背景

我正在研究HackerRank問題Word Order。任務是

1. 閱讀以下輸入 stdin

   4
   bcdef
   abcdefg
   bcde
   bcdef
   

2. 生成反映以下內容的輸出：

   • 第一行的唯一詞數
   • 每個唯一單詞的出現次數 示例：

   3       # Number of unique words
   2 1 1   # count of occurring words, 'bcdef' appears twice = 2
   

問題

我撰寫了兩個解決方案，第二個通過了初始測驗但由于超出時間限制而失敗。第一個也可以，但我對輸出進行了不必要的排序（雖然會出現時間限制問題）。

筆記

• 在第一個解決方案中，我對值進行了不必要的排序，這在第二個解決方案中已修復
• 我渴望更好地（正確地）使用標準 Python 資料結構、串列/字典理解——我特別渴望收到一個不匯入任何附加模塊的解決方案，import os除非需要。

代碼

import os

def word_order(words):
    # Output no of distinct words
    distinct_words = set(words)
    n_distinct_words = len(distinct_words)
    print(str(n_distinct_words))
    
    # Count occurrences of each word
    occurrences = []
    
    for distinct_word in distinct_words:
        n_word_appearances = 0
        for word in words:
            if word == distinct_word:
                n_word_appearances  = 1
        occurrences.append(n_word_appearances)
    occurrences.sort(reverse=True)
    print(*occurrences, sep=' ')
    # for o in occurrences:
    #     print(o, end=' ')
    
def word_order_two(words):
    '''
    Run through all words and only count multiple occurrences, do the maths
    to calculate unique words, etc. Attempt to construct a dictionary to make
    the operation more memory efficient.
    '''
    # Construct a count of word occurrences
    dictionary_words = {word:words.count(word) for word in words}
    
    # Unique words are equivalent to dictionary keys
    unique_words = len(dictionary_words)
    
    # Obtain sorted dictionary values
    # sorted_values = sorted(dictionary_words.values(), reverse=True)
    result_values = " ".join(str(value) for value in dictionary_words.values())
    # Output results
    print(str(unique_words))
    print(result_values)
    return 0

if __name__ == '__main__':
    q = int(input().strip())

    inputs = []
    for q_itr in range(q):
        s = input()
        inputs.append(s)
        
    # word_order(words=inputs)
    word_order_two(words=inputs)

uj5u.com熱心網友回復：

沒有任何匯入，您可以通過以下方式計算唯一元素

len(set(words))

并計算它們的出現次數

def counter(words):
    count = dict()
    for word in words:
        if word in count:
            count[word]  = 1
        else:
            count[word] = 1
    return count.values()

uj5u.com熱心網友回復：

那些嵌套回圈在性能方面非常糟糕（它們使您的演算法成為二次型）并且非常不必要。您可以在單次迭代中獲得所有計數。您可以使用普通dict或專用的collections.Counter：

from collections import Counter

def word_order(words):
    c = Counter(words)
    print(len(c))
    print(" ".join(str(v) for _, v in c.most_common()))

顯示 Counter 及其方法作業原理的“手動”實作：

def word_order(words):
    c = {}
    for word in words:
        c[word] = c.get(word, 0)   1
    print(len(c))
    print(" ".join(str(v) for v in sorted(c.values(), reverse=True)))
    # print(" ".join(map(str, sorted(c.values(), reverse=True))))

uj5u.com熱心網友回復：

您可以使用Counterthen 列印輸出，如下所示：

>>> from collections import Counter
>>> def counter_words(words):
...   cnt = Counter(words)
...   print(len(cnt))
...   print(*[str(v) for k,v in c.items()] , sep=' ')

>>> inputs = ['bcdef' , 'abcdefg' , 'bcde' , 'bcdef']
>>> counter_words(inputs)
3
2 1 1

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