获取不连续的id区间

get_uncontinue_id_range.py

In [25]: from itertools import groupby

In [26]: f = open('Fattrid.csv')

In [27]: lines = f.readlines()

In [28]: f.close()

In [29]: ids = sorted([int(line.split()[0]) for line in lines])

In [30]: fun = lambda (i, v): v - i

In [31]: c_ids_list = []

In [32]: for k, g in groupby(enumerate(ids), fun):
    ...:     c_ids = [v for i, v in g]
    ...:     c_ids_list.append(c_ids)
    ...:
In [35]: for i, l in enumerate(c_ids_list):
    ...:     if i >= len(c_ids_list)-1:break
    ...:     if c_ids_list[i+1][0] - c_ids_list[i][-1] > 1000:
    ...:         print '[%s, %s]' %(c_ids_list[i][-1]+1,  c_ids_list[i+1][0]-1)
    ...:
[922, 9043]
[16182, 36999]
[38000, 49999]
[680119, 1999999]
[2236283, 2339999]

文件中有很多id和name的行，现在要新增一些id，id不能与已有的id重复，需要找到一段有1000个位置以上空余区间来新增这些id

通过值减下标的方式分组

In [40]: for k, g in groupby(enumerate(lst), fun):
    ...:     print k, g, [v for i, v in g]
    ...:
    ...:
1 <itertools._grouper object at 0x0BDED0F0> [1, 2, 3]
2 <itertools._grouper object at 0x03C05AF0> [5, 6, 7, 8]
4 <itertools._grouper object at 0x03BFC910> [11, 12, 13]