Histogram Bins

The problem, from first principles

A histogram is the first plot anyone draws of a numeric column, and everything it shows depends on a choice made before any ink is spent: how wide the bins are. Too few bins and a two-humped distribution looks like one hump. Too many and the picture is a comb of single-row spikes that shows noise instead of shape. A chart is a drawing of numbers, so what can be tested is the numbers behind it: the bin edges and the count in each bin. This question builds both, and two standard rules for choosing how many bins there should be.

From theory to code

Implement sturges_bins(n), the bin count from Sturges' rule, then freedman_diaconis_bins(x), the bin count from the Freedman-Diaconis rule, then bin_edges(x, n_bins), the equal-width edges spanning the data, then histogram_counts(x, edges), how many values fall in each bin. The signatures and docstrings are already in the editor.

Constraints

• x is a 1D float array with at least two values and at least two distinct values. n is the number of values.
• sturges_bins(n) returns ceil(log2(n)) + 1 as an int, for n >= 1.
• freedman_diaconis_bins(x) uses bin width 2 * IQR / n ** (1/3), where IQR is the difference between the 75th and 25th percentiles (linear interpolation, as np.percentile does by default). It returns ceil((max - min) / width) as an int, and at least 1. If the IQR is zero it falls back to sturges_bins(len(x)).
• bin_edges(x, n_bins) returns n_bins + 1 equally spaced edges from min(x) to max(x) inclusive.
• histogram_counts(x, edges) returns an integer array of length len(edges) - 1. A value v belongs to bin i when edges[i] <= v < edges[i + 1], except that the last bin is closed on the right, so v == edges[-1] counts in the last bin. Values outside [edges[0], edges[-1]] are ignored. Do not call np.histogram.

Hints