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I have been trying to add a bell curve to my histogram an outline it with a color so that it is more pleasing. enter image description here
I have added what my histogram looks like to give someone an idea on what I am working with, also here is my code thus far, thank you in advance.
ggplot(data = mammal.data.22.select2)+
geom_histogram(aes(x=Time, fill=Species))+
scale_fill_manual(values=c("paleturquoise4", "turquoise2"))+
facet_wrap(~Species, nrow=1)+
ylab("Observations")+
xlab("Time of Day")+
theme(strip.text.x = element_blank())
Let's build a histogram with a build-in dataset that seems similar-ish to your data structure.
library(ggplot2)
binwidth <- 0.25
p <- ggplot(iris, aes(Petal.Length)) +
geom_histogram(
aes(fill = Species),
binwidth = binwidth,
alpha = 0.5
) +
facet_wrap(~ Species)
You can use stat_bin() + geom_step() to give an outline to the histogram, without colouring the edge of every rectangle in the histogram. The only downside is that the first and last bins don't touch the x-axis.
p + stat_bin(
geom = "step", direction = "mid",
aes(colour = Species), binwidth = binwidth
)
To overlay a density function with a histogram, you could calculate the relevant parameters yourself and use stat_function() with fun = dnorm repeatedly. Alternatively, you can use ggh4x::stat_theodensity() to achieve a similar thing. Note that whether you use stat_function() or stat_theodensity(), you should scale the density back to the counts that your histogram uses (or scale histogram to density). In the example below, we do that by using after_stat(count * binwidth).
p + ggh4x::stat_theodensity(
aes(colour = Species,
y = after_stat(count * binwidth))
)
Created on 2022-04-15 by the reprex package (v2.0.1)
(disclaimer: I'm the author of ggh4x)
I want to convert GPS location (latitude, longitude) into x,y coordinates.
I found many links about this topic and applied it, but it doesn't give me the correct answer!
I am following these steps to test the answer:
(1) firstly, i take two positions and calculate the distance between them using maps.
(2) then convert the two positions into x,y coordinates.
(3) then again calculate distance between the two points in the x,y coordinates
and see if it give me the same result in point(1) or not.
one of the solution i found the following, but it doesn't give me correct answer!
latitude = Math.PI * latitude / 180;
longitude = Math.PI * longitude / 180;
// adjust position by radians
latitude -= 1.570795765134; // subtract 90 degrees (in radians)
// and switch z and y
xPos = (app.radius) * Math.sin(latitude) * Math.cos(longitude);
zPos = (app.radius) * Math.sin(latitude) * Math.sin(longitude);
yPos = (app.radius) * Math.cos(latitude);
also i tried this link but still not work with me well!
any help how to convert from(latitude, longitude) to (x,y) ?
Thanks,
No exact solution exists
There is no isometric map from the sphere to the plane. When you convert lat/lon coordinates from the sphere to x/y coordinates in the plane, you cannot hope that all lengths will be preserved by this operation. You have to accept some kind of deformation. Many different map projections do exist, which can achieve different compromises between preservations of lengths, angles and areas. For smallish parts of earth's surface, transverse Mercator is quite common. You might have heard about UTM. But there are many more.
The formulas you quote compute x/y/z, i.e. a point in 3D space. But even there you'd not get correct distances automatically. The shortest distance between two points on the surface of the sphere would go through that sphere, whereas distances on the earth are mostly geodesic lengths following the surface. So they will be longer.
Approximation for small areas
If the part of the surface of the earth which you want to draw is relatively small, then you can use a very simple approximation. You can simply use the horizontal axis x to denote longitude λ, the vertical axis y to denote latitude φ. The ratio between these should not be 1:1, though. Instead you should use cos(φ0) as the aspect ratio, where φ0 denotes a latitude close to the center of your map. Furthermore, to convert from angles (measured in radians) to lengths, you multiply by the radius of the earth (which in this model is assumed to be a sphere).
x = r λ cos(φ0)
y = r φ
This is simple equirectangular projection. In most cases, you'll be able to compute cos(φ0) only once, which makes subsequent computations of large numbers of points really cheap.
I want to share with you how I managed the problem. I've used the equirectangular projection just like #MvG said, but this method gives you X and Y positions related to the globe (or the entire map), this means that you get global positions. In my case, I wanted to convert coordinates in a small area (about 500m square), so I related the projection point to another 2 points, getting the global positions and relating to local (on screen) positions, just like this:
First, I choose 2 points (top-left and bottom-right) around the area where I want to project, just like this picture:
Once I have the global reference area in lat and lng, I do the same for screen positions. The objects containing this data are shown below.
