【发布时间】：2018-11-01 04:02:07
【问题描述】：
我正在创建一个直方图，显示文本文件中的单词和每个单词的频率。
我查看了我之前的代码，该代码有效并试图使其模块化。这是课堂练习，因为我们将来会创建一个推文生成器。 我在某处做错了事，但我看不到这对我的生活有什么影响。
我是从纯文本文件创建的：
- 列表列表
- 元组列表
- 键值对字典
这是我目前所拥有的：
import re
import sys
import string
def read_file(file):
# first_list = [] ### Unsure if I should actually keep these in here.
# second_list = []###
document_text = open(file, 'r')
text_string = document_text.read().lower()
match_pattern = re.findall(r'\b[a-z]{1, 15}\b', text_string)
return match_pattern
#----------LIST_OF_LISTS---------------
def list_of_lists(match_pattern):
read_file(file)
match_pattern.sort()
list_array = []
count = 0
index = None
for word in match_pattern:
if word == index:
count += 1
else:
list_array.append([index, count])
index = word
count = 1
else:
list_array([index, count])
list_array.pop(0)
return str(list_array)
#------END OF LIST_OF_LISTS-------------
#----------LIST_OF_TUPLES---------------
def list_of_tuples(match_pattern):
read_file(file)
frequency = {}
first_list = []
second_list = []
unique_count = 0
for word in match_pattern:
count = frequency.get(word, 0)
frequency[word] = count + 1
first_list.append(word)
if int(frequency[word]) == 1:
unique_count += 1
for word in match_pattern:
second_list.append(int(frequency[word]))
zipped = zip(first_list, second_list)
return list(set((zipped)))
return str("There are " + str(unique_count) + " words in this file")
#----------END OF LIST_OF_TUPLES---------
#----------DICTIONARY FUNCTION-----------
def dictionary_histogram(match_pattern):
dict_histo = {}
for word in match_pattern:
if word not in dict_histo:
dict_histo[word] = 1
else:
dict_histo[word] += 1
return str(dict_histo)
def unique_word_dict(histogram):
''' Takes the histogram and returns the amount of unique words withi it.'''
return len(histogram.keys())
def frequency(histogram, word):
'''takes in the histogram and a word, then returns a value of the word if the
key exists within the dictionary, else return 0'''
if word in histogram:
return str(histogram[word])
else:
return str(0)
#------------End of Dictionary-----------------
#
# def unique_word(histogram):
# ''' Takes the histogram and returns the amount of unique words withi it.'''
# return len(histogram)
#
# def frequency(word, histogram):
# '''takes a histogram and a word, then returns the value of the word.'''
# return histogram[word]
if __name__ == '__main__':
file = str(sys.argv[1])
read_file(file)
list_of_tuples(match_pattern)
虽然，我确实相信我的 如果 name == 'main'： 是错误的，但我确实尝试了几种不同的变体，但似乎没有什么对我有用。
我也尝试过改变一些东西，但这也没有用。
import re
import sys
import string
def read_file(file):
document_text = open(file, 'r')
text_string = document_text.read().lower()
# match_pattern = re.findall(r'\b[a-z]{1, 15}\b', text_string) ### Think I should move this to the actual list function maybe???
### I originally had it return match_pattern and then I used match_pattern in my list functions i.e list_of_lists(match_pattern)
document_text.close()
return text_string
#----------LIST_OF_LISTS---------------
def list_of_lists(text_string):
match_pattern = re.findall(r'\b[a-z]{1, 15}\b', text_string)
# match_pattern.sort() #Maybe use this
list_array = []
count = 0
index = None
for word in match_pattern:
if word == index:
count += 1
else:
list_array.append([index, count])
index = word
count = 1
else:
list_array.append([index, count])
list_array.pop(0)
return str(list_array)
#------END OF LIST_OF_LISTS-------------
#----------LIST_OF_TUPLES---------------
def list_of_tuples(text_string):
match_pattern = re.findall(r'\b[a-z]{1, 15}\b', text_string)
frequency = {}
first_list = []
second_list = []
unique_count = 0
for word in match_pattern:
count = frequency.get(word, 0)
frequency[word] = count + 1
first_list.append(word)
if int(frequency[word]) == 1:
unique_count += 1
for word in match_pattern:
second_list.append(int(frequency[word]))
zipped = zip(first_list, second_list)
# return list(set((zipped)))
return str(list(set(zipped)))
# return str("There are " + str(unique_count) + " words in this file")
#----------END OF LIST_OF_TUPLES---------
#----------DICTIONARY FUNCTION-----------
def dictionary_histogram(text_string):
dict_histo = {}
for word in match_pattern:
if word not in dict_histo:
dict_histo[word] = 1
else:
dict_histo[word] += 1
return str(dict_histo)
def unique_word_dict(histogram):
''' Takes the histogram and returns the amount of unique words withi it.'''
return len(histogram.keys())
def frequency(histogram, word):
'''takes in the histogram and a word, then returns a value of the word if the
key exists within the dictionary, else return 0'''
if word in histogram:
return str(histogram[word])
else:
return str(0)
#------------End of Dictionary-----------------
#
# def unique_word(histogram):
# ''' Takes the histogram and returns the amount of unique words withi it.'''
# return len(histogram)
#
# def frequency(word, histogram):
# '''takes a histogram and a word, then returns the value of the word.'''
# return histogram[word]
# read_file(file)
# list_of_tuples(read_file(file))
if __name__ == '__main__':
file = str(sys.argv[1])
# print(list_of_lists(read_file(file)))
【问题讨论】：
标签： python-3.x function oop histogram modular