0

我在 postgres 表中有以下形式的数据：

Col1   Col2    Col3              Col4
id1    a       b                 c 
id2    id1     timeBegin         1###-##-##
id2    id1     timeEnd           22##-##-##
id3    id4     id5               id6
id6    id3     timeBegin         2##-##-##
id7    id3     timeEnd           200-3-## 
id13   id8     id14              id15
id8    id9     timeBegin         -2-1-1
id10   id11    id12              id13

这里 1###-##-## 表示从 (1000-01-01 到 1999-12-31) 的时间不确定性

和 22##-##-## 表示从 (2200-01-01 到 2200-12-31) 的时间不确定性

和 2##-##-## 表示从 (200-01-01 到 200-12-31) 的时间不确定性

和 200-3-## 暗示从 (200-3-01 到 200-3-31) 的时间不确定性

和 20-3-## 表示从 (20-3-01 到 20-3-31) 的时间不确定性

和 200-3-## 表示从 (200-3-01 到 200-3-31) 的时间不确定性

和 -200-3-## 表示从 (-200-3-31 到 -200-3-01) 的时间不确定性

现在我想将 col1==col2 的 3 行合并为以下形式之一：

Col1   Col2    Col3              Col4       timeBegin      timeEnd
id1    a       b                 c          1000-01-01     2200-12-31 
id3    id4     id5               id6        200-01-02      200-3-31
id10   id11    id12              id13       NULL           NULL
id13   id8     id14              id15       2-1-1 BC       9999-12-12

如果未给出 col1==col2 的 timeEnd，则假定 9999-12-12 为 timeEnd

如果未给出 col1==col2 的 timeBegin，则假定 01-01-01 为 timeBegin

即我想在合并时花费最少的 timeBegin 和最多的 timeEnd。

是否可以在 postgres 中执行此连接操作。即我可以把它写成一个SQL 连接查询吗？

如果我也可以使用 python 之类的编程语言（以有效的方式）实现所需的功能：那将是很棒的。

4

3 回答 3

1

我不相信这是最好的方法，但这里有一对 Postgres 函数可以完成将模式转换为最小和最大日期的困难：

Example Fiddle

Create Function preprocessPattern(pat varchar(11), out cpat varchar(10), out neg boolean) as $$
declare
  y varchar(4);
  m varchar(2);
  d varchar(2);
  i int;
begin
  neg = false;
  if left(pat, 1) = '-' then
    neg = true;
    pat = right(pat, -1);
  end if;

  i = position('-' in pat);

  y = right('000' || left(pat, i - 1), 4);
  pat = right(pat, -i);
  i = position('-' in pat);
  m = right('0' || left(pat, i - 1), 2);
  pat = right(pat, -i);
  d = right('0' || pat, 2);
  cpat = y || '-' || m || '-' || d;
end;
$$ Language plpgsql;

Create Function dateFromFmt(fmt varchar(10), neg boolean) returns date as $$
begin
  if neg then
    return to_date(fmt || ' BC', 'yyyy-mm-dd BC');
  else
    return to_date(fmt, 'yyyy-mm-dd');
  end if;
end;
$$ Language plpgsql;

Create Function minDateFromPattern(pat varchar(11)) returns date as $$
declare
    i int;
    neg boolean;
    n varchar(10);
begin
    select * into pat, neg from preprocessPattern(pat);
    i = position('#' in pat);
    if i = 0 then
      return dateFromFmt(pat, neg);
    else
      n = left(pat, i - 1) || right('0000-00-00', 0 - position('#' in pat) + 1);
      n = replace(n, '-00', '-01');
      return dateFromFmt(n, neg);
    end if;
end;
$$ Language plpgsql;


