DAI SQL FAQ
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列名和字符串如何区分
字符串用单引号（'）来包围，而列名通常不需要引号。如果列名中包含特殊字符（例如空格、连字符、其他符号等），您需要使用双引号（"）来引用列名。如下：
%%sql
select date, instrument, close/open as "con1: close/open"
from cn_stock_bar1d
where date = '2024-08-12' and "con1: close/open" > 1
如何实现自定义的带有窗口的 macro 函数
创建 macro 函数的语法可参见 create macro.
DAI 提供的滚动窗口函数 m_aggregate_func (e.g. m_avg, m_median, etc.) 有这样的模型：
create macro m_aggregate_func(args, win_sz, pb:=instrument, ob:=date) as
aggregate_func(args) over (partition by pb order by ob rows win_sz-1 preceding)
注意：args 如果是多个参数，定义 macro 函数时需要分开指明。默认参数 pb, ob 如果有多个参数也需要分开指明。也就是说每个标识符只能传入一个参数。
类似的，时间截面窗口函数 c_rank, c_avg 实现：
create macro c_rank_asc(arg, pb:=date) as
rank() over (partition by pb order by arg);
create macro c_rank_desc(arg, pb:=date) as
rank() over (partition by pb order by arg desc);
-- 不能传入关键字作为参数，asc/desc 是关键字
-- create macro c_rank(arg, pb:=date, order_type:=asc) as
-- rank() over (partition by pb order by arg order_type);
-- workaround
create macro c_rank(arg, pb:=date, ascending:=true) as
if (ascending, rank() over (partition by pb order by arg),
rank() over (partition by pb order by arg desc));
create macro c_avg(arg, pb:=date) as
avg(arg) over (partition by pb);
窗口函数宏传入多个 partition by/order by 项
有时候我们会需要按多个列做分区，比如求每日同一行业内所有的股票的平均收盘价：
select
instrument,
date,
close,
c_avg(close, pb:='date, industry_level1_code') as avg1,
-- avg1 is equivalent to avg2
avg(close) over (partition by date, industry_level1_code) as avg2,
-- some factor: f(close, avg)
from cn_stock_bar1d
join cn_stock_industry_component -- industry_level1_code
using(date, instrument)
where date >= '2023-01-01' and date < '2024-01-01' and industry = 'sw2021'
我们可以传入以逗号,分隔的字符串（单引号括起）给 pb参数, e.g., pb:=’A, B, C, D’，它会自动解析成 partition by A, B, C, D。如果某列列名含有空格，e.g., pb:=’date, industry level name’ ，它相当于 partition by date, “industry level name“，即该项除去两边的空白符。order by可以传入类似的字符串，但由于 order 有升降序之分，可以传入前置 -来表示降序，否则为升序。例如，ob:=’A,-B,C’ 则表示 order by A asc, B desc, C asc（亦可省略asc, 默认升序）。
df = pd.DataFrame({'A': ['a', 'b', 'c', 'a', 'b', 'c', 'a', 'b', 'c'],
'B': [1, 1, 1, np.nan, 2, 2, 3, 3, 3],
'C': [1, 2, 3, 4, 5, 6, 7, 8, 9],
'D': ['X', 'X', 'X', 'Y', 'Y', 'Y', 'Y', 'X', 'X']})
df2 = pd.DataFrame({'A': ['a', 'b', 'c'],
'E': ['apple', 'banana', 'coconut']})
select *,
rank('A, -B') as x, -- equivalent to rAB
rank() over (order by A, B desc) as rAB,
rank_by(D, '-A, C') as y, -- equivalent to rDAC
rank() over (partition by D order by A desc, C) as rDAC
from df
join df2 using (A)
-- output
A B C D E x rAB y rDAC
0 c 1.0 3 X coconut 8.0 8.0 1.0 1.0
1 c 3.0 9 X coconut 6.0 6.0 2.0 2.0
2 b 1.0 2 X banana 5.0 5.0 3.0 3.0
3 b 3.0 8 X banana 3.0 3.0 4.0 4.0
4 a 1.0 1 X apple 2.0 2.0 5.0 5.0
5 c 2.0 6 Y coconut 7.0 7.0 1.0 1.0
6 b 2.0 5 Y banana 4.0 4.0 2.0 2.0
7 a NaN 4 Y apple NaN NaN 3.0 3.0
8 a 3.0 7 Y apple 1.0 1.0 4.0 4.0
当有歧义时（例如包含 join 语句），可以列名前可添加表名来消除歧义。比如上面的例子可以写成 rank_by(D, '-A, df.C') 如果 df2也含有列C。
