像umbellar =傘都是相同的單詞。找到所有字符與python中的其他詞匹配的單詞
Input = [「umbellar」，「goa」，「傘」，「ago」，「aery」，「alem」，「ayre」，「gnu」，「eyra」，「egma」 ，「leam」，「amel」，「year」，「meal」，「yare」，「gun」，「alme」，「ung」，「male」，「lame」，「mela」，「mage」]
所以輸出應爲：
輸出= [ [ 「umbellar」， 「傘」]， [ 「前」， 「果阿」]， [ 「丙烯酸酯」， 「艾爾」， 「eyra」 ，「yare」，「year」]， [「alem」，「alme」，「amel」，「lame」，「leam」，「male」，「meal」，「mela」] [「gnu」 ，「gun」，「ung」] [「egma」，「遊戲」，「法師」]， ]
他們不是平等的話，他們是anagrams。
字謎可以通過文字整理髮現：
sorted('umbellar') == sorted('umbrella')
from itertools import groupby
def group_words(word_list):
sorted_words = sorted(word_list, key=sorted)
grouped_words = groupby(sorted_words, sorted)
for key, words in grouped_words:
group = list(words)
if len(group) > 1:
yield group
例子：
>>> group_words(["umbellar","goa","umbrella","ago","aery","alem","ayre","gnu","eyra","egma","game","leam","amel","year","meal","yare","gun","alme","ung","male","lame","mela","mage" ])
<generator object group_words at 0x0297B5F8>
>>> list(_)
[['umbellar', 'umbrella'], ['egma', 'game', 'mage'], ['alem', 'leam', 'amel', 'meal', 'alme', 'male', 'lame', 'mela'], ['aery', 'ayre', 'eyra', 'year', 'yare'], ['goa', 'ago'], ['gnu', 'gun', 'ung']]
正如其他人指出你是在爲你的列表尋找字謎的所有組話。這裏你有一個可能的解決方案。該算法查找候選對象並選擇一個（第一個元素）作爲規範詞，並將其餘的詞作爲可能詞刪除，因爲anagrams是可傳遞的，並且一旦您發現某個詞屬於anagram組，則不需要重新計算它。
input = ["umbellar","goa","umbrella","ago","aery","alem","ayre","gnu",
"eyra","egma","game","leam","amel","year","meal","yare","gun",
"alme","ung","male","lame","mela","mage" ]
res = dict()
for word in input:
res[word]=[word]
for word in input:
#the len test is just to avoid sorting and comparing words of different len
candidates = filter(lambda x: len(x) == len(word) and\
sorted(x) == sorted(word),res.keys())
if len(candidates):
canonical = candidates[0]
for c in candidates[1:]:
#we delete all candidates expect the canonical/
del res[c]
#we add the others to the canonical member
res[canonical].append(c)
print res.values()
這algth輸出...
[['year', 'ayre', 'aery', 'yare', 'eyra'], ['umbellar', 'umbrella'],
['lame', 'leam', 'mela', 'amel', 'alme', 'alem', 'male', 'meal'],
['goa', 'ago'], ['game', 'mage', 'egma'], ['gnu', 'gun', 'ung']]
collections.defaultdict就派上用場了：
from collections import defaultdict
input = ["umbellar","goa","umbrella","ago","aery","alem","ayre","gnu",
"eyra","egma","game","leam","amel","year","meal","yare","gun",
"alme","ung","male","lame","mela","mage" ]
D = defaultdict(list)
for i in input:
key = ''.join(sorted(input))
D[key].append(i)
output = D.values()
和輸出是[['umbellar', 'umbrella'], ['goa', 'ago'], ['gnu', 'gun', 'ung'], ['alem', 'leam', 'amel', 'meal', 'alme', 'male', 'lame', 'mela'], ['egma', 'game', 'mage'], ['aery', 'ayre', 'eyra', 'year', 'yare']]
上的答案是正確的.... ..但我已經被挑戰做同樣的事情，而不使用....'groupby（）'....... 這..... 添加打印語句將幫助你在調試代碼和運行時輸出....
def group_words(word_list):
global new_list
list1 = []
_list0 = []
_list1 = []
new_list = []
for elm in word_list:
list_elm = list(elm)
list1.append(list(list_elm))
for ee in list1:
ee = sorted(ee)
ee = ''.join(ee)
_list1.append(ee)
_list1 = list(set(_list1))
for _e1 in _list1:
for e0 in word_list:
if len(e0) == len(_e1):
list_e0 = ''.join(sorted(e0))
if _e1 == list_e0:
_list0.append(e0)
_list0 = list(_list0)
new_list.append(_list0)
_list0 = []
return new_list
和輸出
[['umbellar', 'umbrella'], ['goa', 'ago'], ['gnu', 'gun', 'ung'], ['alem', 'leam', 'amel', 'meal', 'alme', 'male', 'lame', 'mela'], ['egma', 'game', 'mage'], ['aery', 'ayre', 'eyra', 'year', 'yare']]
它這個功課？如果是這樣，那麼標記它。 –
假設一個相同的單詞必須具有相同的長度，然後對列表中的每個字符串進行排序並檢查匹配。 –