解释代码:words_list = [five_high_word,four_high_word,three_high_word,two_high_word,one_high_word] score_dict = dict() frequency_dict = dict() score = 2 for words in words_list: for word in words: if word in score_dict: score_dict[word] += score frequency_dict[word] += 1 else: score_dict[word] = score frequency_dict[word] = 1 score -= 1 del score,five_high_word,four_high_word,three_high_word,two_high_word,one_high_word,words_list
时间: 2024-04-14 09:31:40 浏览: 9
这段代码的作用是将多个单词列表中的单词进行计分和统计频率,并将结果存储在两个字典中。
首先,定义了一个包含五个单词列表的`words_list`。接下来,创建了两个空字典`score_dict`和`frequency_dict`,用于存储计分和频率信息。
然后,定义了一个初始分数`score`为2。通过两层循环遍历`words_list`中的每个单词。对于每个单词,首先检查它是否已经存在于`score_dict`字典中。如果存在,则将其对应的分数和频率分别增加`score`和1;如果不存在,则将其添加到字典中,并设置初始分数为`score`和频率为1。
在每次内层循环结束后,将分数`score`减1,以便下一个单词的分数降低。
最后,使用`del`语句删除了代码中不再使用的变量,以释放内存空间。
总结起来,该代码通过循环遍历多个单词列表,对单词进行计分和统计频率,并将结果存储在字典中。
相关问题
优化下列python代码中的递归函数:def words_divi(words, n): division_words = words / n print(division_words) if division_words <= words_max: if isinstance(division_words, int): for i in range(n): sram_list.append(str(division_words) + 'x' + str(bits_max)) else: round(division_words) if division_words in words_list: for i in range(n): sram_list.append(str(division_words) + 'x' + str(bits_max)) else: while division_words < words_max: division_words = division_words + 1 if division_words in words_list: for i in range(n): sram_list.append(str(division_words) + 'x' + str(bits_max)) else: n = n + 1 print(n) words_divi(words, n) if __name__ == "__main__": words_list = [] bits_list = [] words_list.append(list(range(32, 1025, 16)) + list(range(1056, 8193, 16))) bits_list.append(list(range(16, 145, 1))) words = 16388 bits = 148 words_max = max(words_list[0]) bits_max = max(bits_list[0]) words_min = min(words_list[0]) bits_min = min(bits_list[0]) sram_list = [] num = 2 words_divi(words, num) print(sram_list)
可以使用尾递归优化来优化该递归函数,避免栈溢出的问题。具体实现方法是将递归函数中的参数改为可变参数,并将递归调用改为参数的更新,直到满足递归终止条件。以下是优化后的代码:
def words_divi(words, n, sram_list):
division_words = words / n
print(division_words)
if division_words <= words_max:
if isinstance(division_words, int):
for i in range(n):
sram_list.append(str(division_words) + 'x' + str(bits_max))
else:
round(division_words)
if division_words in words_list:
for i in range(n):
sram_list.append(str(division_words) + 'x' + str(bits_max))
else:
while division_words < words_max:
division_words = division_words + 1
if division_words in words_list:
for i in range(n):
sram_list.append(str(division_words) + 'x' + str(bits_max))
return sram_list
else:
return words_divi(words, n+1, sram_list)
if __name__ == "__main__":
words_list = []
bits_list = []
words_list.append(list(range(32, 1025, 16)) + list(range(1056, 8193, 16)))
bits_list.append(list(range(16, 145, 1)))
words = 16388
bits = 148
words_max = max(words_list[0])
bits_max = max(bits_list[0])
words_min = min(words_list[0])
bits_min = min(bits_list[0])
sram_list = []
num = 2
sram_list = words_divi(words, num, sram_list)
print(sram_list)
def is_plaintext3(text, word1, word2 ,word3): words_found = 0 for word in text.split(): if word.lower() in word1: words_found += 10 if word.lower() in word2: words_found += 3 if word.lower() in word3: words_found += 1 return (words_found / len(text.split()))这段代码一直输出0可能的原因是?
这段代码可能一直输出0的原因是因为在参数word1、word2、word3中传入的是字符串,而不是列表。因此,if语句中的条件无法满足,导致words_found变量一直为0。要修复这个问题,应该在传入参数word1、word2、word3时,将它们转换为列表类型。例如,可以使用以下代码传递参数:
```
is_plaintext3("text string", ["word1"], ["word2"], ["word3"])
```