1. 斐波那契 from itertools import islice def fib(): a, b = 0, 1 while True: yield a a, b = b, a+b print list(islice(fib(), 5)) # [0, 1, 1, 2, 3] 2. for……else……用法(以查找素数为例) 正常版本: def print_prime(n): for i in xrange(2, n): found = True for j in xrange(2, i)
Linq to sql 使用group by 统计多个字段,然后返回多个字段的值,话不多说,直接上例子: where u.fy_no == fy_no orderby u.we_no group u by new { weno = u.we_no, wename = u.we_name } into g select new { g.Key.weno, g.Key.wename }; 结果就是根据we_no和we_name的统计结果,返回这两个字段的数据.
mysql count group by统计条数方法 mysql 分组之后如何统计记录条数? gourp by 之后的 count,把group by查询结果当成一个表再count一次select count(*) as count from(SELECT count(*) FROM 表名 WHERE 条件 GROUP BY id ) a; 实战例子:select count(*) as total from (select count(*) from users group by user_i
--查询月销售量小于20的人员销售情况 SELECT C.Category, SUM(c.[Count]) AS CategorySum FROM test.dbo.Category c --WHERE c.Category IN ('A','C') GROUP BY C.Category ORDER BY CategorySum --查询每门课的最低分 SELECT C.Category, MIN(C.[Count]) AS CategoryMinCount FROM test.dbo.Cat
group by 最后一个时间是多少按多少分组 select count(1), trunc(a.refund_insert_time, 'hh24') + case when to_char(refund_insert_time,'mi') >= '30' then numtodsinterval(30,'minute') else numtodsinterval(0,'minute') end from refund_record a where a.refund_complete_time
137. Single Number II Given an array of integers, every element appears three times except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 记录32个bit每个bit出现的
