从单词列表创建字谜列表 [英] Create lists of anagrams from a list of words

问题描述

我想从单词列表中找到字谜的创建列表.我应该在代码或递归中使用另一个循环吗?

I want to find create lists of anagrams from a list of words. Should I use another loop in my code or recursion?

some_list = ['bad', 'app', 'sad', 'mad', 'dab','pge', 'bda', 'ppa', 'das', 'dba']

new_list = [some_list[0]]
i = 0
while i+1 < len(some_list):
    if (''.join(sorted(some_list[0]))) == (''.join(sorted(some_list[i+1]))):
        new_list.append(some_list[i+1])
        i = i+1
    else:
        i = i+1

print(new_list)

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• 我的输出是['bad', 'dab', 'bda', 'dba'].但我也想要更多清单 some_list中的其他字谜.

---

• My output is ['bad', 'dab', 'bda', 'dba']. But I also want more lists of other anagrams from some_list.

我希望输出为: -['app', 'ppa'] -['bad', 'dab', 'bda', 'dba'] -['sad', 'das']

I want the output to be: - ['app', 'ppa'] - ['bad', 'dab', 'bda', 'dba'] - ['sad', 'das']

推荐答案

我建议您编写Python，而不是Java或您在此处仿真的任何其他语言.这是您的Python核心代码，具有正常的循环且没有所有不必要的内容:

I recommend you write Python, not Java or whatever other language you're emulating there. Here's your core code in Python, with normal looping and without all the unnecessary stuff:

new_list = [some_list[0]]
for word in some_list[1:]:
    if sorted(some_list[0]) == sorted(word):
        new_list.append(word)

我看不到用于递归的方法，但是是的，您可以在其周围包裹一个外循环以找到其他字谜组.

I don't see use for recursion, but yes, you could wrap an outer loop around this to find the other anagram groups.

尽管这是我的方法，但使用有用的 itertools .groupby :

Though this is how I'd do it, using the helpful itertools.groupby:

for _, group in groupby(sorted(some_list, key=sorted), sorted):
    group = list(group)
    if len(group) > 1:
        print(group)

打印:

['bad', 'dab', 'bda', 'dba']
['sad', 'das']
['app', 'ppa']

---

已更改问题的替代解决方案，其中对组进行了排序:

---

Alternative solution for the changed question with sorting the groups:

groups = (list(group) for _, group in groupby(sorted(some_list, key=sorted), sorted))
print([group for group in sorted(groups) if len(group) > 1])

输出:

[['app', 'ppa'], ['bad', 'dab', 'bda', 'dba'], ['sad', 'das']]

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