我想检查两个字符串是否是字谜。例如，如果我的单词是“halo”，我想检查这些字母是否出现在“loha”中。它应该匹配，因为它是一个字谜。
我的尝试失败了，我不知道为什么。我的代码和输出如下。我有一个单词列表，我想查看列表中哪些元素是字谜。
def anagram(myList):
for elem in myList:
chars = set(elem)
if all((c in chars) for c in myList):
print "Yes, anagram ", elem, chars
else:
print "NOT anagram ", elem, chars
wordsList = ["halo", "loha", "ahlo", "sully"]
anagram(wordsList)
这是我的输出
NOT anagram halo set(['a', 'h', 'l', 'o'])
NOT anagram loha set(['a', 'h', 'l', 'o'])
NOT anagram ahlo set(['a', 'h', 'l', 'o'])
NOT anagram sully set(['y', 's', 'u', 'l'])
答案 0 :(得分：1)
试试这个
def isanagram(ele1,ele2):
ele1=list(ele1)
ele2=list(ele2)
return(sorted(ele1)==sorted(ele2))
print(isanagram("ahlo", "halo"))
Out put
真
对于多元素检查
print(map(isanagram,["ahlo", "halo"],[ "alho", "sully"])
Out put
[True，False]
答案 1 :(得分：0)
这应该证明相对较快，因为它保留了一个排序列表，因此每次迭代时它只需要创建一个新列表。
from itertools import islice
def anagram(myList):
s1 = sorted(myList[0])
for i,elem in enumerate(islice(myList,1,len(myList))):
s2 = sorted(elem)
if s1 == s2:
print ("Yes anagram: ",myList[i],elem)
else:
print ("Not anagram: ",myList[i],elem)
s1 = s2
wordsList = ["halo", "loha", "ahlo", "sully"]
anagram(wordsList)
结果：
Yes anagram: halo loha
Yes anagram: loha ahlo
Not anagram: ahlo sully
不同的结构选项：
from itertools import islice
def is_anagram(sorted_1,sorted_2):
if sorted_1 == sorted_2:
return True
return False
wordsList = ["halo", "loha", "ahlo", "sully"]
s1 = sorted(wordsList[0])
for i,elem in enumerate(islice(wordsList,1,len(wordsList))):
s2 = sorted(elem)
if is_anagram(s1,s2):
print ("Yes anagram: ",wordsList[i],elem)
else:
print ("Not anagram: ",wordsList[i],elem)
s1 = s2
答案 2 :(得分：0)
您可以使用counter输出单词中每个字母的字母计数字典
from Collections import Counter
word1='halo'
word2='hola'
if Counter(word1) == Counter(word2):
print 'Yes Anagram {} {}'.format(word1,word2)
else :
print 'Not Anagram {} {}'.format(word1,word2)
答案 3 :(得分：0)
只需在两个字符串上使用sorted()并进行比较即可。
def is_anagram(str1, str2):
return sorted(str1) == sorted(str2)
k = is_anagram('angel', 'glean')
if k == True:
print('Strings are anagrams')
else:
print('Strings are not anagrams')
此处，sorted('angel')输出['a', 'e', 'g', 'l', 'n']和sorted('glean')也输出['a', 'e', 'g', 'l', 'n']。两者都相等，因此是字谜。
说清楚：
>>> sorted('angel')
['a', 'e', 'g', 'l', 'n']
>>> sorted('glean')
['a', 'e', 'g', 'l', 'n']
>>> sorted('angel') == sorted('glean')
True
解决您的问题：
def is_anagram(str1, str2):
return sorted(str1) == sorted(str2)
wordsList = ["halo", "loha", "ahlo", "sully"]
each = wordsList[0]
for another in wordsList:
if each != another:
k = is_anagram(each, another)
if k == True:
print('{} and {} are anagrams'.format(each, another))
else:
print('{} and {} are not anagrams'.format(each, another))
<强>输出：
halo and loha are anagrams
halo and ahlo are anagrams
halo and sully are not anagrams
答案 4 :(得分：0)
如果要查找列表中的所有angrams，可能需要改为：
# Checks if two strings are anagrams
def isAnagram(str1, str2):
return sorted(str1) == sorted(str2)
# Iterates over all items in your list and compares it with all other items to check if they are anagrams.
def anagrams(myList):
for i in range(0,len(myList)):
for j in range(i+1, len(myList)):
if isAnagram(myList[i], myList[j]):
print "Yes anagram: ", myList[i], myList[j]
else:
print "Not anagram: ", myList[i], myList[j]
wordsList = ["halo", "loha", "ahlo", "sully"]
anagram(wordsList)
这将返回以下内容：
Yes anagram: halo loha
Yes anagram: halo ahlo
Not anagram: halo sully
Yes anagram: loha ahlo
Not anagram: loha sully
Not anagram: ahlo sully
这可能不是最有效的解决方案，但它可以完成工作。
答案 5 :(得分：0)
使用套装：
`def anagram(string1,string2):
if string1=='' or string2== '':
if string1!='' or string2!='':
return False
else:
return True
if len(string1)!=len(string2) or set(string1)!=set(string2):
return False
stemp1 = list(string1)
stemp2 = list(string2)
stemp2.remove(stemp1[0])
return anagram(''.join(stemp1[1:]),''.join(stemp2))`
输出True
如评论中所述，即使附加了len参数，这也不起作用。 这是一个递归的anagram测试来赎回自己
SELECT t1.ID
FROM Table1 t1
JOIN Table2 t2 ON
t2.ID = t1.ID
WHERE t2.Usernames IS NULL
答案 6 :(得分：0)
下面的代码将一个单词列表作为输入并将它们分组为anagrams
def is_grouped(word, anagram_groups):
is_grouped = False
for group in anagram_groups:
if word in group:
print "{0} is already part of anagram group={1}".format(word, group)
is_grouped = True
break
return is_grouped
def get_anagram_group(word_to_check, word_to_check_index, wordlist):
word_to_check_as_list = list(word_to_check)
word_to_check_as_list.sort()
group = [word_to_check]
for i in range(0, len(wordlist)):
if not i == word_to_check_index:
word = wordlist[i]
word_as_list = list(word)
word_as_list.sort()
if word_as_list == word_to_check_as_list:
group.append(word)
return group
def group_anagrams(wordlist):
anagram_groups = []
for i in range(0, len(wordlist)):
word_under_test = wordlist[i]
# check if the word's anagrams already identified as part of anagram group
anagramed = is_grouped(word_under_test, anagram_groups)
if not anagramed:
# if word not already a part of anagram group then find find all anagrams for the word
anagram_group = get_anagram_group(word_under_test, i, wordlist)
if len(anagram_group) == 1:
print "no anagrams found for word {0}".format(word_under_test)
else:
print "anagrams found for word {0}={1}".format(word_under_test, anagram_group)
anagram_groups.append(anagram_group)
return anagram_groups
wlist = ['aloha' , 'hoala', 'loaha', '123', '321', 'xya']
print group_anagrams(wlist)
输出：
anagrams found for word aloha=['aloha', 'hoala', 'loaha']
hoala is already part of anagram group=['aloha', 'hoala', 'loaha']
loaha is already part of anagram group=['aloha', 'hoala', 'loaha']
anagrams found for word 123=['123', '321']
321 is already part of anagram group=['123', '321']
no anagrams found for word xya
[['aloha', 'hoala', 'loaha'], ['123', '321']]