anagram finder / dict mapping问题 [英] anagram finder / dict mapping question
问题描述
大家好，
我写了一个程序，它接受一个字符串序列并在文本文件中找到所有单词
（每行一个单词）并打印出来：
def anagfind（字母）：#find这些字母的字谜
fin = open（''text.txt''）#one每行文字
wordbox = []＃这是单词行的地方
for fin：
word = line.strip （）
count = 0
for char in letters：
如果char不在单词中：
break
else：
count + = 1
if count == len（word）：
wordbox.append（单词）
返回wordbox
现在我想修改代码，自然地找到
单词表中的所有字谜。最好的方法是什么？使用提示？是否可以迭代dict键来获得
？我怎样才能制作一个字母，从一组字母映射到一个字母列表，从那些
字母拼写出来？难道我不需要把这套字母作为字典中的一个关键吗？
一如既往 - 感谢帮助有人试图学习...
Dave
Hi All,
I wrote a program that takes a string sequence and finds all the words
inside a text file (one word per line) and prints them:
def anagfind(letters): #find anagrams of these letters
fin = open(''text.txt'') #one word per line file
wordbox = [] #this is where the words will go
for line in fin:
word = line.strip()
count = 0
for char in letters:
if char not in word:
break
else:
count += 1
if count == len(word):
wordbox.append(word)
return wordbox
Now I''d like to modify the code to naturally find all anagrams inside a
wordlist. What would be the best way to do this? Using Hints? Is it
possible to iterate over dict keys? How can I make a dict that maps
from a set of letters to a list of words that are spelled from those
letters? Wouldn''t I need to make the set of letters a key in a dict?
As always - Thanks for helping someone trying to learn...
Dave
推荐答案
2008年5月8日星期四11:02:12 +1000，dave< sq * ************@1ya2hoo3.net
写道：
On Thu, 08 May 2008 11:02:12 +1000, dave <sq*************@1ya2hoo3.net
wrote:
大家好，
我编写了一个程序，它接受一个字符串序列并在文本文件中找到所有单词
（每行一个单词）并打印出来：
def anagfind（字母）：#find这些字母的字谜
fin = open（''text.txt''）每行文件#one word
wordbox = []＃这就是单词将去的地方
for fin：
word = line.strip（）
count = 0
for char in letters：
如果char不在单词中：
break
else：
count + = 1
if count == len（word）：
wordbox.append（单词）
返回wordbox
现在我想修改代码以自然地找到
wordlist。最好的方法是什么？使用提示？是否可以迭代dict键来获得
？我怎样才能制作一个字母，从一组字母映射到一个字母列表，从那些
字母拼写出来？难道我不需要把这套字母作为字典中的一个关键吗？
一如既往 - 感谢帮助有人试图学习...
Dave
Hi All,
I wrote a program that takes a string sequence and finds all the words
inside a text file (one word per line) and prints them:
def anagfind(letters): #find anagrams of these letters
fin = open(''text.txt'') #one word per line file
wordbox = [] #this is where the words will go
for line in fin:
word = line.strip()
count = 0
for char in letters:
if char not in word:
break
else:
count += 1
if count == len(word):
wordbox.append(word)
return wordbox
Now I''d like to modify the code to naturally find all anagrams inside a
wordlist. What would be the best way to do this? Using Hints? Is it
possible to iterate over dict keys? How can I make a dict that maps
from a set of letters to a list of words that are spelled from those
letters? Wouldn''t I need to make the set of letters a key in a dict?
As always - Thanks for helping someone trying to learn...
Dave
建议：对于每个单词，对其字符进行排序并将其用作
字典键。如果两个单词具有相同的字符组合，那么它们就是字谜。例如：edam和制造是字谜
，因为他们有'a'，'d''，'e''和'm'字母。
请参阅编程珍珠。作者：Jon Bentley。
-
Kam-Hung Soh< a href =" http://kamhungsoh.com/blog">软件Salariman< / a>
Suggestion: for each word, sort their characters and use them as the
dictionary key. If two words have the same combination of characters,
then they are anagrams. For example: "edam" and "made" are anagrams
because they have the letters ''a'', ''d'', ''e'' and ''m''.
Refer "Programming Pearls" by Jon Bentley.
--
Kam-Hung Soh <a href="http://kamhungsoh.com/blog">Software Salariman</a>
这就是我想出来的。我的问题是，我无法让他们正确评估..当比较词（）运行它发现自己...
