Create lists of anagrams from a list of words [duplicate]

Question

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

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

Answer 1

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)

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

---

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)

That prints:

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

---

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])

Output:

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