python自动调整单调分箱代码

很高兴能为您提供帮助！以下是一份Python自动调整单调分箱的代码示例：

import numpy as np
import pandas as pd
from sklearn.tree import DecisionTreeRegressor

def monotonic_binning(x, y, max_bins=10, min_samples=100):
    # create initial bins
    initial_bins = pd.qcut(x, q=max_bins, duplicates='drop').unique().categories
    bins = initial_bins
    # loop until bins are monotonic
    while True:
        # calculate mean y for each bin
        df = pd.DataFrame({'x': x, 'y': y, 'bins': pd.cut(x, bins=bins)})
        mean_y = df.groupby('bins')['y'].mean()
        # check if bins are monotonic
        if np.all(mean_y.diff().fillna(0) >= 0) or np.all(mean_y.diff().fillna(0) <= 0):
            break
        # if bins are not monotonic, merge adjacent bins with closest mean y
        merge_idx = (mean_y - mean_y.shift()).abs().argmin()
        bins = sorted(set(list(bins) + [mean_y.index[merge_idx], mean_y.index[merge_idx+1]]))
    # check if each bin has enough samples
    df = pd.DataFrame({'x': x, 'y': y, 'bins': pd.cut(x, bins=bins)})
    bin_counts = df.groupby('bins')['y'].count()
    if (bin_counts < min_samples).any():
        # if any bin has less than min_samples, merge it with the adjacent bin with closest mean y
        mean_y = df.groupby('bins')['y'].mean()
        merge_idx = (mean_y - mean_y.shift()).abs().argmin()
        bins = sorted(set(list(bins) + [mean_y.index[merge_idx], mean_y.index[merge_idx+1]]))
        # recursively call monotonic_binning with updated bins
        return monotonic_binning(x, y, max_bins=max_bins, min_samples=min_samples)
    else:
        return bins

# example usage
x = np.random.normal(size=1000)
y = x**2 + np.random.normal(size=1000)
bins = monotonic_binning(x, y, max_bins=10, min_samples=100)
tree = DecisionTreeRegressor(max_depth=3)
tree.fit(pd.cut(x, bins=bins).cat.codes.values.reshape(-1, 1), y)

请注意，这只是一个示例代码，您可能需要根据您的数据和需求进行调整。如果您有任何问题或需要进一步的帮助，请随时联系我！