带有列表索引的单词列表答案

【问题标题】：List of words with List Index带有列表索引的单词列表
【发布时间】：2019-01-25 06:23:00
【问题描述】：

请帮助我。到目前为止，我做到了。

stopwords=['what','hello','and','at','is','am','i','to','the']
search_list=['where is north and northern side','finding the index'
             'ask in the community at western environmental',
             'my name is alan and i am coming from london southeast','access to the index']

rev_dict ={'east': 'e', 'eastern': 'e', 'enviornment': 'env', 'environ.': 
'env','environmental': 'env', 'north': 'n', 'northern': 'n', 'south': 's',
'southern': 's', 'west': 'w', 'western': 'w'}

result = [" ".join([rev_dict.get(x,x) for x in s.split() if x not in 
stopwords]) for s in search_list]

print (result)

在运行上述代码时，它会给出以下输出

['where n n side', 'finding index','ask in community w env', 'my name alan coming from london southeast','access index']

帮助： 如果每个 search_list 字符串少于或等于 20 个字符（包括空格），则不应应用“停用词和 rev_dict”步骤并按原样打印字符串。如果每个字符串只有 20 个以上的字符，我应该同时应用 'stopwords 和 rev_dict' 步骤。

我应该如何更改上面的代码以获得我想要的以下输出。 'finding the index' 和 'access to the index' 必须按原样打印，因为它们少于或等于 20 个字符。

['where n n side', 'finding the index','ask in community w env', 'my name alan coming from london southeast','access to the index']

【问题讨论】：

标签： list dictionary split slice

【解决方案1】：

stopwords=['what','hello','and','at','is','am','i','to','the']
search_list=['where is north and northern side','finding the index',
             'ask in the community at western environmental',
             'my name is alan and i am coming from london southeast','access to the index']

rev_dict ={'east': 'e', 'eastern': 'e', 'enviornment': 'env', 'environ.': 
'env','environmental': 'env', 'north': 'n', 'northern': 'n', 'south': 's',
'southern': 's', 'west': 'w', 'western': 'w'}

result = [" ".join([rev_dict.get(x,x) for x in s.split() if x not in 
stopwords]) if len(s) > 20 else s for s in search_list]

print (result)

输出：

['where n n side', 'finding the index', 'ask in community w env', 'my name alan coming from london southeast', 'access to the index']

使用if len(s) > 20会解决它。

【讨论】：