Create Function maxDateFromPattern(pat varchar(11)) returns date as $$
declare
    i int;
    y int;
    m int;
    d int;
    x varchar(10);
    neg boolean;
    res date;
begin
    select * into pat, neg from preprocessPattern(pat);
    i = position('#' in pat);
    if i = 0 then
        return dateFromFmt(pat, neg);
    elsif i = 1 then
        return date '9999-12-31';
        -- from here down, pick the next highest mask, convert to min date then subtract one day
    elsif i <= 6 then -- just add 1 to year
        if i = 6 then i = 5; end if; -- skip - char
        x = cast(cast(left(pat, i - 1) as int) + 1 as varchar) || right(pat, 0 - i + 1);
    else
      y = cast(left(pat, 4) as int);
      if i = 7 then
          m = cast(substr(pat, 6, 1) as int) + 1;
          if m = 2 then
              m = 0;
              y = y + 1;
          end if;
          x = left(to_char(y, 'FM0000'), 4) || '-' || to_char(m, 'FM0') || '#-##';
      elsif i = 9 then
          m = cast(substr(pat, 6, 2) as int) + 1;
          if m > 12 then
              m = 1;
              y = y + 1;
          end if;
          x = left(to_char(y, 'FM0000'), 4) || '-' || to_char(m, 'FM00') || '-##';
      elseif i = 10 then
          m = cast(substr(pat, 6, 2) as int);
          d = cast(substr(pat, 9, 1) as int) + 1;
          if (m = 2 and d = 3) or d = 4 then
              m = m + 1;
              d = 0;
              if m > 12 then
                  m = 1;
                  y = y + 1;
              end if;
          end if;
          x = left(to_char(y, 'FM0000'), 4) || '-' || to_char(m, 'FM00') || '-' || to_char(d, 'FM0') || '#';
      end if;
    end if;
    -- the original logic looks a little silly now as we're preprocessing twice
    res = minDateFromPattern(x) - interval '1 day';
    if neg then
      return dateFromFmt(to_char(res, 'yyyy-mm-dd'), neg);
    else
      return res;
    end if;
end;
$$ Language  plpgsql;

这已经经历了足够多的迭代，可以使用一些重构

于 2013-10-06T22:30:18.380 回答
0

SELECT下面的方法使用带有语句的单个 SQL ，这些CASE语句通过后处理子查询来应用规则。可以进行进一步的调整，但它给出了总体思路。为复杂性道歉 - 这开始相当简单，仅基于帖子中的规则，但经过修改以处理一般情况，结果非常繁琐！

已知限制：此方法目前不能很好地处理闰年，并且假设 2 月的最后一天始终是 28 日，因此在安全方面会出错。这可以修复，但闰年计算并非完全微不足道，所以我故意将这一点排除在外，以免过于复杂。