//top-left reference point
var p0 = {
scrX: 23.69, // Minimum X position on screen
scrY: -0.5, // Minimum Y position on screen
lat: -22.814895, // Latitude
lng: -47.072892 // Longitude
}
//bottom-right reference point
var p1 = {
scrX: 276, // Maximum X position on screen
scrY: 178.9, // Maximum Y position on screen
lat: -22.816419, // Latitude
lng: -47.070563 // Longitude
}
var radius = 6371; //Earth Radius in Km
//## Now I can calculate the global X and Y for each reference point ##\\
// This function converts lat and lng coordinates to GLOBAL X and Y positions
function latlngToGlobalXY(lat, lng){
//Calculates x based on cos of average of the latitudes
let x = radius*lng*Math.cos((p0.lat + p1.lat)/2);
//Calculates y based on latitude
let y = radius*lat;
return {x: x, y: y}
}
// Calculate global X and Y for top-left reference point
p0.pos = latlngToGlobalXY(p0.lat, p0.lng);
// Calculate global X and Y for bottom-right reference point
p1.pos = latlngToGlobalXY(p1.lat, p1.lng);
/*
* This gives me the X and Y in relation to map for the 2 reference points.
* Now we have the global AND screen areas and then we can relate both for the projection point.
*/
// This function converts lat and lng coordinates to SCREEN X and Y positions
function latlngToScreenXY(lat, lng){
//Calculate global X and Y for projection point
let pos = latlngToGlobalXY(lat, lng);
//Calculate the percentage of Global X position in relation to total global width
pos.perX = ((pos.x-p0.pos.x)/(p1.pos.x - p0.pos.x));
//Calculate the percentage of Global Y position in relation to total global height
pos.perY = ((pos.y-p0.pos.y)/(p1.pos.y - p0.pos.y));
//Returns the screen position based on reference points
return {
x: p0.scrX + (p1.scrX - p0.scrX)*pos.perX,
y: p0.scrY + (p1.scrY - p0.scrY)*pos.perY
}
}
//# The usage is like this #\\
var pos = latlngToScreenXY(-22.815319, -47.071718);
$point = $("#point-to-project");
$point.css("left", pos.x+"em");
$point.css("top", pos.y+"em");
As you can see, I made this in javascript, but the calculations can be translated to any language.
P.S. I'm applying the converted positions to an HTML element whose id is "point-to-project". To use this piece of code on your project, you shall create this element (styled as position absolute) or change the "usage" block.
Since this page shows up on top of google while i searched for this same problem, I would like to provide a more practical answers. The answer by MVG is correct but rather theoratical.
I have made a track plotting app for the fitbit ionic in javascript. The code below is how I tackled the problem.
//LOCATION PROVIDER
index.js
var gpsFix = false;
var circumferenceAtLat = 0;
function locationSuccess(pos){
if(!gpsFix){
gpsFix = true;
circumferenceAtLat = Math.cos(pos.coords.latitude*0.01745329251)*111305;
}
pos.x:Math.round(pos.coords.longitude*circumferenceAtLat),
pos.y:Math.round(pos.coords.latitude*110919),
plotTrack(pos);
}
plotting.js
plotTrack(position){
let x = Math.round((this.segments[i].start.x - this.bounds.minX)*this.scale);
let y = Math.round(this.bounds.maxY - this.segments[i].start.y)*this.scale; //heights needs to be inverted
//redraw?
let redraw = false;
//x or y bounds?
if(position.x>this.bounds.maxX){
this.bounds.maxX = (position.x-this.bounds.minX)*1.1+this.bounds.minX; //increase by 10%
redraw = true;
}
if(position.x<this.bounds.minX){
this.bounds.minX = this.bounds.maxX-(this.bounds.maxX-position.x)*1.1;
redraw = true;
};
if(position.y>this.bounds.maxY){
this.bounds.maxY = (position.y-this.bounds.minY)*1.1+this.bounds.minY; //increase by 10%
redraw = true;
}
if(position.y<this.bounds.minY){
this.bounds.minY = this.bounds.maxY-(this.bounds.maxY-position.y)*1.1;
redraw = true;
}
if(redraw){
reDraw();
}
}
function reDraw(){
let xScale = device.screen.width / (this.bounds.maxX-this.bounds.minX);
let yScale = device.screen.height / (this.bounds.maxY-this.bounds.minY);
if(xScale<yScale) this.scale = xScale;
else this.scale = yScale;
//Loop trough your object to redraw all of them
}
For completeness I like to add my python adaption of #allexrm code which worked really well. Thanks again!
