histogram

Suppose I need to plot a dataset like below: set.seed(1) dataset <- sample(1:7, 1000, replace=T) hist(dataset) As you can see in the plot below, the two leftmost bins do not have any space between them unlike the rest of the bins. I tried changing xlim, but it didn't work. Basically I would like to have each number (1 to 7) represented as a bin, and additionally, I would like any two adjacent bins to have space beween them...Thanks! The best way is to set the breaks argument manually. Using the data from your code, hist(dataset,breaks=rep(1:7,each=2)+c(-.4,.4)) gives the following plot: The

I want to plot with ggplot the frequency of values from a numeric vector. With plot() is quite straight forward but I can't get the same result with ggplot . library(ggplot2) dice_results <- c(1,3,2,4,5,6,5,3,2,1,6,2,6,5,6,4) hist(dice_results) ggplot(dice_results) + geom_bar() # Error: ggplot2 doesn't know how to deal with data of class numeric Should I create a dataframe for ggplot() to plot my vector? Please look at the help page ?geom_histogram . From the first example you may find that this works. qplot(as.factor(dice_results), geom="histogram") Also look at ?ggplot . You will find that

OK so I have six possible values for data to be which are '32', '22', '12', '31', '21' and '11'. I have these stored as strings. Is it possible for python to sort through the data and just make six bins and show how many of each I have? Or do the inputs to a histogram HAVE to be numerical? data = ['32', '22', '12', '32', '22', '12', '31', '21', '11'] dict((x, data.count(x)) for x in data) Result {'11': 1, '12': 2, '21': 1, '22': 2, '31': 1, '32': 2} Did you consider using collections.Counter ? # python 2.7 >>> l = ['32', '22', '12', '31', '21', '11', '32'] >>> import collections >>>

I'm trying to make a histogram in [R], and the normal curve that describes the histogram as follows: w<-rnorm(1000) hist(w,col="red",freq=F,xlim=c(-5,5)) curve(dnorm(w),-5,5,add=T,col="blue") But when I try to plot the normal curve by curve function shows me the following error: Error en curve(dnorm(w), -5, 5, add = T, col = "blue") : 'expr' must be a function, or a call or an expression containing 'x' What am I doing wrong? You just need to drop the "w" argument to dnorm in curve : w<-rnorm(1000) hist(w,col="red",freq=F,xlim=c(-5,5)) curve(dnorm,-5,5,add=T,col="blue") To use something other

An array of frequency is created. int arr[10]; arr[0] = 24; arr[1] = 28; arr[2] = 23; arr[3] = 29; //and so on How do i draw a histogram using OpenCV in C++ based on the array? Miki OpenCV is not really suited for plotting. However, for simple histograms, you can draw a rectangle for each column (eventually with alternating colors) This is the code: #include <opencv2\opencv.hpp> #include <algorithm> using namespace std; using namespace cv; void drawHist(const vector<int>& data, Mat3b& dst, int binSize = 3, int height = 0) { int max_value = *max_element(data.begin(), data.end()); int rows = 0;

I want to visualise the proportional data (Nij/n) about the sinus (independent) and arr/AHB (dependent variable) cases in females and males by R. ggplot2 approach and any other is welcome! Pseudocode histogram of the second and third columns for the groups N11.1, ..., N32.1 Code N11.1 N22.1 N33.1 N44.1 N21.1 N31.1 N32.1 Sinus 1.0 0.0 0.0 0.0 0.0 0.0 12.0 Arr/AHB 1.0 0.0 0.0 0.1 0.0 0.0 20.9 N11.1 N22.1 N33.1 N44.1 N21.1 N31.1 N32.1 Sinus 1.0 0.0 0.0 0.0 0.0 0.0 4.0 Arr/AHB 1.0 0.0 0.0 0.0 0.0 0.0 24.0 The first column has the row.names. Code with the data library("ggplot2") data.female <-

I have the following data: 1 3215 2 321 ... 31_60 59 60+ 32 I would like to generate histogram using gnuplot and put the value of bar on top of it. Here is the gnuplot command I tried to create histogram: set style data histogram set xtics rotate plot 'file.dat' using 2:xtic(1) Can someone tell me how to add values on top of the bars generated? I found the following link related histogram ( http://gnuplot-tricks.blogspot.com/2009/10/more-on-histograms.html ), but didnt get what its doing exactly. With this as some sample data file Data.dat : 1 10 2 20 3 15 4 16 5 19 6 5 You could run this