我正在制作一个如下所示的小程序
"""Count words."""
# TODO: Count the number of occurences of each word in s
# TODO: Sort the occurences in descending order (alphabetically in case of ties)
# TODO: Return the top n words as a list of tuples (<word>, <count>)
from operator import itemgetter
def count_words(s, n):
"""Return the n most frequently occuring words in s."""
t1=[]
t2=[]
temp={}
top_n={}
words=s.split()
for word in words:
if word not in temp:
t1.append(word)
temp[word]=1
else:
temp[word]+=1
t1 = sorted(temp,key=temp.get,reverse=True) # to get sorted keys
t2 = sorted(temp.values(),reverse=True) # to get sorted values
top_n = dict(zip(t1,t2))
print top_n
return
def test_run():
"""Test count_words() with some inputs."""
count_words("cat bat mat cat bat cat", 3)
count_words("betty bought a bit of butter but the butter was bitter", 3)
if __name__ == '__main__':
test_run()
我只是想对键值对进行排序。我有以下问题:
答案 0 :(得分:3)
根据顶部的注释,您希望返回键/值元组的列表。因此,您希望按值
对字典的项进行排序sorted(temp.items(), key=itemgetter(1), reverse=True)
请注意,您分别对键和值进行排序的策略不起作用 - 您最终会将键与不属于一起的值进行匹配。
另请注意,collections.Counter
已针对完成此任务进行了优化(请参阅.most_common
)
答案 1 :(得分:1)
sorted([(value,key) for (key,value) in temp.items()])
答案 2 :(得分:0)
您可以按以下值对字典进行排序:
sorted(temp.items(), key = lambda x: x[1], reverse = True)