亞零

2013-10-15 34 views
0

我想編寫代碼來獲得這種形式的列表的子序列:亞零

l = [240,200,160,4,0,0,0,0,4,4,4,0,0,0,1,1,1,1]

基本上,我需要的是不爲零的子序列,所以我需要在這個形式的輸出:

output = [[255,200,160,4] , [4,4,4] , [1,1,1,1]]

謝謝。

回答

10

連續的子序列?尋呼GROUPBY博士,博士itertools.groupby

>>> from itertools import groupby 
>>> l = [240,200,160,4,0,0,0,0,4,4,4,0,0,0,1,1,1,1] 
>>> [list(g) for k,g in groupby(l, lambda x: x != 0) if k] 
[[240, 200, 160, 4], [4, 4, 4], [1, 1, 1, 1]] 

或者即使我們採取的事實bool(0)False優勢和bool(any other integer)True

>>> [list(g) for k,g in groupby(l, bool) if k] 
[[240, 200, 160, 4], [4, 4, 4], [1, 1, 1, 1]] 
+3

+1'尋呼GROUPBY博士,博士itertools.groupby' – inspectorG4dget

+0

...和使用'bool'而不是我醜陋的'lambda'。你當然是更好的解決方案 – inspectorG4dget

3
In [117]: l = [240,200,160,4,0,0,0,0,4,4,4,0,0,0,1,1,1,1] 

In [118]: [list(vals) for mask,vals in itertools.groupby(l, key=lambda n:n!=0) if mask] 
Out[118]: [[240, 200, 160, 4], [4, 4, 4], [1, 1, 1, 1]]