Question

“海洋”一词由五个连续的，重叠的州邮政缩写组成：马萨诸塞州（MA），阿肯色州（AR），罗德岛（RI），印第安纳州（IN）和内布拉斯加州（NE）。找到一个具有相同属性的七个字母的单词。

我使用Python打开一个大约5000个单词的列表。我想先找到一个包含5个州缩写的单词。

def puzzleH(word):
    states = ['al', 'ak', 'az', 'ar', 'ca', 'co', 'ct', 'dc', 'de', 'fl', 'ga', 
              'hi', 'id', 'il', 'in', 'ia', 'ks', 'ky', 'la', 'me', 'md', 
              'ma', 'mi', 'mn', 'ms', 'mo', 'mt', 'ne', 'nv', 'nh', 'nj', 
              'nm', 'ny', 'nc', 'nd', 'oh', 'ok', 'or', 'pa', 'ri', 'sc', 
              'sd', 'tn', 'tx', 'ut', 'vt', 'va', 'wa', 'wv', 'wi', 'wy']
    checker = 0;
    for st in states:
        if st in word:
            checker+=1
    if checker==5:
        # ...still thinking...
        #pos = (i for i,st in enumerate(word) if st in states)
        #for i in pos: print(i)
        #return word

# Main program
ListH = []
for word in wordList:
    if puzzleH(word)!=None:
        ListH.append(puzzleH(word))

找到包含5个状态缩写的单词后，我将找到每个状态缩写的索引。并将这些索引的列表与[0,1,2,3,4]或[1,2,3,4,5]或[2,3,4,5,6]进行比较。但是我不知道该怎么做。欢迎使用任何有效的新算法。谢谢你。

                                                          --Tuan--

Answer 1

为什么不使用st in word，而不使用word.find( st )，它会返回匹配的索引，即-1。然后只存储找到的索引

def puzzleH( word ):
    states = ['al', 'ak', 'az', 'ar', 'ca', 'co', 'ct', 'dc', 'de', 'fl', 'ga',
              'hi', 'id', 'il', 'in', 'ia', 'ks', 'ky', 'la', 'me', 'md',
              'ma', 'mi', 'mn', 'ms', 'mo', 'mt', 'ne', 'nv', 'nh', 'nj',
              'nm', 'ny', 'nc', 'nd', 'oh', 'ok', 'or', 'pa', 'ri', 'sc',
              'sd', 'tn', 'tx', 'ut', 'vt', 'va', 'wa', 'wv', 'wi', 'wy']
    found_list = []
    for st in states:
        position = word.find( st )
        if ( position != -1 ):
            found_list.append( ( st, position ) )  # <-- Keep the word + position
    if ( len( found_list ) >= 5 ):
        print("[%s]: " % ( word ) )
        for state, position in found_list:
            print( "   \"%s\" at %d" % ( state, position ) )


for word in [ 'marine', 'desert', 'dessert', 'icecream', 'chocolate', 'ohmmeter', 'comically' ]:
    puzzleH( word )

哪个给：

$ python3 ./state_find.py 
[marine]: 
   "ar" at 1
   "in" at 3
   "ma" at 0
   "ne" at 4
   "ri" at 2

编辑：针对Linux词典文件进行测试：

words = open( '/usr/share/dict/words.pre-dictionaries-common', 'rt' ).read().split('\n')
for word in words:
    if ( word.find( "'" ) == -1 ):
        puzzleH( word )

给出很多结果：

# (just the tail ...)
[windowpane]: 
   "in" at 1
   "ne" at 8
   "nd" at 2
   "pa" at 6
   "wi" at 0
[windowpanes]: 
   "in" at 1
   "ne" at 8
   "nd" at 2
   "pa" at 6
   "wi" at 0
[windstorms]: 
   "in" at 1
   "ms" at 8
   "nd" at 2
   "or" at 6
   "wi" at 0
[windward]: 
   "ar" at 5
   "in" at 1
   "nd" at 2
   "wa" at 4
   "wi" at 0