SELECT Col1, Col2, Col3, Col4,
       CASE WHEN timeBegin IS NULL AND timeEnd IS NOT NULL
            THEN '01-01-01'
            WHEN timeBegin LIKE '-%' -- Handle negative (= BC) dates separately
            THEN CASE WHEN SUBSTRING(timeBegin, 2) LIKE '%-[0-1]#'
                      THEN SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 2) || '9 BC'
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-1-##'
                      THEN SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 4) || '-31 BC'
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-2-##'
                      THEN SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 4) || '-28 BC' -- No leap year calculation
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-3-##'
                      THEN SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 4) || '-31 BC'
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-4-##'
                      THEN SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 4) || '-30 BC'
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-5-##'
                      THEN SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 4) || '-31 BC'
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-6-##'
                      THEN SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 4) || '-30 BC'
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-7-##'
                      THEN SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 4) || '-31 BC'
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-8-##'
                      THEN SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 4) || '-31 BC'
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-9-##'
                      THEN SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 4) || '-30 BC'
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-10-##'
                      THEN SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 4) || '-31 BC'
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-11-##'
                      THEN SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 4) || '-30 BC'
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-0#-##'
                      THEN REPLACE(SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 7), '#', '9') || '-09-30 BC'
                      WHEN SUBSTRING(timeBegin, 2) LIKE '%-12-##' OR SUBSTRING(timeBegin, 2) LIKE '%-_#-##'
                      THEN REPLACE(SUBSTRING(timeBegin, 2, CHAR_LENGTH(timeBegin) - 7), '#', '9') || '-12-31 BC'
                      ELSE REPLACE(SUBSTRING(timeBegin, 2), '#', '9')
                 END
            ELSE REPLACE(REPLACE(REPLACE(timeBegin, '-0#', '-01'), '-##', '-01'), '#', '0')
       END AS timeBegin,
       CASE WHEN timeEnd IS NULL AND timeBegin IS NOT NULL
            THEN '9999-12-12'
            WHEN timeEnd LIKE '-%' -- Handle negative (= BC) dates separately
            THEN REPLACE(REPLACE(REPLACE(SUBSTRING(timeEnd, 2), '-0#', '-01'), '-##', '-01'), '#', '0') || ' BC'
            WHEN timeEnd LIKE '%-[0-1]#'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 1) || '9'
            WHEN timeEnd LIKE '%-1-3#' OR timeEnd LIKE '%-1-##'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 2) || '31'
            WHEN timeEnd LIKE '%-2-2#' OR timeEnd LIKE '%-2-##'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 2)|| '28' -- No leap year calculation
            WHEN timeEnd LIKE '%-3-3#' OR timeEnd LIKE '%-3-##'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 2)|| '31'
            WHEN timeEnd LIKE '%-4-3#' OR timeEnd LIKE '%-4-##'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 2) || '30'
            WHEN timeEnd LIKE '%-5-3#' OR timeEnd LIKE '%-5-##'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 2) || '31'
            WHEN timeEnd LIKE '%-6-3#' OR timeEnd LIKE '%-6-##'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 2) || '30'
            WHEN timeEnd LIKE '%-7-3#' OR timeEnd LIKE '%-7-##'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 2) || '31'
            WHEN timeEnd LIKE '%-8-3#' OR timeEnd LIKE '%-8-##'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 2) || '31'
            WHEN timeEnd LIKE '%-9-3#' OR timeEnd LIKE '%-9-##'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 2) || '30'
            WHEN timeEnd LIKE '%-10-3#' OR timeEnd LIKE '%-10-##'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 2) || '31'
            WHEN timeEnd LIKE '%-11-3#' OR timeEnd LIKE '%-11-##'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 2) || '30'
            WHEN timeEnd LIKE '%-0#-##'
            THEN SUBSTRING(timeEnd, 1, CHAR_LENGTH(timeEnd) - 5) || '09-30'
            WHEN timeEnd LIKE '%-1#-##' OR timeEnd LIKE '%-##-##'
            THEN SUBSTRING(REPLACE(timeEnd, '#', '9'), 1, CHAR_LENGTH(timeEnd) - 5) || '12-31'
            ELSE REPLACE(timeEnd, '#', '9')
       END AS timeEnd
FROM
(SELECT t1.*,
        (SELECT Col4 FROM Tbl t2 WHERE (t1.Col1 = t2.Col2 OR t1.Col2 = t2.Col1)
                                   AND t2.Col3 = 'timeBegin') AS timeBegin,
        (SELECT Col4 FROM Tbl t2 WHERE (t1.Col1 = t2.Col2 OR t1.Col2 = t2.Col1)
                                   AND t2.Col3 = 'timeEnd') AS timeEnd
 FROM Tbl t1
 WHERE t1.Col3 NOT IN ('timeBegin', 'timeEnd')) subquery
ORDER BY CAST(SUBSTRING(Col1, 3) AS INT)

