从列表中查找元素并将附近的元素存储在列表中-Python

我有一个标记化单词的列表，我正在搜索其中的一些单词，并将附近的3个元素存储到找到的单词中。代码是：

要查找的单词--要查找的单词列表

代币——我必须从一个大列表中查找单词

for x in words_to_find:
        if x in tokens:
            print "Matched word is", x
            indexing = tokens.index(x)
            print "This is index :", indexing
            count = 0
            lower_limit = indexing - 3
            upper_limit = indexing + 3
            print "Limits are", lower_limit,upper_limit 
            for i in tokens:
                if count >= lower_limit and count <= upper_limit:
                    print "I have entered the if condition"
                    print "Count is : ",count
                    wording = tokens[count]
                    neighbours.append(wording)
                else:
                    count +=1
                    break
                count +=1
           final_neighbour.append(neighbours)
    print "I am in access here", final_neighbour

用于x的单词查找：
如果令牌中有x：
打印“匹配字为”，x
索引=标记。索引（x）
打印“这是索引：”，索引
计数=0
下限=索引-3
上限=索引+3
打印“限制为”、下限、上限
对于令牌中的i：
如果count>=下限且count，则每个单词的邻居都会改变。因此，让它每一个字都无效。而且count也应该分配给indexing-3，如果大于等于0，则为下限，否则为0，因为找到的单词中的前三个单词和后三个单词是您需要的

for x in words_to_find:
    neighbours=[] # the neighbour for the new word will change, therefore make it null!
    if x in tokens:
        print "Matched word is", x
        indexing = tokens.index(x)
        print "This is index :", indexing
        lower_limit = indexing - 3
        upper_limit = indexing + 3
        count = lower_limit if lower_limit >=0 else 0# lower_limit starts from the index-3 of the word found!
        print "Limits are", lower_limit,upper_limit,count
        for i in tokens:
            if count >= lower_limit and count <= upper_limit:
                print "I have entered the if condition"
                print "Count is : ",count
                wording = tokens[count]
                neighbours.append(wording)
            else:
                count +=1
                break
            count +=1
        final_neighbour.append(neighbours)
    print "I am in access here", final_neighbour

建议

您可以使用列表切片来获取匹配单词前后的3个单词。这也将提供所需的输出

lower_limit = lower_limit if lower_limit >=0 else 0
neighbours.append(tokens[lower_limit:upper_limit+1])

就是

final_neighbour=[]
for x in words_to_find:
    neighbours=[] # the neighbour for the new word will change, therefore make it null!
    if x in tokens:
        print "Matched word is", x
        indexing = tokens.index(x)
        print "This is index :", indexing
        lower_limit = indexing - 3
        upper_limit = indexing + 3
        lower_limit = lower_limit if lower_limit >=0 else 0# lower_limit starts from the index-3 of the word found!
        print "Limits are", lower_limit,upper_limit
        neighbours.append(tokens[lower_limit:upper_limit+1])
        final_neighbour.append(neighbours)
    print "I am in access here", final_neighbour

希望有帮助
 在for循环中有一行在下面

neighbours.append(wording)

什么是“邻居”

您应该在append语句之前对其进行初始化（特别是在循环之外…更喜欢在代码的开头使用，在这里您定义了标记和单词\u to\u find），如下所示

neighbours[]

我们可以使用切片来获取邻居，而不是使用计数进行迭代

tokens = [u'प्रीमियम',u'एंड',u'गिव',u'फ्रॉम',u'महाराष्ट्रा',u'मुंबई',u'इंश्योरेंस',u'कंपन‌​ी',u'फॉर',u'दिस']
words_to_find = [u'फ्रॉम',u'महाराष्ट्रा']
final_neighbours = {}
for i in words_to_find:
    if i in tokens:
       print "Matched word : ",i
       idx = tokens.index(i)
       print "this is index : ",idx
       idx_lb = idx-3
       idx_ub = idx+4
       print "Limits : ",idx_lb,idx_ub
       only_neighbours =  tokens[idx_lb : idx_ub]
       only_neighbours.remove(i)
       final_neighbours[i]= only_neighbours

for k,v in final_neighbours.items():
    print "\nKey:",k
    print "Values:"
    for i in v:
       print  i,

 Output:
  Matched word :  फ्रॉम
  this is index :  3
  Limits :  0 7
  Matched word :  महाराष्ट्रा
  this is index :  4
  Limits :  1 8

  Key: महाराष्ट्रा
  Values:
  एंड गिव फ्रॉम मुंबई इंश्योरेंस कंपन‌​ी 
  Key: फ्रॉम
  Values:
  प्रीमियम एंड गिव महाराष्ट्रा मुंबई इंश्योरेंस

当你说“问题”时，问题是什么？你没有告诉我们问题是什么。什么东西在这里不起作用？你得到了什么结果？@JennerFelton，我得到了附近元素的上下限的正确索引，但最后的邻域列表是一个空列表。我试图为邻居列表中的每个find单词添加附近的单词，并为找到的所有这些单词创建一个列表。请帮我做这个或任何更好的方式做这个样本IO将是有帮助的<代码>要查找的单词

，

标记

？要查找的单词=[u'प्रीमियम',u'एंड',u'गिव',u'फ्रॉम',u'महाराष्ट्रा',u'मुंबई',u'इंश्योरेंस',u'कंपनी',u'फॉर',u'दिस']请为该示例输入提供一个示例输入和所需的输出。列表理解将其转换为列表列表：[[u'\u091f\u0942'，u'\u0930\u093f\u0915\u0949\u0930\u094d\u0921'，u'\u092f\u094b\u0930'，u'\u092a\u094d\u0930\u0940\u092e\u093f\u092e'，u'\u092f\u0921'，u'\u092f\u0935']另外，因为我处理的是印地语单词，它存储在优尼科。任何直接方法都可以直接查看印地语单词。当我逐个访问元素时，它会以印地语打印，但正在寻找更好的方法。该方法工作良好。谢谢。我唯一关心的是，我有印地语单词，我必须将它们转换为unicode才能使用它。现在我的输出这看起来像是这样的一个（u’\u’\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu’\uuu’\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu’\uuu’\uuuu’\uuuuuuu’\uuuuuuuuuu’\uuuuuuuuuuuuuuuuuuuuuuuuuu0930\uuuuuu0930\u0930\uu0930\u0930\u0930\uu0930\u0930\uu0930\uuuuuuuuuuuuuuuuuu0930\uuuuuuuuuuuuuu0930\u0907\u091f'，u'\u0938\u0930'，u'\u092f\u094b\u0930']“有没有办法看到印地语单词如果你单独打印这些单词，你就可以看到印地语单词。你已经编辑了我的代码。谢谢你的帮助。你能解释一下背后的逻辑吗？单独打印时是否将unicode转换成ASCII。但是AsciiunCode中不支持印地语字母是ASCII的超集。要了解unicode，选中此项，印地语字符属于Devanagiri unicode块，

tokens = [u'प्रीमियम',u'एंड',u'गिव',u'फ्रॉम',u'महाराष्ट्रा',u'मुंबई',u'इंश्योरेंस',u'कंपन‌​ी',u'फॉर',u'दिस']
words_to_find = [u'फ्रॉम',u'महाराष्ट्रा']
final_neighbours = {}
for i in words_to_find:
    if i in tokens:
       print "Matched word : ",i
       idx = tokens.index(i)
       print "this is index : ",idx
       idx_lb = idx-3
       idx_ub = idx+4
       print "Limits : ",idx_lb,idx_ub
       only_neighbours =  tokens[idx_lb : idx_ub]
       only_neighbours.remove(i)
       final_neighbours[i]= only_neighbours

for k,v in final_neighbours.items():
    print "\nKey:",k
    print "Values:"
    for i in v:
       print  i,

 Output:
  Matched word :  फ्रॉम
  this is index :  3
  Limits :  0 7
  Matched word :  महाराष्ट्रा
  this is index :  4
  Limits :  1 8

  Key: महाराष्ट्रा
  Values:
  एंड गिव फ्रॉम मुंबई इंश्योरेंस कंपन‌​ी 
  Key: फ्रॉम
  Values:
  प्रीमियम एंड गिव महाराष्ट्रा मुंबई इंश्योरेंस