哦，“爱护别人”是一个很好的选择：

[philandering]: 
   "de" at 6
   "hi" at 1
   "il" at 2
   "in" at 9
   "la" at 3
   "nd" at 5
   "ri" at 8

编辑：似乎没有像我应该的那样阅读规范。单词必须完全由重叠状态码组成。

这是一个修复该问题的版本。它从输入的单词创建成对的字母，寻找要匹配的状态码，如果找到，则记录位置和状态码（与之前相同）。

def puzzleH( word ):
    states = ['al', 'ak', 'az', 'ar', 'ca', 'co', 'ct', 'dc', 'de', 'fl', 'ga',
              'hi', 'id', 'il', 'in', 'ia', 'ks', 'ky', 'la', 'me', 'md',
              'ma', 'mi', 'mn', 'ms', 'mo', 'mt', 'ne', 'nv', 'nh', 'nj',
              'nm', 'ny', 'nc', 'nd', 'oh', 'ok', 'or', 'pa', 'ri', 'sc',
              'sd', 'tn', 'tx', 'ut', 'vt', 'va', 'wa', 'wv', 'wi', 'wy']
    found_list = []
    word_position = 0
    for i in range( len( word ) - 1 ):
        two_letters = word[i] + word[i+1]
        if ( two_letters in states ):
            found_list.append( ( two_letters, i ) )
        else:
            found_list = []
            break # word needs to be made of all state-codes

    if ( len( found_list ) >= 5 ):
        print("[%s]: " % ( word ) )
        for state, position in found_list:
            print( "   \"%s\" at %d" % ( state, position ) )

words = open( '/usr/share/dict/words.pre-dictionaries-common', 'rt' ).read().split('\n')

for word in words:
    if ( word.find( "'" ) == -1 ):
        puzzleH( word )

发现的最长的是：

[malarial]: 
   "ma" at 0
   "al" at 1
   "la" at 2
   "ar" at 3
   "ri" at 4
   "ia" at 5
   "al" at 6

有趣的是，在整个73,000个单词的词典中，只有4个单词（> = 5个代码）。

Answer 2

您可以创建一个将状态缩写映射到索引的字典，然后使用给定单词的自身字母（但偏移量为1）进行压缩来遍历给定单词的相邻字母对，然后在字典中查找一对字母，然后如果找到，则将相应的索引添加到输出列表中：

state_indices = {state: index for index, state in enumerate(states)}
def puzzleH(word):
    indices = []
    for pair in zip(word, word[1:]):
        candidate = ''.join(pair)
        if candidate not in state_indices:
            break
        indices.append(state_indices[candidate])
    else:
    return indices

for word in wordList:
    indices = puzzleH(word)
    if indices is not None:
        print(word, indices)

Answer 3

如果目的是仅识别包含五个连续的，重叠的州邮政缩写的单词，则可以尝试以下操作：

states = ['al', 'ak', 'az', 'ar', 'ca', 'co', 'ct', 'dc', 'de', 'fl', 'ga',
          'hi', 'id', 'il', 'in', 'ia', 'ks', 'ky', 'la', 'me', 'md',
          'ma', 'mi', 'mn', 'ms', 'mo', 'mt', 'ne', 'nv', 'nh', 'nj',
          'nm', 'ny', 'nc', 'nd', 'oh', 'ok', 'or', 'pa', 'ri', 'sc',
          'sd', 'tn', 'tx', 'ut', 'vt', 'va', 'wa', 'wv', 'wi', 'wy']

def check_word(word):
    if len(word) != 7:
        return False
    counter = 0
    for i in range(0, len(word) - 1):
        abv = word[i] + word[i + 1]
        if abv in states:
            counter += 1
            if counter == 5:
                return True
        else:
            counter = 0
    return False


for w in wordList:
    print("{0} : {1}".format(w, check_word(w)))

在另一个列表中查找包含5个连续字符串的单词[Python]

3 个答案: