下面是题目要求:
用 Python 实现函数 count_words(),该函数输入字符串 s 和数字 n,返回 s 中 n 个出现频率最高的单词。返回值是一个元组列表,包含出现次数最高的 n 个单词及其次数,即 [(<单词1>, <次数1>), (<单词2>, <次数2>), ... ],按出现次数降序排列。
您可以假设所有输入都是小写形式,并且不含标点符号或其他字符(只包含字母和单个空格)。如果出现次数相同,则按字母顺序排列。
例如:
print count_words("betty bought a bit of butter but the butter was bitter",3)
输出:
[('butter', 2), ('a', 1), ('betty', 1)]
我已经按照单词出现频率排好了顺序,但是怎么在这基础上再实现字母排序?
from operator import itemgetter
sorted([('butter', 2), ('a', 1), ('betty', 1)], key = itemgetter(1,0))
import collections
def count_words(s, n):
return collections.Counter(s.split()).most_common(n)
print count_words("betty bought a bit of butter but the butter was bitter", 3)
下面这个是不借助collections模块
def count_words(s, n):
dic = {}
for key in s.split():
dic[key] = dic[key] + 1 if key in dic else 1
items = sorted(dic.items(), key=lambda d: d[1], reverse=True)
return items[:n]
print count_words("betty bought a bit of butter but the butter was bitter", 3)
這應該滿足你的條件,只是我自認為寫得很醜,等有時間再優化吧:
(改了一下 code 這樣做比較好!)
import collections
def count_words(s, n):
lst = collections.Counter(s.split()).most_common()
lst.sort(key=lambda t: t[0])
lst.sort(key=lambda t: t[1], reverse=True)
return lst[0:n]
print(count_words("betty bought a bit of butter but the butter was bitter", 3))
想了一想,覺得下面更好:
import collections
def count_words(s, n):
lst = collections.Counter(s.split()).most_common()
lst.sort(key=lambda t: (-t[1], t[0]))
return lst[0:n]
print(count_words("betty bought a bit of butter but the butter was bitter", 3))
评论里不能贴图片,我就放这算了。