Problems specifying number of bins in Histogram

Question

So I wanted to create a Histogram with a custom number of bins. Luckily there is a simple function for this in the form of Histogram[data, n] where n specifies the number of bins. There is even an example of using it in the documentation:

data = RandomVariate[NormalDistribution[0, 1], 200];
Histogram[data, 5]

Problem is that this doesn't produce a Histogram with five bins but with seven bins:

Histogram[data, 6]still gives seven bins:

Histogram[data, 7] also gives seven bins:

But Histogram[data, 10] gives thirteen bins:

Now I am utterly confused. Is this a bug or have I misunderstood something completely? How would I go about producing a Histogram with five bins?

Answer 1

Under mma 8 you can use the undocumented {"Raw", n} bin specification to get exactly the number of bins you would like. Otherwise the bin widths and boundaries are chosen to be "nice" numbers.

Here is an example:

data = RandomVariate[NormalDistribution[0, 1], 200];
Histogram[data, {"Raw", 5}]

(I saw this first in a comment by Brett Champion to the answer here.)

Answer 2

This is known and it seems it is intentional. Even on the doc page of Histogram you can find examples of this behavior. It looks like the number specification is only seen as an order of magnitude indicator. A workaround would be to specify bin lists yourself.

Histogram[data, {-2, 3, 1}]

Answer 3

Looking through the internal code for Histogram and following the rabbit hole, this behaviour is clearly intentional and the documentation is definitely misleading. The code for Histogram has a "main" function that looks something like the following:

mainFunction[args__, o : OptionsPattern[]] := 
 Block[{data, stuff, width, height},
    data = First@{args};
    stuff = Rest@{args};
    {width, height} = Switch[Length@stuff,
            0, {Automatic, Automatic},
            1, {First@stuff, Automatic},
            2, stuff
        ];
    ...
  ]

If you continue down other related functions, you'll observe that nowhere do they interpret the argument as number of bins, but only as widths, which are then smoothed with a default smoothing function.

Hence, even though the documentation says that using n for bspec uses n bins, the implementation does not reflect it.

Answer 4

As an alternative to Sjoerd C. de Vries answer, you could construct your own histogram with the numbers of bins you want to have with BinCounts Just as an easy example (I am sure it can be better written and a lot of things can be done better):

data = RandomVariate[NormalDistribution[0, 1], 200];      
startValue=-2.5;
endValue=3.5;
binWidth=1;
scale=Table[startValue+i*binWidth,{i,0,(endValue-startValue)/binWidth-1}](*shift the bin center as you like!*);
binnedData=BinCounts[data,{startValue,endValue,binWidth}];
forPlot=Transpose[{scale,binnedData}];

By using

ListPlot[forPlot,InterpolationOrder->0,Filling->Axis,Joined→True]

You will get: