蟒集團整數的列表，就近值

Question

我有一個列表：

d = [23,67,110,25,69,24,102,109] 

我怎樣才能與動態間隙組最近的值，並創建這樣一個元組，什麼是最快的方法是什麼？ ：

[(23,24,25),(67,69),(102,109,110)] 

Answer 1

像

d = [23,67,110,25,69,24,102,109] 

d.sort() 

diff = [y - x for x, y in zip(*[iter(d)] * 2)] 
avg = sum(diff)/len(diff) 

m = [[d[0]]] 

for x in d[1:]: 
    if x - m[-1][0] < avg: 
     m[-1].append(x) 
    else: 
     m.append([x]) 


print m 
## [[23, 24, 25], [67, 69], [102, 109, 110]] 

拳我們計算時序元件，然後一起組元件，其差小於平均值之間的平均差值。