Question

for word in list6:
    if word = "TRUMP":

所以，我在辩论记录中列出了每个单词。当特朗普说话时，它以“TRUMP”开头。我需要接受他的话并把它们分成一个单独的列表。如果list6中的单词是“TRUMP”，那么我需要将所有内容放入列表中，直到它显示另一个人的姓名。他不止一次说话。

我只需要帮助完成这个循环。

Answer 1

list6 = ['TRUMP','I','am','good', 'HILLARY','I','am','good','too','TRUMP','But','How?']
person_words = {'TRUMP':[], 'HILLARY':[]}

person_names = person_words.keys()

one_person_onetime_words = []

for word in list6:
    if word in person_names:
        if len(one_person_onetime_words):
            person_words[this_person].append(one_person_onetime_words)
            one_person_onetime_words = []
        this_person = word
    else:
        one_person_onetime_words.append(word)

person_words[this_person].append(one_person_onetime_words)

print person_words

给出

{'HILLARY': [['I', 'am', 'good', 'too']], 'TRUMP': [['I', 'am', 'good'], ['But', 'How?']]}

所以，这只是一次性提供所有人的所有不同的谈话。

正如您在对自己问题的评论中所提到的，如果您只想获得一个人的话，可以使用以下内容：

from copy import copy

list6 = ['TRUMP','I','am','good', 'HILLARY','I','am','good','too','TRUMP','But','How?']
person_words = []
all_persons = ['TRUMP', 'HILLARY']
person_looking_for = 'TRUMP'

filter_out_persons = copy(all_persons)
filter_out_persons.remove(person_looking_for)

person_onetime_words = []

capture_words = False
for word in list6:
    if word == person_looking_for:
        capture_words = True
        if len(person_onetime_words):
            person_words.append(person_onetime_words)
            person_onetime_words = []
    elif word not in filter_out_persons and capture_words:
        person_onetime_words.append(word)
    else:
        capture_words = False

person_words.append(person_onetime_words)
print "{}'s words".format(person_looking_for)
print person_words

这给了

TRUMP's words
[['I', 'am', 'good'], ['But', 'How?']]

并且，以下将给出一个单词作为键的字典，该值将再次成为一个字典，每个人的频率为该单词。

import pprint

list6 = ['TRUMP','I','am','good', 'HILLARY','I','am','good','too','TRUMP','But','How?']

person_names = ['TRUMP','HILLARY']

word_frequency = {}
for word in list6:
    if word in person_names:
        person = word
    else:
        word = word.lower()
        if word in word_frequency:
            if person in word_frequency[word]:
                word_frequency[word][person] += 1
            else:
                word_frequency[word][person] = 1
        else:
            word_frequency[word] = {person: 1}

pprint.pprint(word_frequency)

给出

{'am': {'HILLARY': 1, 'TRUMP': 1},
 'but': {'TRUMP': 1},
 'good': {'HILLARY': 1, 'TRUMP': 1},
 'how?': {'TRUMP': 1},
 'i': {'HILLARY': 1, 'TRUMP': 1},
 'too': {'HILLARY': 1}}

蟒蛇。如何列出较大列表的一小部分？

1 个答案: