面向对象设计新思路_数值分组

d1 = 3
d2 = 4
d3 = 5
d4 = 6
d5 = 100
d6 = 101
d7 = 102

d8 = 1001
d9 = 1002
d10 = 1003
s = [d1,d2,d3,d4,d5,d6,d7,d8,d9,d10]
s = [132.15521177766695, 149.62021922186852, 161.26816176790754, 172.12968947860213, 174.92238278733799, 177.67118505824178, 183.31260185813738, 185.77599414348452, 198.57978245531439, 212.36793543282374, 223.65453270613585, 230.36397287770498, 234.76277388035777, 245.01993796423994, 254.86463073561225]
ss = s.sort()
print s
gap = 0
strip_gap_value = 5
gap = strip_gap_value
last = s[0]
sub_b = [s[0]]
bucket = [sub_b]
for i in s[1:]:
    feet = i - last
    if feet > gap:
        sub_b = [i]
        bucket.append(sub_b)
        gap = strip_gap_value
    else:
        sub_b.append(i)
        gap = gap + feet
    last = i

print bucket