[英]Get all combinations of list elements not ignoring position in Python
[英]Python: Get all combinations of sequential elements of list
給定一個數組,說x = ['A','I','R']
我想將輸出作為
[['A','I','R'],['A','I'],['I','R'],['A'],['I'],['R']]
我不想作為輸出是:
[['A','I','R'],['A','I'],['I','R'],['A','R'],['A'],['I'],['R']] # extra ['A','R'] which is not in sequence .
下面是提供不需要的輸出的代碼:
letter_list = [a for a in str]
all_word = []
for i in xrange(0,len(letter_list)):
all_word = all_word + (map(list, itertools.combinations(letter_list,i))) # dont use append. gives wrong result.
all_word = filter(None,all_word) # remove empty combination
all_word = all_word + [letter_list] # add original list
我的觀點是我只想要序列的組合。 有什么方法可以使用itertools
還是應該編寫自定義函數?
嘗試使用yield
:
x = ['A','I','R']
def groupme(x):
s = tuple(x)
for size in range(1, len(s) + 1):
for index in range(len(s) + 1 - size):
yield list(x[index:index + size])
list(groupme(x))
>>> [['A'], ['I'], ['R'], ['A', 'I'], ['I', 'R'], ['A', 'I', 'R']]
是的,您可以使用itertools
:
>>> x = ['A', 'I', 'R']
>>> xs = [x[i:j] for i, j in itertools.combinations(range(len(x)+1), 2)]
>>> xs
[['A'], ['A', 'I'], ['A', 'I', 'R'], ['I'], ['I', 'R'], ['R']]
>>> sorted(xs, key=len, reverse=True)
[['A', 'I', 'R'], ['A', 'I'], ['I', 'R'], ['A'], ['I'], ['R']]
信用: hochl回答
不要試圖變得如此神奇:兩個循環可以完成您想要的; 一個超過可能的序列開始,內部超過可能的序列長度:
x = "AIR" # strings are iterables/sequences, too!
all_words = []
for begin in xrange(len(x)):
for length in xrange(1,len(x) - begin+1):
all_words.append(x[begin:begin+length])
使用列表理解:
letters=['A', 'I', 'R']
[letters[start:end+1]
for start in xrange(len(letters))
for end in xrange(start, len(letters))]
[['A'], ['A', 'I'], ['A', 'I', 'R'], ['I'], ['I', 'R'], ['R']]
如果擁有建議的順序很重要(從最長到最短,並且起始位置的長度相同),則可以改為:
[letters[start:start+l+1]
for l in range(len(letters))[::-1]
for start in xrange(len(letters)-l)]
[['A', 'I', 'R'], ['A', 'I'], ['I', 'R'], ['A'], ['I'], ['R']]
只是為了解決Holroy的評論。 如果不是使用列表推導,而是使用生成器表達式(僅將外部[]
替換為()
),則所需的內存將大大減少。 但是在這種情況下,您必須小心不要多次使用結果,或者例如不要嘗試在結果上使用列表方法(例如len或刪除元素)。
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