我应该有mapdict的键吗？迭代自己？那是
可能吗？
def annafind（）：
fin = open（''text.txt''）＃文件每行有一个单词
mapdic = {}＃每个单词都被排序&进入这里
for line in fin：
rawword = line.strip（）
word = list（rawword）
word.sort（）
mapdic [''''。join（word）] = 0
返回mapdic
def comparewords （）：***没有按预期工作
fin = open（''text.txt''）
for line in fin：
line = line.strip（）
word = list（line）
word.sort（）
sortedword =（'''' .join（word））
如果在mapdic中排序：
打印行
2008-05- 07 19:25:53 -0600，Kam-Hung Soh < ka ********* @ gmail.comsaid：
This is what i''ve came up with. My problem is that I can''t get them to
properly evaluate.. when comparewords() runs it finds itself...
Should I have the keys of mapdict iterate over itself? Is that
possible?
def annafind():
fin = open(''text.txt'') # file has one word per line
mapdic = {} # each word gets sorted & goes in here
for line in fin:
rawword = line.strip()
word = list(rawword)
word.sort()
mapdic[''''.join(word)] = 0
return mapdic
def comparewords(): ***not working as intended
fin = open(''text.txt'')
for line in fin:
line = line.strip()
word = list(line)
word.sort()
sortedword = (''''.join(word))
if sortedword in mapdic:
print line
On 2008-05-07 19:25:53 -0600, "Kam-Hung Soh" <ka*********@gmail.comsaid:
On Thu，2008年5月8日11:02:12 +1000，dave< sq*************@@ya2hoo3.net>
写道：
On Thu, 08 May 2008 11:02:12 +1000, dave <sq*************@1ya2hoo3.net>
wrote:
>大家好，全部，
我写了一个程序，它接受一个字符串序列并在文本文件中找到所有单词
>Hi All,
I wrote a program that takes a string sequence and finds all the words
>每行一字）并打印出来：
def anagfind（字母）：#find这些字母的字谜
fin = open（''text.txt''）#one word per line file
wordbox = []＃这是单词将用于行中的单词：
word = line.strip（）
count = 0
for char in信件：
如果char不在字中：
break
否则：
count + = 1
如果count == len（字）：
wordbox。追加（word）
返回wordbox
现在我想修改代码，自然找到
里面的所有字谜
>inside a text file (one word per line) and prints them:
def anagfind(letters): #find anagrams of these letters
fin = open(''text.txt'') #one word per line file
wordbox = [] #this is where the words will go
for line in fin:
word = line.strip()
count = 0
for char in letters:
if char not in word:
break
else:
count += 1
if count == len(word):
wordbox.append(word)
return wordbox
Now I''d like to modify the code to naturally find all anagrams inside
a
a
> wordlist。最好的方法是什么？使用提示？是
>wordlist. What would be the best way to do this? Using Hints? Is it
>可以迭代dict键吗？如何制作一个dict，将
>possible to iterate over dict keys? How can I make a dict that maps
>从一组字母映射到从
<拼写的单词列表
>from a set of letters to a list of words that are spelled from those
>字母？难道我不需要把这套字母作为字典中的关键吗？
一如既往 - 感谢帮助有人试图学习...
戴夫
>letters? Wouldn''t I need to make the set of letters a key in a dict?
As always - Thanks for helping someone trying to learn...
Dave
建议：对于每个单词，对其字符进行排序并将其用作
字典键。如果两个单词具有相同的字符组合，那么它们就是字谜。例如：edam和制造是字谜
，因为他们有'a'，'d''，'e''和'm'字母。
请参阅编程珍珠。作者：Jon Bentley。
-
Kam-Hung Soh< a href =" http://kamhungsoh.com/blog">软件萨拉里曼< / a>
Suggestion: for each word, sort their characters and use them as the
dictionary key. If two words have the same combination of characters,
then they are anagrams. For example: "edam" and "made" are anagrams
because they have the letters ''a'', ''d'', ''e'' and ''m''.
Refer "Programming Pearls" by Jon Bentley.
--
Kam-Hung Soh <a href="http://kamhungsoh.com/blog">Software Salariman</ a>
2008年5月8日星期四15:42:07 +1000，dave< sq ********* ****@1ya2hoo3.net
写道：
On Thu, 08 May 2008 15:42:07 +1000, dave <sq*************@1ya2hoo3.net
wrote:
这就是我想出来的。我的问题是，我无法让他们
正确评估..当comparewords（）运行它发现自己...应该
我有mapdict的键迭代自己？这可能吗？
def annafind（）：
fin = open（''text.txt''）#file每行一个字
mapdic = {}＃每个单词都被排序&进入这里
for line in fin：
rawword = line.strip（）
word = list（rawword）
word.sort（）
mapdic [''''。join（word）] = 0
返回mapdic
def comparewords（）：***无法正常工作
fin = open（''text.txt''）
for line in fin：
line = line.strip（）
word = list（line）
word.sort（）
sortedword =（ ''''。join（word））
如果在mapdic中排序：
打印行
开2008-05-07 19:25:53 -0600，Kam-Hung Soh < ka ********* @ gmail.com
说：
This is what i''ve came up with. My problem is that I can''t get them to
properly evaluate.. when comparewords() runs it finds itself... Should
I have the keys of mapdict iterate over itself? Is that possible?