radius = 6371 #Earth Radius in KM
class referencePoint:
def __init__(self, scrX, scrY, lat, lng):
self.scrX = scrX
self.scrY = scrY
self.lat = lat
self.lng = lng
# Calculate global X and Y for top-left reference point
p0 = referencePoint(0, 0, 52.526470, 13.403215)
# Calculate global X and Y for bottom-right reference point
p1 = referencePoint(2244, 2060, 52.525035, 13.405809)
# This function converts lat and lng coordinates to GLOBAL X and Y positions
def latlngToGlobalXY(lat, lng):
# Calculates x based on cos of average of the latitudes
x = radius*lng*math.cos((p0.lat + p1.lat)/2)
# Calculates y based on latitude
y = radius*lat
return {'x': x, 'y': y}
# This function converts lat and lng coordinates to SCREEN X and Y positions
def latlngToScreenXY(lat, lng):
# Calculate global X and Y for projection point
pos = latlngToGlobalXY(lat, lng)
# Calculate the percentage of Global X position in relation to total global width
perX = ((pos['x']-p0.pos['x'])/(p1.pos['x'] - p0.pos['x']))
# Calculate the percentage of Global Y position in relation to total global height
perY = ((pos['y']-p0.pos['y'])/(p1.pos['y'] - p0.pos['y']))
# Returns the screen position based on reference points
return {
'x': p0.scrX + (p1.scrX - p0.scrX)*perX,
'y': p0.scrY + (p1.scrY - p0.scrY)*perY
}
pos = latlngToScreenXY(52.525607, 13.404572);
pos['x] and pos['y] contain the translated x & y coordinates of the lat & lng (52.525607, 13.404572)
I hope this is helpful for anyone looking like me for the proper solution to the problem of translating lat lng into a local reference coordinate system.
Best
Its better to convert to utm coordinates, and treat that as x and y.
import utm
u = utm.from_latlon(12.917091, 77.573586)
The result will be (779260.623156606, 1429369.8665238516, 43, 'P')
The first two can be treated as x,y coordinates, the 43P is the UTM Zone, which can be ignored for small areas (width upto 668 km).
I set the x-axis as logarithmic scale. The maximum value is 10000 and the minimum value is 1.
GraphPane mypane = zedgraphcontrol.GraphPane;
mypane.XAxis.Type = AxisType.Log;
myPane.XAxis.Scale.Min = 1;
myPane.XAxis.Scale.Max = 10000;
But then the graph looks like this :
The values : 10^0, 10^1, 10^2, 10^3, 10^4 repeat at least 2 times. It looks like 2 x-axis overlap each other.
Can anyone tell me what i did wrong ?
Maybe it's just a glitch, I found a workaround by calling AxisChange() before setting the Min and Max:
mypane.XAxis.Type = AxisType.Log;
mypane.AxisChange();
mypane.XAxis.Scale.Min = 1;
mypane.XAxis.Scale.Max = 10000;
I'm using ZedGraph.
I have 2 curves to draw, the first curve is based on the scale of YAxis, while the second one is based on the Y2Axis, the value in first curve is far bigger than the second value.
In my project, both curves are based on YAxis, which makes the graph ugly.
Does anyone has experience to draw second curve based on the Y2Axis?
Here is my code: (What should I change?)
PointPairList p1 = new PointPairList(),
p2 = new PointPairList();
//code to add data into p1 and p2
GraphPane gp = new GraphPane();
gp.AddCurve(p1, "", Color.Black);
gp.AddCurve(p2, "", Color.Blue);
gp.XAxis.Scale.Min = v1;
gp.Y2Axis.Scale.Max = v2;
gp.AxisChange();
gp.XAxis.Scale.IsUseTenPower = false;
gp.Y2Axis.Scale.IsUseTenPower=false;
Thank you.
If I want to set the Y2Axis share the same grid of Y1Axis, after:
LineItem curveY2 = gp.AddCurve(p2, "", Color.Blue);
...
curveY2 .IsY2Axis = true;
i.e., the grid is based on Y1Axis, then Y2Axis has the same grid but with different lable.
For example, Y1Axis is from 1 to 300, and have 7 rows, however Y2Axis has 1 to 20, I want the Y2Axis also have 7 rows (same as the Y1Axis), which function should I use?
LineItem curveY2 = gp.AddCurve(p2, "", Color.Blue);
...
curveY2 .IsY2Axis = true;
//If you have more than one axis on the related side, you have to assign the index of the axis
curveY2 .YAxisIndex = 0;
I asked this question yesterday about storing a plot within an object. I tried implementing the first approach (aware that I did not specify that I was using qplot() in my original question) and noticed that it did not work as expected.