select *,
m_max(B, 2, pb:=D, ob:='C') as max_asc,
m_max(B, 2, pb:=D, ob:='-C') as max_desc,
from df
-- output
A B C D max_asc max_desc
0 c 3.0 9 X 3.0 NaN
1 b 3.0 8 X 3.0 3.0
2 c 1.0 3 X 1.0 3.0
3 b 1.0 2 X 1.0 1.0
4 a 1.0 1 X NaN 1.0
5 a 3.0 7 Y 3.0 NaN
6 c 2.0 6 Y 2.0 3.0
7 b 2.0 5 Y NaN 2.0
8 a NaN 4 Y NaN NaN
行业市值中性化 c_neutralize 的使用
-
为什么求出来基本全是 NaN值呢？
答：
c_neutralize(y, industry_level1_code, market_cap)中传入的因子值是y(右端向量），通过industry_level1_code获取的dummy矩阵及取对数后的市值（以及为constant添加的常向量1）共同组成X，然后做的线性回归。当y,X中有任何元素为NaN时求线性方程时就会全部为NaN。所以算之前需要先把 NaN 过滤掉或者填充，一般y很有可能含有 NaN，比如y = m_avg(close, 5)。 -
c_neutralize采用的算法：def process_factor(factor): median = np.median(factor) mad = np.median(abs(factor - median)) mad *= 3*1.4826 clipped = factor.clip(median - mad, median + mad) # pandas std() has ddof=1, while numpy std() has ddof=0 return (clipped - clipped.mean()) / clipped.std(ddof=1) def sm_resid(data): X = pd.get_dummies(data['industry_level1_code']).astype('float64') # X = X.reindex(columns=dummy_cols) X['log_marketcap'] = np.log(data['total_market_cap']) y = process_factor(data['pb']) X = sm.add_constant(X) # intercept term model = sm.OLS(y, X) results = model.fit(method='pinv') return pd.DataFrame(results.resid, index=data.index) def np_resid(data): X = pd.get_dummies(data['industry_level1_code']).astype('float64') # X = X.reindex(columns=dummy_cols) X['log_marketcap'] = np.log(data['total_market_cap']) y = process_factor(data['pb']) beta, _, _, _ = np.linalg.lstsq(X, y, rcond=None) y_pred = X.dot(beta) return pd.DataFrame(y - y_pred, index=data.index) t1 = time.time() df['sm_resid'] = df.groupby('date').apply(sm_resid).reset_index(level=0, drop=True)使用例子和对比参见: https://bigquant.com/codeshare/5739e696-fd64-480c-95ec-793ea9ff889c
-
如果想自己处理 factor 后再只算个残差， 可以使用
c_neutralize_resid(y, industry_level1_code, log_marketcap)函数 （y = processed_factor）。
缺失值填充/替换
当分区后的数据含有NA值时，填充其值为前一个非NA值：
select
instrument, date,
last(columns(* exclude (date, instrument)) ignore nulls)
over (partition by instrument order by date rows between
unbounded preceding and current row) as 'columns(*)'
from cn_stock_bar1d
where date >= '2023-01-01'
order by instrument, date
columns(* exclude (date, instrument)) 会依次扩展成每一个非 date 和 instrument 的列，最好减少不需要操作的数据。as 'columns(*)’ 会保存处理后的数据成原始的列名，as 'columns(*)_suffix_name’ 的话会把处理后的数据依次存成原列名加上后缀名（_suffix_name）的列名。
类似的如果想批量替换NA值为其他值，比如 0，可以把 last 函数替换成 ifnull 函数：
select
instrument, date,
ifnull(columns(* exclude (date, instrument, name)), 0) as 'columns(*)'
from cn_stock_bar1d
where date >= '2023-01-01'
order by instrument, date
把当天open, close为空的值填充为当天所有股票的中位数：
select
instrument, date,
ifnull(columns(['close', 'open']),
c_median(columns(['close', 'open']))) as 'columns(*)'
from cn_stock_bar1d
where date >= '2023-01-01'
order by instrument, date
计算连续上涨天数，最近一次“xxx”至今天数
m_consecutive_true_count(expr)可以用于计算该类问题。该函数会计算 expr 从当前行起往前取连续为 true 的数量, expr 为 NULL 的行其对应结果为 NULL.
