問題描述

python中怎么對列表以區(qū)間進行統(tǒng)計？假設list=[1,1,1,2,3,4,4,5,5,6,7,7,7,7,8,9,9,9,10……99，99，99，100，100]

怎么寫程序可以以10為一個區(qū)間分別統(tǒng)計，如統(tǒng)計出小于10的數(shù)字頻率，大于10小于20的頻率，大于20小于30的頻率……大于90小于100的頻率？抱歉題目描述的不好

問題解答

# code for python3from itertools import groupbylst = [1, 1, 1, 2, 3, 4, 4, 5, 5, 6, 7, 7, 7, 7, 8, 9, 9, 9, 10, 99, 99, 99, 100, 100]dic = {}for k, g in groupby(lst, key=lambda x: (x-1)//10): dic[’{}-{}’.format(k*10+1, (k+1)*10)] = len(list(g)) print(dic)

結(jié)果:

{’91-100’: 5, ’1-10’: 19}

我回答過的問題: Python-QA

# coding: utf-8lst = [1, 1, 1, 2, 3, 4, 4, 5, 5, 6, 7, 7, 7, 7, 8, 9, 9, 9, 10, 99, 99, 99, 100, 100]intervals = {’{0}-{1}’.format(10 * x + 1, 10 * (x + 1)): 0 for x in range(10)}for _ in lst: for interval in intervals:start, end = tuple(interval.split(’-’))if int(start) <= _ <= int(end): intervals[interval] += 1print intervals