How does numpy.histogram works with the normal=True option

Question

I got one question about the numpy.histogram option normed, the function is:

numpy.histogram(a, bins=10, range=None, normed=False, weights=None, density=None)

normed : bool, optional

This keyword is deprecated in Numpy 1.6 due to confusing/buggy behavior. It will be removed in Numpy 2.0. Use the density keyword instead. If False, the result will contain the number of samples in each bin. If True, the result is the value of the probability density function at the bin, normalized such that the integral over the range is 1. Note that this latter behavior is known to be buggy with unequal bin widths; use density instead. weights : array_like, optional

I try it with this code:

   imhist, bins = histogram([0,1,2,3], bins=4, normed=True)
   print "normed=True:", imhist
   print "bins:", bins
   Output:
   normal=True: [ 0.33333333  0.33333333  0.33333333  0.33333333]
   bins: [ 0.    0.75  1.5   2.25  3.  ]

   imhist, bins = histogram([0,1,2,3], bins=4)
   print "normed=None:", imhist
   print "bins:", bins
   Output:
   normal=None: [1 1 1 1]
   bins: [ 0.    0.75  1.5   2.25  3.  ]

What I am feel confused is about the when normed=True, "the result is the value of the probability density function at the bin, normalized such that the integral over the range is 1." Because I thought the imhist should be this:

   normal=True: [0.25  0.25  0.25  0.25]

The 4 values equally drop in 4 bins, and that is why "normal=None: [1 1 1 1]"

   Value:[ 0         1      2      3   ]
   bins: [ 0.    0.75  1.5   2.25  3.  ]

I have refer to this How does numpy.histogram() work? post, but it does use the normed=True option.

@DSM I am confused with the imhist it return, it should be [0.25 0.25 0.25 0.25] but not [ 0.33333333 0.33333333 0.33333333 0.33333333], How come it returns this? 0.33333333 = 1/3, that looks like a probability for one value out of 3 values. — Liao Zhuodi

DSM DSM · Accepted Answer · 2014-09-08T12:46:02

The docs didn't say that it would return values that summed to 1, they said

If True, the result is the value of the probability density function at the bin, normalized such that the integral over the range is 1.

So in your case, it's not that imhist should be [0.25]*4, but:

>>> imhist
array([ 0.33333333,  0.33333333,  0.33333333,  0.33333333])
>>> imhist * np.diff(bins)
array([ 0.25,  0.25,  0.25,  0.25])
>>> (imhist * np.diff(bins)).sum()
1.0

That's the invariant you get. Whenever you change the bins you'll change those values.

How does numpy.histogram works with the normal=True option

2 Answers