histogram

I am trying to plot a several histogram on the same plot but I figured out that some colors are assigned to different series, which bother me a little. Is there a way of forcing color bars to be unique ? That works for small data set, but when I use a lot of data, I see this problem coming back here is an example, the blue color is assigned twice to two different data samples All the examples and the solutions to attribute colors to histograms in matplotlib (at least those I found) are suggesting to normalize x axis between 0 and 1 like this example , but this is not what I want to have

I need to plot a weighted histogram of density rather than frequency. I know that freq = FALSE is available in hist() but you can't specify weights. In ggplot2 I can do this: library(ggplot2) w <- seq(1,1000) w <-w/sum(w) v <- sort(runif(1000)) foo <- data.frame(v, w) ggplot(foo, aes(v, weight = w)) + geom_histogram() But where is the equivalent of freq = FALSE ? By default, geom_histogram() will use frequency rather than density on the y-axis. However, you can change this by setting your y aesthetic to ..density.. like so: ggplot(foo, aes(x = v, y = ..density.., weight = w)) + geom_histogram(

I have a pre-binned frequency table for a rather large dataset. That is, a single column vector of bins and a single column vector of counts associated with those bins. I'd like R to plot a histogram of this data by doing further binning and summing the existing counts. For example, if in the pre-binned data I have something like [(0.01, 5000), (0.02, 231), (0.03, 948)], where the first number is the bin and the second is the count, and I choose 0.04 as the new bin width, I'd expect to get [(0.04, 6179)]. What's the fastest and or easiest way to do this in R? Looks like ggplot2 has the answer.

For a given Array[Double] , for instance val a = Array.tabulate(100){ _ => Random.nextDouble * 10 } what is a simple approach to calculate a histogram with n bins ? A very similar preparation of values as in @om-nom-nom 's answer, yet the histogram method quite small by using partition , case class Distribution(nBins: Int, data: List[Double]) { require(data.length > nBins) val Epsilon = 0.000001 val (max,min) = (data.max,data.min) val binWidth = (max - min) / nBins + Epsilon val bounds = (1 to nBins).map { x => min + binWidth * x }.toList def histo(bounds: List[Double], data: List[Double]):

Suppose I have a dataframe (df) (Pandas) or RDD (Spark) with the following two columns: timestamp, data 12345.0 10 12346.0 12 In Pandas, I can create a binned histogram of different bin lengths pretty easily. For example, to create a histogram over 1 hr, I do the following: df = df[ ['timestamp', 'data'] ].set_index('timestamp') df.resample('1H',how=sum).dropna() Moving to Pandas df from Spark RDD is pretty expensive for me (considering the dataset). Consequently, I prefer to stay within the Spark domain as much as possible. Is there a way to do the equivalent in Spark RDD or dataframes?

Right so I have a table such as this in PostgreSQL: timestamp duration 2013-04-03 15:44:58 4 2013-04-03 15:56:12 2 2013-04-03 16:13:17 9 2013-04-03 16:16:30 3 2013-04-03 16:29:52 1 2013-04-03 16:38:25 1 2013-04-03 16:41:37 9 2013-04-03 16:44:49 1 2013-04-03 17:01:07 9 2013-04-03 17:07:48 1 2013-04-03 17:11:00 2 2013-04-03 17:11:16 2 2013-04-03 17:15:17 1 2013-04-03 17:16:53 4 2013-04-03 17:20:37 9 2013-04-03 17:20:53 3 2013-04-03 17:25:48 3 2013-04-03 17:29:26 1 2013-04-03 17:32:38 9 2013-04-03 17:36:55 4 And I would like to get the following output: timestampwindowstart = 2013-04-03 15:44:58