-dimensional vectors with roughly 11000 values) on the same plot together to get an idea of the overlap/difference but it would be more useful to me if I could see a plot of the difference.
something along the lines of the following (though this obviously doesn't work):
> plot(density(data1)-density(data2))
I'm relatively new to R and have been unable to find what I'm looking for on any of the forums.
Thanks in advance
This should work:
plot(x =density(data1, from= range(c(data1, data2))[1],
to=range(c(data1, data2))[2] )$x,
y= density(data1, from= range(c(data1, data2))[1],
to=range(c(data1, data2))[2] )$y-
density(data2, from= range(c(data1, data2))[1],
to=range(c(data1, data2))[2] )$y )
The trick is to make sure the densities have the same limits. Then you can plot their differences at the same locations.My understanding of the need for the identical limits comes from having made the error of not taking that step in answering a similar question on Rhelp several years ago. Too bad I couldn't remember the right arguments.
It looks like you need to spend a little time learning how to use R (or any other language, for that matter). Help files are your friend.
From the output of ?density :
Value [i.e. the data returned by the function]
If give.Rkern is true, the number R(K), otherwise an object with class
"density" whose underlying structure is a list containing the
following components.
x the n coordinates of the points where the density is estimated.
y the estimated density values. These will be non-negative, but can
be zero [remainder of "value" deleted for brevity]
So, do:
foo<- density(data1)
bar<- density(data2)
plot(foo$y-bar$y)
Your comments, suggestions, or solutions are/will be greatly appreciated, thank you.
I'm using the fpc package in R to do a dbscan analysis of some very dense data (3 sets of 40,000 points between the range -3, 6).
I've found some clusters, and I need to graph just the significant ones. The problem is that I have a single cluster (the first) with about 39,000 points in it. I need to graph all other clusters but this one.
The dbscan() creates a special data type to store all of this cluster data in. It's not indexed like a data frame would be (but maybe there is a way to represent it as such?).
I can graph the dbscan type using a basic plot() call. But, like I said, this will graph the irrelevant 39,000 points.
tl;dr:
how do I graph only specific clusters of a dbscan data type?
If you look at the help page (?dbscan) it is organized like all others into sections labeled Description, Usage, Arguments, Details and Value. The Value section describes what the function dbscan returns. In this case it is simply a list (a standard R data type) with a few components.
The cluster component is simply an integer vector whose length it equal to the number of rows in your data that indicates which cluster each observation is a member of. So you can use this vector to subset your data to extract only those clusters you'd like and then plot just those data points.
For example, if we use the first example from the help page:
set.seed(665544)
n <- 600
x <- cbind(runif(10, 0, 10)+rnorm(n, sd=0.2), runif(10, 0, 10)+rnorm(n,
sd=0.2))
ds <- dbscan(x, 0.2)
we can then use the result, ds to plot only the points in clusters 1-3:
#Plot only clusters 1, 2 and 3
plot(x[ds$cluster %in% 1:3,])
Without knowing the specifics of dbscan, I can recommend that you look at the function smoothScatter. It it very useful for examining the main patterns in a scatterplot when you otherwise would have too many points to make sense of the data.
The probably most sensible way of plotting DBSCAN results is using alpha shapes, with the radius set to the epsilon value. Alpha shapes are closely related to convex hulls, but they are not necessarily convex. The alpha radius controls the amount of non-convexity allowed.
This is quite closely related to the DBSCAN cluster model of density connected objects, and as such will give you a useful interpretation of the set.
As I'm not using R, I don't know about the alpha shape capabilities of R. There supposedly is a package called alphahull, from a quick check on Google.
I need to get a plot of a Lorentz curve of a cumulative variable as a function of the number of observations. I want both axes to be displayed on a percentage basis (e.g. say observations are the number of buyers and the y variable is the amount they bought, buyers are already ranked in descending order, I want to get the plot that says "The top 10% buyers purchased 90% of the total bought"). My dataset is a couple million observations.
What is the best way to do this? Sub-questions:
If I need to add two variables for the quantiles of total observations and total $ bought (so as to use them to plot), what is the object that returns the row number? I tried:
user_quantile <- row(df)/nrow(df)
but I get a matrix of identical columns (user_quantile.1, user_quantile.2) of which I only need one column.
Is there instead any way to skip adding percentages as variables and only have them for axes values?
The plot has way to many points than I need to get the line. What is the best approach to minimize the computational effort and get a nice graph?
Thanks.
You may want to acquaint yourself with the excellent RSeek search engine for R content. One quick query for Lorentz curve (and Lorenz curve) lead to these packages:
ineq: Measuring inequality, concentration, and poverty
reldist: Relative Distribution Methods
GeoXp: Interactive exploratory spatial data analysis
lawstat: An R package for biostatistics, public policy and law
all of which seem to supply a Lorenz curve function.
In order to get the plot done you need first to arrange the raw data.
1) You can use the cut2() function from the Hmisc package to cut the data in quantiles. Check the documentation, it's not hard. It's similar to the cut() from the base package.
2) After using the cut2() function with the income data, you need to compute the frequency of each decile. Use table() for that. Then calculate percentages of income for each decile.
3) Now you should have a very small table with the following columns:
Decile, cumulative % of total income.
Add another column with the 45 degree line. Just add a constant cumulative % of income.
finaltable$cumulative_equality_line = seq(0.1, 1, by = 0.1)
4) You can use base graphics or ggplot2 for plotting. I guess you can do it with the info of step 3 or perhaps check out specific plotting questions.
I'll have to do it soon, but i already have the final table. I'll post the code for plotting once i do it.
Good luck!