library(ggplot2) # add ggplot2
string = "C:/example.pdf" # Setup pdf
pdf(string,height=6,width=9)
x_range <- range(1,50) # Specify Range
# Create a list to hold the plot objects.
pltList <- list()
pltList[]
for(i in 1 : 16){
# Organise data
y = (1:50) * i * 1000 # Get y col
x = (1:50) # get x col
y = log(y) # Use natural log
# Regression
lm.0 = lm(formula = y ~ x) # make linear model
inter = summary(lm.0)$coefficients[1,1] # Get intercept
slop = summary(lm.0)$coefficients[2,1] # Get slope
# Make plot name
pltName <- paste( 'a', i, sep = '' )
# make plot object
p <- qplot(
x, y,
xlab = "Radius [km]",
ylab = "Services [log]",
xlim = x_range,
main = paste("Sample",i)
) + geom_abline(intercept = inter, slope = slop, colour = "red", size = 1)
print(p)
pltList[[pltName]] = p
}
# close the PDF file
dev.off()
I have used sample numbers in this case so the code runs if it is just copied. I did spend a few hours puzzling over this but I cannot figure out what is going wrong. It writes the first set of pdfs without problem, so I have 16 pdfs with the correct plots.
Then when I use this piece of code:
string = "C:/test_tabloid.pdf"
pdf(string, height = 11, width = 17)
grid.newpage()
pushViewport( viewport( layout = grid.layout(3, 3) ) )
vplayout <- function(x, y){viewport(layout.pos.row = x, layout.pos.col = y)}
counter = 1
# Page 1
for (i in 1:3){
for (j in 1:3){
pltName <- paste( 'a', counter, sep = '' )
print( pltList[[pltName]], vp = vplayout(i,j) )
counter = counter + 1
}
}
dev.off()
the result I get is the last linear model line (abline) on every graph, but the data does not change. When I check my list of plots, it seems that all of them become overwritten by the most recent plot (with the exception of the abline object).
A less important secondary question was how to generate a muli-page pdf with several plots on each page, but the main goal of my code was to store the plots in a list that I could access at a later date.
Ok, so if your plot command is changed to
p <- qplot(data = data.frame(x = x, y = y),
x, y,
xlab = "Radius [km]",
ylab = "Services [log]",
xlim = x_range,
ylim = c(0,10),
main = paste("Sample",i)
) + geom_abline(intercept = inter, slope = slop, colour = "red", size = 1)
then everything works as expected. Here's what I suspect is happening (although Hadley could probably clarify things). When ggplot2 "saves" the data, what it actually does is save a data frame, and the names of the parameters. So for the command as I have given it, you get
> summary(pltList[["a1"]])
data: x, y [50x2]
mapping: x = x, y = y
scales: x, y
faceting: facet_grid(. ~ ., FALSE)
-----------------------------------
geom_point:
stat_identity:
position_identity: (width = NULL, height = NULL)
mapping: group = 1
geom_abline: colour = red, size = 1
stat_abline: intercept = 2.55595281266726, slope = 0.05543539319091
position_identity: (width = NULL, height = NULL)
However, if you don't specify a data parameter in qplot, all the variables get evaluated in the current scope, because there is no attached (read: saved) data frame.
data: [0x0]
mapping: x = x, y = y
scales: x, y
faceting: facet_grid(. ~ ., FALSE)
-----------------------------------
geom_point:
stat_identity:
position_identity: (width = NULL, height = NULL)
mapping: group = 1
geom_abline: colour = red, size = 1
stat_abline: intercept = 2.55595281266726, slope = 0.05543539319091
position_identity: (width = NULL, height = NULL)
So when the plot is generated the second time around, rather than using the original values, it uses the current values of x and y.
I think you should use the data argument in qplot, i.e., store your vectors in a data frame.
See Hadley's book, Section 4.4:
The restriction on the data is simple: it must be a data frame. This is restrictive, and unlike other graphics packages in R. Lattice functions can take an optional data frame or use vectors directly from the global environment. ...
The data is stored in the plot object as a copy, not a reference. This has two
important consequences: if your data changes, the plot will not; and ggplot2 objects are entirely self-contained so that they can be save()d to disk and later load()ed and plotted without needing anything else from that session.
There is a bug in your code concerning list subscripting. It should be
pltList[[pltName]]
not
pltList[pltName]
Note:
class(pltList[1])
[1] "list"
pltList[1] is a list containing the first element of pltList.
class(pltList[[1]])
[1] "ggplot"
pltList[[1]] is the first element of pltList.
For your second question: Multi-page pdfs are easy -- see help(pdf):
onefile: logical: if true (the default) allow multiple figures in one
file. If false, generate a file with name containing the
page number for each page. Defaults to ‘TRUE’.
For your main question, I don't understand if you want to store the plot inputs in a list for later processing, or the plot outputs. If it is the latter, I am not sure that plot() returns an object you can store and retrieve.
Another suggestion regarding your second question would be to use either Sweave or Brew as they will give you complete control over how you display your multi-page pdf.
Have a look at this related question.