Histogram based on frequency or count data

frequency_histogram.py

import numpy as np
import pandas as pd


def frequency_histogram(
    data: pd.DataFrame,
    n_bins=20,
    bins=None,
    log_bins=False,
    normalize=False,
    for_bar=True):

    x, f, *_ = data.columns

    xmin, xmax = data[x].min(), data[x].max()

    if log_bins and bins is None:
        bins = np.logspace(np.floor(np.log10(xmin)), np.log10(xmax), num=n_bins+1)
    elif bins is None:
        bins = np.linspace(xmin, xmax, num=n_bins+1)

    assignments = pd.cut(data[x], bins, include_lowest=True)

    totals = data.groupby(assignments)[f].sum()

    edges = (
        totals.index  # `totals` has a Categorical index
        .categories   # the categories are represented as an ``IntervalIndex``
        .to_tuples()  # convert the ``IntervalIndex`` to a regular index of tuples
        .tolist()     # convert the regular index to a list of tuples
    )
    edges = pd.DataFrame(edges, columns=['edge_left', 'edge_right'])

    result = edges.join(totals.reset_index(drop=True))

    if for_bar:
        # Give a result that's useful for ``matplotlib.pyplot.bar`` that
        # has left edges and bin widths.
        #
        # CAUTION You still need to use `plt.bar( … , align='edge')`
        result['edge_right'] = result['edge_right'] - result['edge_left']
        result.rename(columns={'edge_right': 'edge_width'}, inplace=True)

    if normalize:
        # TODO CAUTION This won't produce a properly normalized density - we
        # actually want sum(bin width * bin height) to be 1, but this only ensures
        # sum(bin height) is 1
        result[f] = result[f] / result[f].sum()

    return result


def histogram_logarithmic(a, w, bins=10):
    amin, amax = a.min(), a.max()
    edge = np.logspace(np.floor(np.log10(amin)), np.log10(amax), num=bins+1)
    height, edge = np.histogram(a, bins=edge, weights=w, density=True)
    width = np.diff(edge)
    left_edge = edge[:-1]
    return height, left_edge, width