Numbers 表保存数字的值及其频率。
+----------+-------------+
| Number | Frequency |
+----------+-------------|
| 0 | 7 |
| 1 | 1 |
| 2 | 3 |
| 3 | 1 |
+----------+-------------+
在此表中，数字为 0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 3，所以中位数是 (0 + 0) / 2 = 0。
+--------+
| median |
+--------|
| 0.0000 |
+--------+
请编写一个查询来查找所有数字的中位数并将结果命名为 median 。
--感觉做得太复杂了
select
sum(number)/count(distinct number) as median
from(
select
IF(sum_number%2=0,
case when (sum_number/2>=@lastcount and sum_number/2<=countn) or (sum_number/2+1>=@lastcount and sum_number/2+1<=countn) then number else null end,
case when ceil(sum_number/2)>=@lastcount and ceil(sum_number/2)<=countn then number else null end) as number,
--分奇偶，奇数返回一个就好，偶数考虑散步到两个数字的情况，关键是sum_number/2在该数累计频数和上个数累计频数的区间里
@lastcount:=countn+1 as lastcount --算上个累计频数结束后，从哪里开始
from(
select
number,
frequency,
@countn:= case when @countn=0 then frequency else @countn+frequency end as countn, --记录累计频数
sum_number --总共多少数
from
numbers,(select @countn:=0) b,(select sum(frequency) as sum_number from numbers) s
order by
number) t, (select @lastcount:=0) t1) t2
