减少循环的运行时间并提高效率

我有一个需要大量单词的循环，将每个单词分解为字母并将它们附加到一个大列表中。减少循环的运行时间并提高效率

然后我检查出现最信，如果未出现在字符串中，我把它存储在具有两个空间的列表：

list[0] =字母出现的最

list[1] =多少次出现

这个循环是超级低效。它可以工作，但返回一个值需要大约25-30秒。在此之前，它会继续下去，不会返回任何值。

如何提高我编写的代码的效率？

def choose_letter(words, pattern): 
    list_of_letters = [] 
    first_letter = [] # first spot is the letter, second is how many times it appears 
    second_letter =[] # first spot is letter, second how many times it appears 
    max_appearances = ["letter", 0] 
    for i in range(len(words)): # splits up every word into letters 
     list_of_letters.append(list(words[i])) 
    list_of_letters = sum(list_of_letters, []) # concatenates the lists within the list 
    first_letter = list_of_letters.count(0) 
    for j in list_of_letters: 
     second_letter = list_of_letters.count(j) 
     if second_letter >= max_appearances[1] and j not in pattern: 
      max_appearances[0] = j 
      max_appearances[1] = second_letter 
     else: 
      list_of_letters.remove(j) 
    return max_appearances[0]

来源

2016-12-06 RonaldB

这可能是http://codereview.stackexchange.com/ – Blorgbeard

更好的候选人，当你移到codereview时，他们会要求查看你对该代码运行的分析器的输出。 –

看起来像['collections.Counter']的工作（https://docs.python.org/3.5/library/collections.html#collections.Counter）。 – bereal

使其更快的一种方法是选择更好的数据结构。下面是使用collections.Counter一个例子：

from collections import Counter 

def choose_letter(words, pattern): 
    pattern = set(pattern) 
    letters = (letter 
       for word in words 
       for letter in word 
       if letter not in pattern) 
    letters = Counter(letters) 
    return letters.most_common(1)[0][0] 


mywords = 'a man a plan a canal panama'.split() 
vowels = 'aeiou' 
assert choose_letter(mywords, vowels) == 'n'

下面是一个使用collections.defaultdict：

from collections import defaultdict 

def choose_letter(words, pattern): 
    pattern = set(pattern) 
    counts = defaultdict(int) 
    for word in words: 
     for letter in word: 
      if letter not in pattern: 
       counts[letter] += 1 
    return max(counts, key=counts.get) 

mywords = 'a man a plan a canal panama'.split() 
vowels = 'aeiou' 
assert choose_letter(mywords, vowels) == 'n'

来源

2016-12-06 21:54:23

你做了很多循环&，你不需要操作列表。每当您执行count或not in时，您都会强制程序循环查看列表/字符串以找到要查找的内容。从列表中删除所有这些项目也非常昂贵。一个更优雅的解决方案是循环访问一次单词/字母列表，然后使用字典来计算每个字母的出现次数。从那里，你有一个字符/字符对&你可以从那里得到键/值，排序清单&看前两个值。

from collections import defaultdict 
from itertools import chain 

def choose_letter(words, pattern=""): 
    count_dict = defaultdict(int) # all unknown values default to 0 
    for c in chain(*words): 
     count_dict[c] += 1 
    # you could replace this "not in" with something more efficient 
    filtered = [(char, count) for (char,count) in count_dict.items() if char not in pattern] 
    filtered.sort(lambda a,b: -cmp(a[0], b[0])) 
    print filtered 
    return filtered[0][0]

如果你不希望深入论证拆包，itertools & defaultdicts，你可以只说：

count_dict = {} 
for word in words: 
    for char in word: 
     count_dict[char] = count_dict.get(char, 0) + 1

...如果你不想尝试挖掘到的参数拆包然而。

来源

2016-12-06 22:14:30

减少循环的运行时间并提高效率

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