Python中的字谜检查

Question

我正在尝试编写一个程序来比较两个单词列表并检查单词以查看它们是否是字谜。

例如，

输入：['cinema','host','aab','train'], ['iceman', 'shot', 'bab', 'rain']

我正在使用以下代码：

#!/usr/bin/env python
anagram_dict = {}

def anagram_solver(first_words,second_words):
    for word in first_words:
         first_word = list(word)
         second_word = list(second_words[first_words.index(word)])
         first_copy = first_word
         second_copy = second-word
         if len(first_word) != len(second_word):
              anagram_dict[first_words.index(word)] = 0
         else:
              for char in first_word:
                  second_word = second_copy
                  if char in second_word:
                       first_copy.remove(char)
                       second_copy.remove(char)
                  else:
                      pass
              if len(first_copy) == len(second_copy):
                  print first_copy
                  print second_copy
                  anagram_dict[first_words.index(word)] = 1
              else:
                  anagram_dict[first_words.index(word)] = 0
    for k,v in anagram_dict.items():
    print "%d : %d" %(k,v)

if __name__ == "__main__":
     anagram_solver(['cinema','host','aab','train'],['iceman','shot','bab','rain'])

当我执行此脚本时，在 for 循环 for char in first_word: 中，循环被一个列表项跳过。例如，如果它正在处理列表['c','i','n','e','m','a'] 它只处理'c','n','m' 并忽略其他项目。如果我删除 list.remove()，那么它不会跳过项目。

人们可以执行这个脚本来更好地理解我在这里试图解释的内容。

只是想知道为什么会出现这种行为以及如何克服这个问题？

Answer 1

您可以简单地对单词进行排序并检查它们是否相等：

def anagram_solver(first_words, second_words):
    result = []
    for i in xrange(len(first_words)):
        a = list(first_words[i])
        b = list(second_words[i])
        a.sort()
        b.sort()
        
        result.append(a == b)
    return result

例子：

>>> a = ['cinema','host','aab','train']
>>> b = ['iceman', 'shot', 'bab', 'rain']
>>> anagram_solver(a, b)

[True, True, False, False]

Answer 2

Python 通过引用处理列表，因此当您设置first_copy = first_word 时，实际上只是让first_copy 和first_word 指向同一个列表。您可以使用

来克服这种行为（实际上是复制列表）

first_copy = first_word[:]
second_copy = second_word[:]

Answer 3

根据标题回答您的问题：“Anagram check in Python”

您可以在 one 三行中做到这一点：

first_words = ['cinema','host','aab','train']
second_words = ['iceman', 'shot', 'bab', 'rain']

print [sorted(a) == sorted(b) for (a,b) in zip(first_words,second_words)]

制作：

[True, True, False, False]

Answer 4

您可以将enumerate 与 sorted 一起使用：

[sorted(a[ind]) == sorted(ele) for ind, ele in enumerate(b)]

Answer 5

有两种方法可以做到这一点。一个很简单，另一个有点复杂，但是是最优的。 第一种方法

def anagram1(s1,s2):
# We need to get rid of the empty spaces and
# lower case the string

s1 = s1.replace(' ', '').lower()
s2 = s2.replace(' ', '').lower()

# Now we will return boolean for sorted match.
return sorted(s1) == sorted(s2)

下一个方法有点长：

def anagram2(s1, s2):
# We will remove spaces and will lower case the string
s1 = s1.replace(' ', '').lower()
s2 = s2.replace(' ', '').lower()

# We will do the edge case to check if both strings have same number of letters
if len(s1) != len(s2):
    return False

# will creat an empty dictionary.
count = {}

for letter in s1:
    if letter in count:
        # We are assigning value 1 for every letter in s1
        count[letter] += 1
    # if it is the start of loop u just want to assign one into it.
    else:
        count[letter] = 1

对于 s2，我们将做相反的事情。

for letter in s2:
    if letter in count:
        # We are making every value of the letters from 1 to zero
        count[letter] -= 1
    else:
        count[letter] = 1

for k in count:
    if count[k] != 0:
        return False

# other wise just return true
return True

Answer 6

def anagram(string_one, string_two):
    string_one = string_one.replace(' ', '').lower()
    string_two = string_two.replace(' ', '').lower()
    string_list_one = []
    string_list_two = []
    for letters in string_one:
        string_list_one.append(letters)

    for letters_t in string_two:
        string_list_two.append(letters_t)

    string_list_one.sort()
    string_list_two.sort()

    if(string_list_one == string_list_two):
        return True
    else:
        return False