def annafind():
fin = open(''text.txt'') # file has one word per line
mapdic = {} # each word gets sorted & goes in here
for line in fin:
rawword = line.strip()
word = list(rawword)
word.sort()
mapdic[''''.join(word)] = 0
return mapdic
def comparewords(): ***not working as intended
fin = open(''text.txt'')
for line in fin:
line = line.strip()
word = list(line)
word.sort()
sortedword = (''''.join(word))
if sortedword in mapdic:
print line
On 2008-05-07 19:25:53 -0600, "Kam-Hung Soh" <ka*********@gmail.com
said:
> 2008年5月8日星期四11:02:12 +1000，dave< sq ************* @ 1ya2hoo3.net>
写道：
>On Thu, 08 May 2008 11:02:12 +1000, dave <sq*************@1ya2hoo3.net>
wrote:
>>大家好，
我写了一个程序，它接受一个字符串序列，找到所有单词
>>Hi All,
I wrote a program that takes a string sequence and finds all the words
>> ;在文本文件中（每行一个单词）并打印出来：
def anagfind（字母）：#find这些字母的字谜
fin = open（''text.txt''）#one word每行文件
wordbox = []＃这是单词行的地方
for fin：
word = line.strip（）
count = 0
对于字母中的字符：
如果字母不在字中：
中断
否则：
计数+ = 1
如果count == len（字）：
wordbox.append（word）
返回wordbox
现在我想修改代码，自然找到
里面的所有字谜
>>inside a text file (one word per line) and prints them:
def anagfind(letters): #find anagrams of these letters
fin = open(''text.txt'') #one word per line file
wordbox = [] #this is where the words will go
for line in fin:
word = line.strip()
count = 0
for char in letters:
if char not in word:
break
else:
count += 1
if count == len(word):
wordbox.append(word)
return wordbox
Now I''d like to modify the code to naturally find all anagrams inside
a
a
>> wordlist。最好的方法是什么？使用提示？是
>>wordlist. What would be the best way to do this? Using Hints? Is it
>>可以迭代dict键吗？如何制作一个dict，将
>>possible to iterate over dict keys? How can I make a dict that maps
>>从一组字母映射到拼写为$ b $的单词列表b
>>from a set of letters to a list of words that are spelled from those
>>字母？难道我不需要让这套字母成为字典中的关键吗？
一如既往 - 感谢帮助有人试图学习......
Dave
>>letters? Wouldn''t I need to make the set of letters a key in a dict?
As always - Thanks for helping someone trying to learn...
Dave
建议：对于每个单词，对其字符进行排序并将其用作
字典键。如果两个单词具有相同的字符组合，那么它们就是字谜。例如：edam和制造是字谜
因为他们有字母a，d，e和m。
请参阅编程珍珠。作者：Jon Bentley。
-
Kam-Hung Soh< a href =" http://kamhungsoh.com/blog">软件Salariman< /
a>
Suggestion: for each word, sort their characters and use them as the
dictionary key. If two words have the same combination of characters,
then they are anagrams. For example: "edam" and "made" are anagrams
because they have the letters ''a'', ''d'', ''e'' and ''m''.
Refer "Programming Pearls" by Jon Bentley.
--
Kam-Hung Soh <a href="http://kamhungsoh.com/blog">Software Salariman</
a>
你的代码总是会返回相同的列表，因为每个单词都是一个
anagram本身。
提示：为每个字典键创建一个列表，然后在列表中没有该字的情况下向列表中添加一个单词
。所以：
mapdic（''adem''） - [" edam"，" made"]
P.S.回复时，约定是将您的回复添加到底部，
不是消息的顶部。
-
Kam-Hung Soh< a href =" http://kamhungsoh.com/blog">软件Salariman< / a>
Your code is always going to return the same list because every word is an
anagram of itself.
Tip: Create a list for each dictionary key, then add a word to the list if
that word is not in the list. So:
mapdic(''adem'') --["edam", "made"]
P.S. When replying, the convention is to add your response to the bottom,
not top of the message.
--
Kam-Hung Soh <a href="http://kamhungsoh.com/blog">Software Salariman</a>
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