python中,怎么做个字典,数句子中单词出现的次数? python中，怎么做个字典，数句子中单词出
python\u4e2d\uff0c\u600e\u4e48\u505a\u4e2a\u5b57\u5178\uff0c\u6570\u53e5\u5b50\u4e2d\u5355\u8bcd\u51fa\u73b0\u7684\u6b21\u6570\u4ee5\u4e0b\u662f\u4e00\u79cd\u53ef\u80fd\u7684\u8ba1\u7b97\u65b9\u6cd5\u3002
sentence='''by: William Shakespeare From fairest creatures we desire increase,That thereby beauty's rose might never die,But as the riper should by time decease,His tender heir might bear his memory;But thou, contracted to thine own bright eyes,Feed'st thy light's flame with self-substantial fuel,Making a famine where abundance lies,Thyself thy foe, to thy sweet self too cruel.Thout that are now the world's fresh ornamentAnd only herald to the gaudy spring,Within thine own bud buriest thy contentAnd, tender churl, mak'st waste in niggarding.Pity the world, or else this glutton be,To eat the world's due, by the grave and thee.'''words=sentence.split()from string import punctuationwords=[w.strip() for w in words]for p in punctuation: words=[w.strip(p) for w in words]diff_words=set(words)stat={}for w in diff_words: stat[w]=words.count(w)print(stat)
def get_words(sentence):
return sentence.split(" ")
def get_word(word):
if word.isalpha():
return word
else:
return word[:-1]
if __name__ == "__main__":
test = "this is a test, thanks! this is another test, thanks!"
test_words = get_words(test)
test_words_clear = list()
for word in test_words:
test_words_clear.append(get_word(word))
result = dict()
for word in test_words_clear:
if result.get(word, None) is None:
result[word] = 1
else:
result[word] += 1
print(result)
words = text.split()
wordscount = [words.count(elem) for elem in words]
worddict={map(None,words,wordscount)}
要完成你的目标：
我要
计算每个词语出现的次数，
把词语出现的次数和词语列表组合成字典；
伪代码出来了，程序也就有了....
python有着强大的列表解析，内建模块使用C完成，他们很快，所以能够调用内建模块你就不要自己多事。
尽量按照伪代码去完成程序，除非必须细化，否则让python为你操作低层吧，他很聪明
第三句释义：
对于列表元素计数有很简单的list.count(）方法。
这个语句就是利用了这个，statement for element in list fileter expression这是list的解析式。
通过这个你可以方便的将一个list解析为另一个。这一句就对list中所有的元素运用count()方法，并且建立新的list。
另外一个例子：a=[2,3,4,7,8]找到这个list中的偶数，并得到它们的平方列表
这样写：[elem*elem for elem in a if elem%2==0]
第四句释义：
list有个map方法，它可以提供list的映射。map(statement,list1,list2)就是将list1,list2按照statement规则映射。我的表达式是none所以得到的结果就是简单的[('this', 3), ('is', 4), ('a', 1)]这样的二元tuple列表。
dict函数则是针对上述list的。它能把上述list转化为字典。
这在你做数据处理时很有好处，你可以很轻松的建立一个hash表。
python的强大在于简明，遇到程序尝试着用最简单地方法去解决，你会很快喜欢上的。
以上^乐于与您交流
words 已经得到了，用一个 word_dict 当作 map 统计频度就可以了:
text = raw_input("Enter a sentence:")
words = text.split()
word_dict = {}
for w in words:
if w not in word_dict:
word_dict[w] = 1
else:
word_dict[w] = word_dict[w] + 1
print word_dict
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