我是新手Python3开发者,我有一个关于解决问题的不同方法的问题,涉及使用不同的数据结构。我的问题是如何比较不同采样技术之间的权衡。在我的程序中,我首先使用字典数据结构来解决这个问题。然后我尝试只使用列表数据结构来重写它。我尝试考虑排序的好处,但我无法确定两种方法之间的区别。似乎两种方法之间并没有太大的区别。
方法1:我使用字典创建直方图键和值对。
方法2:它接受字符串格式的源文本,并返回一个由列表组成的列表,其中每个子列表的第一个元素是单词,第二个元素是它在源文本中的频率。
方法1:我使用字典创建直方图键和值对。
方法2:它接受字符串格式的源文本,并返回一个由列表组成的列表,其中每个子列表的第一个元素是单词,第二个元素是它在源文本中的频率。
# This program Analyze word frequency in a histogram
# sample words according to their observed frequencies
# takes in a source text in string format and returns a dictionary
# in which each key is a unique word and its value is that word's
# frequency in the source text
import sys
import re
import random
import time
def histogram(source_text):
histogram = {}
# removing any sort of string, removing any other special character
for word in source_text.split():
word = re.sub('[.,:;!-[]?', '', word)
if word in histogram:
histogram[word] += 1
else:
histogram[word] = 1
return histogram
def random_word(histogram):
probability = 0
rand_index = random.randint(1, sum(histogram.values()))
# Algorithm 1
for (key, value) in histogram.items():
for num in range(1, value + 1):
if probability == rand_index:
if key in outcome_gram:
outcome_gram[key] += 1
else:
outcome_gram[key] = 1
# return outcome_gram
return key
else:
probability += 1
# Method 2 takes in a source text in string format and returns a list #of lists
# in which the first element in each sublist is the word and the #second element is its frequency in the source texts
# Algorithm 2
# for word in histogram:
# probability += histogram[word]
# if probability >= rand_index:
# if word in outcome_gram:
# outcome_gram[word] += 1
# else:
# outcome_gram[word] = 1
# return word
if __name__ == "__main__":
outcome_gram = {}
dict = open('./fish.txt', 'r')
text = dict.read()
dict.close()
hist_dict = histogram(text)
for number in range(1, 100000):
random_word(hist_dict)
collections.Counter(my_text_corpus)
。 - Joran Beasley