Python优化：在字典或列表列表之间进行选择

2023年11月24日 157次阅读

我是
python3的新手,我对解决这个问题的不同方法提出了疑问.关于使用不同数据结构的问题.

我的问题是如何比较不同采样技术的权衡

我在程序中使用字典数据结构来首先解决这个问题.然后我尝试仅使用列表数据结构重写它.我试着考虑排序的好处,我不知道这两种方法之间有什么区别.它似乎没有在两种方法之间产生那么大的差异.

方法1.我使用字典在直方图中创建直方图键和值对

方法2以字符串格式接收源文本并返回列表列表
其中每个子列表中的第一个元素是单词,第二个元素是单词
元素是源文本中的频率

# 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)

最佳答案 哪个更具可读性？我认为字典版本更容易理解.另请注意,您可以将第二个方法中的2元组列表传递给dict构造函数,以重现第一个方法的输出.这应该让您了解这两种实现如何至少在某些方面大致相同.除非这会导致性能问题,否则我不会太担心它.

Python的优势在于您可以用可读的方式以五行编写相同的代码.

import re, random
from collections import Counter

def histogram(text):
    clean_text = re.sub('[.,:;!-[]?', '', text)
    words = clean_text.split()
    return Counter(words)

def random_word(histogram):
    words, frequencies = zip(*histogram.items())
    return random.choices(words, frequencies, k=1)
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