使用
collections.Counter用于解决第一个问题（计算，结合频率表）以下朱利安Palard的建议下，我实现了第二个问题（从计算频数表百分位数）：
from collections import Counterdef calc_percentiles(cnts_dict, percentiles_to_calc=range(101)): """Returns [(percentile, value)] with nearest rank percentiles. Percentile 0: <min_value>, 100: <max_value>. cnts_dict: { <value>: <count> } percentiles_to_calc: iterable for percentiles to calculate; 0 <= ~ <= 100 """ assert all(0 <= p <= 100 for p in percentiles_to_calc) percentiles = [] num = sum(cnts_dict.values()) cnts = sorted(cnts_dict.items()) curr_cnts_pos = 0 # current position in cnts curr_pos = cnts[0][1] # sum of freqs up to current_cnts_pos for p in sorted(percentiles_to_calc): if p < 100: percentile_pos = p / 100.0 * num while curr_pos <= percentile_pos and curr_cnts_pos < len(cnts): curr_cnts_pos += 1 curr_pos += cnts[curr_cnts_pos][1] percentiles.append((p, cnts[curr_cnts_pos][0])) else: percentiles.append((p, cnts[-1][0])) # we could add a small value return percentilescnts_dict = Counter()for segment in segment_iterator: cnts_dict += Counter(segment)percentiles = calc_percentiles(cnts_dict)