expr m_consecutive_true_count(expr)
0 0
0 0
0 0
1 1
1 2
0 0
1 1
1 2
1 3
0 0
可以看出该特性可以帮助算出连续性因子，比如： 连续上涨天数，最近一次涨停距今天数（通过取 expr = not limit_up）。
select instrument, date, close,
m_consecutive_rise_count(close) as rise_count,
m_consecutive_true_count(close > m_lag(close, 1)) as rise_count2,
if(price_limit_status = 3, 1, 0) as limit_up,
1 - limit_up as not_limit_up,
m_consecutive_true_count(not_limit_up) as days_since_last_limit_up,
from cn_stock_bar1d
join cn_stock_status
using (date, instrument)
where date >= '2024-08-01' and instrument='000062.SZ'
--qualify m_cumsum(limit_up)>0
order by instrument, date
-- output
instrument date close rise_count rise_count2 limit_up not_limit_up days_since_last_limit_up
0 000062.SZ 2024-08-01 71.737065 0 NaN 0 1 1.0
1 000062.SZ 2024-08-02 69.989318 0 0.0 0 1 2.0
2 000062.SZ 2024-08-05 67.129369 0 0.0 0 1 3.0
3 000062.SZ 2024-08-06 68.241571 1 1.0 0 1 4.0
4 000062.SZ 2024-08-07 67.764913 0 0.0 0 1 5.0
5 000062.SZ 2024-08-08 68.321014 1 1.0 0 1 6.0
6 000062.SZ 2024-08-09 68.797673 2 2.0 0 1 7.0
7 000062.SZ 2024-08-12 67.923799 0 0.0 0 1 8.0
8 000062.SZ 2024-08-13 68.321014 1 1.0 0 1 9.0
9 000062.SZ 2024-08-14 69.194888 2 2.0 0 1 10.0
-- 对于我们给定选取的数据集里面，可能不知道上一次涨停是什么时候，此时 days_since_last_limit_up 可以理解为至少隔了这么多交易日
-- 可以添加 qualify m_cumsum(limit_up)>0 来过滤掉这部分“不准确”的结果
10 000062.SZ 2024-08-15 76.106432 3 3.0 1 0 0.0
11 000062.SZ 2024-08-16 83.732964 4 4.0 1 0 0.0
12 000062.SZ 2024-08-19 92.074483 5 5.0 1 0 0.0
13 000062.SZ 2024-08-20 101.289876 6 6.0 1 0 0.0
14 000062.SZ 2024-08-21 111.458585 7 7.0 1 0 0.0
15 000062.SZ 2024-08-22 122.580611 8 8.0 1 0 0.0
16 000062.SZ 2024-08-23 134.814839 9 9.0 1 0 0.0
17 000062.SZ 2024-08-26 148.320156 10 10.0 1 0 0.0
18 000062.SZ 2024-08-27 163.176004 11 11.0 1 0 0.0
19 000062.SZ 2024-08-28 179.461827 12 12.0 1 0 0.0
20 000062.SZ 2024-08-29 186.293929 13 13.0 0 1 1.0
21 000062.SZ 2024-08-30 204.963043 14 14.0 1 0 0.0
22 000062.SZ 2024-09-02 225.459347 15 15.0 1 0 0.0
23 000062.SZ 2024-09-03 248.021171 16 16.0 1 0 0.0
24 000062.SZ 2024-09-04 272.807399 17 17.0 1 0 0.0
25 000062.SZ 2024-09-05 300.056362 18 18.0 1 0 0.0
26 000062.SZ 2024-09-06 330.085831 19 19.0 1 0 0.0
27 000062.SZ 2024-09-09 297.116970 0 0.0 0 1 1.0
28 000062.SZ 2024-09-10 267.405273 0 0.0 0 1 2.0
29 000062.SZ 2024-09-11 272.330741 1 1.0 0 1 3.0
30 000062.SZ 2024-09-12 260.175956 0 0.0 0 1 4.0
31 000062.SZ 2024-09-13 234.198082 0 0.0 0 1 5.0
32 000062.SZ 2024-09-18 245.320107 1 1.0 0 1 6.0
注意到收盘价连续上涨天数因子也可以使用m_consecutive_rise_count(close)计算，但第一天的结果为0。而使用 m_consecutive_true_count(close > m_lag(close, 1)) 计算的因子由于 m_lag(close, 1) 的原因导致第一天 close > m_lag(close, 1) 为NULL，从而其其第一天的结果也为NULL。性能上直接使用m_consecutive_rise_count 会快一些因为不需要算 m_lag(close, 1)。
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