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python中怎么對(duì)列表以區(qū)間進(jìn)行統(tǒng)計(jì)?

瀏覽:178日期:2022-09-09 17:29:27

問(wèn)題描述

python中怎么對(duì)列表以區(qū)間進(jìn)行統(tǒng)計(jì)?假設(shè)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(gè)區(qū)間分別統(tǒng)計(jì),如統(tǒng)計(jì)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(wèn)題解答

回答1:

# 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(guò)的問(wèn)題: Python-QA

回答2:

# 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

標(biāo)簽: Python 編程
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