如何在列表中添加元素的值，这些元素不能作为该字典的另一个键重复？

Question

Suppose I have one list which contains anagram strings. For example,

anList = ['aba','baa','aab','cat','tac','act','sos','oss']

And I want to construct a dictionary which contains element of that list as key and anagram strings of that element will be values of that key as a list, Also elements which will be added into list are not repeated as another key of that dictionary. For example, if 'baa' has added to list, which list is value of key 'aba', then 'baa' can not be added as key further. Output dictionary should be look like,

anDict = {'aba' : ['baa','aab'],'cat' : ['tac','act'],'sos' : ['oss']}

I have tried with many approaches, but problem is that added elements in list are again add as key of dictionary.

How can I done it?

Answer 1

You can group your words by the count of letters using the Counter object:

from collections import Counter
from itertools import groupby

sorted list = sorted(anList, key=Counter)
groups = [list(y) for x, y in groupby(sortedList, key=Counter)]
#[['aba', 'baa', 'aab'], ['cat', 'tac', 'act'], ['sos', 'oss']]

Now, convert the list of lists of anagrams into a dictionary:

{words[0]: words[1:] for words in groups}
#{'aba': ['baa', 'aab'], 'cat': ['tac', 'act'], 'sos': ['oss']}

Answer 2

Here combining both the order of occurrence with the possibility of them not being grouped together:

anagram_list = ['cat','aba','baa','aab','tac','sos','oss','act']

first_anagrams = {}
anagram_dict = {}

for word in anagram_list:
    sorted_word = ''.join(sorted(word))
    if sorted_word in first_anagrams:
        anagram_dict[first_anagrams[sorted_word]].append(word)
    else:
        first_anagrams[sorted_word] = word
        anagram_dict[word] = []

print(anagram_dict)

The output is

{'aba': ['baa', 'aab'], 'sos': ['oss'], 'cat': ['tac', 'act']}

where the key is always the first anagram in order of occurrence, and the algorithm is strictly O(n) for n words of neglible length.

---

Should you want all anagrams in the list including the first one, it becomes much easier:

anagram_list = ['cat','aba','baa','aab','tac','sos','oss','act']

first_anagrams = {}
anagram_dict = defaultdict(list)

for word in anagram_list:
    anagram_dict[first_anagrams.setdefault(''.join(sorted(word)), word)].append(word)

The result is

defaultdict(<type 'list'>, 
    {'aba': ['aba', 'baa', 'aab'], 'sos': ['sos', 'oss'], 'cat': ['cat', 'tac', 'act']})

Answer 3

The answers from @DYZ and @AnttiHaapala handle the expected output posted in the question much better than this one.

Following is an approach that comes with some caveats using collections.defaultdict . Sort each list element to compare it to the anagram key and append any anagrams that are not the same as the key.

from collections import defaultdict

anagrams = ['aba','baa','aab','cat','tac','act','sos','oss']

d = defaultdict(list)
for a in anagrams:
    key = ''.join(sorted(a))
    if key != a:
        d[key].append(a)

print(d)
# {'aab': ['aba', 'baa'], 'act': ['cat', 'tac'], 'oss': ['sos']}

Caveats:

• always uses the ascending sorted version of the anagram as the dict key, which is not an exact match for the example output in the question
• if the ascending sorted version of the anagram is not in the list, this approach will add a previously non-existent anagram as the dict key

Answer 4

You can use the function groupby() on a presorted list. The function sorted (or Counter ) can be used as the key for sorting and grouping:

from itertools import groupby

anList = ['aba', 'baa', 'aab', 'cat', 'tac', 'act', 'sos', 'oss']

{k: v for _, (k, *v) in groupby(sorted(anList, key=sorted), key=sorted)}
# {'aba': ['baa', 'aab'], 'cat': ['tac', 'act'], 'sos': ['oss']}

Answer 5

Here is slow, but working code:

anList = ['aba', 'baa', 'aab', 'cat', 'tac', 'act', 'sos', 'oss']
anDict = {}
for i in anList:
    in_dict = False
    for j in anDict.keys():
        if sorted(i) == sorted(j):
            in_dict = True
            anDict[j].append(i)
            break
    if not in_dict:
        anDict[i] = []

Answer 6

You may use else with a for loop to achieve this:

anList = ['aba','baa','aab','cat','tac','act','sos','oss']
anDict = dict()

for k in anList:
        for ok in anDict:
            if (ok == k): break
            if (sorted(ok) == sorted(k)):
                anDict[ok].append(k)
                break
        else:
            anDict[k] = []

print(anDict)
# {'aba': ['baa', 'aab'], 'cat': ['tac', 'act'], 'sos': ['oss']}

Answer 7

A simple version without itertools.

Create a multimap sorted string -> [anagram string] :

>>> L = ['aba', 'baa', 'aab', 'cat', 'tac', 'act', 'sos', 'oss']
>>> d = {}
>>> for v in L:
...     d.setdefault("".join(sorted(v)), []).append(v)
...
>>> d
{'aab': ['aba', 'baa', 'aab'], 'act': ['cat', 'tac', 'act'], 'oss': ['sos', 'oss']}

Now you've grouped the anagrams, use the first values as key of the return dict:

>>> {v[0]:v[1:] for v in d.values()}
{'aba': ['baa', 'aab'], 'cat': ['tac', 'act'], 'sos': ['oss']}

Answer 8

anList = ['aba', 'baa', 'aab', 'cat', 'tac', 'act', 'sos', 'oss']

anDict = {}
for word in anList:
    sorted_word = ''.join(sorted(word))
    found_key = [key  for key in anDict.keys() if sorted_word  == ''.join(sorted(key))]
    if found_key:
        anDict[found_key[0]].append(word)
    else:
        anDict[word]=[]


>>> anDict
{'aba': ['baa', 'aab'], 'cat': ['tac', 'act'], 'sos': ['oss']}