这是一个SQL Fiddle Demo，它显示它产生与发布的示例类似的结果。

于 2013-10-14T21:36:45.680 回答
0

这个问题实际上有两个部分。一个对应于正确对齐表的数据。另一个是处理日期格式的数不清的混乱。

现在假设有两个 sql 函数 begin_time() 和 end_time()。我将在下面立即讨论它们。

为了对齐数据，左连接表两次：

select t.col1, t.col2, t.col3, t.col4,
       parse_begin_time(bt.col4) as timeBegin,
       parse_end_time(et.col4) as timeEnd
from yourtable t
left join yourtable as bt on begin_t.col2 = t.col1 and bt.col3 = 'timeBegin'
left join yourtable as et on end_t.col2 = t.col1 and et.col3 = 'timeEnd'
where t.col3 not in ('timeBegin', 'timeEnd');

如果由于存在多个条目而产生多个条目，请使用聚合：

select t.col1, t.col2, t.col3, t.col4,
       min(parse_begin_time(bt.col4)) as timeBegin,
       max(parse_end_time(et.col4)) as timeEnd
from yourtable t
left join yourtable as bt on begin_t.col2 = t.col1 and bt.col3 = 'timeBegin'
left join yourtable as et on end_t.col2 = t.col1 and et.col3 = 'timeEnd'
where t.col3 not in ('timeBegin', 'timeEnd')
group by t.col1, t.col2, t.col3, t.col4;

注意：如果您有大量数据，预计上述情况不会特别好。在 create table as ... 语句中运行它们一次并摆脱原始模式，或创建物化视图以供将来使用。

然后，您需要担心如何格式化凌乱的 timeBegin 和 timeEnd 字段，我假设它们存储在文本字段中。这将是这样的：

create or replace function parse_begin_time(text) returns date as $$
declare
  _input  text := $1;
  _output text;

  _bc     boolean := false;
  _y      text;
  _m      text;
  _d      text;

  _tmp    text;
  _i      int;
begin
  _input := trim(both from _input);

  -- PG is fine with '200-01-01 BC' as a date, but not with '-200-01-01'
  if left(_input, 1) = '-'
  then
    _bc    := true;
    _input := right(_input, -1);
  end if;

  -- Extract year, month and day
  _tmp := _input;
  _i   := position(_tmp for '-');
  _y   := substring(_tmp from 1 for i - 1);

  _tmp := substring(_tmp from i);
  _i   := position(_tmp for '-');
  _m   := substring(_tmp from 1 for i - 1);

  _tmp := substring(_tmp from i);
  _i   := position(_tmp for '-');
  _d   := substring(_tmp from 1 for i - 1);

  if _tmp <> '' or left(trim(left '0' from _y), 1) = 'X'
  then
    raise exception 'invalid date input: %', _input;
  end if;

  -- Prevent locale-specific text to date conversion issues with one or two digit years
  -- e.g. rewrite 1-2-3 as 0001-02-03.
  if length(_y) < 4
  then
    _y := lpad(_y, 4, '0');
  end if;

  if length(_m) < 2
  then
    _m := lpad(_m, 2, '0');
  end if;

  if length(_d) < 2
  then
    _d := lpad(_m, 2, '0');
  end if;

  -- Process year, month, day
  -- Add suitable logic here per your specs, using string and date functions
  -- http://www.postgresql.org/docs/current/static/functions-string.html
  -- http://www.postgresql.org/docs/current/static/functions-formatting.html
  -- http://www.postgresql.org/docs/current/static/functions-datetime.html
  -- for end-of-months, use the built-in arithmetics, e.g.:
  -- _date := _date + interval '1 month' - interval '1 day'

  -- Build _output
  _output := _y || '-' || _m || '-' || _d;

  if _bc
    _output := _output || ' BC';
  end if;

  return _output::date;
end;
$$ language plpgsql strict stable;

如果您对它更满意，该语言也可以是 plpython 或 plpythonu。我想你比我更了解这两个，当然有足够的 python 来编写所需的代码。如果您希望在 plpgsql 中有东西，Laurence 的代码是另一个很好的起点。

该strict语句告诉 Postgres 不要费心在 null 输入上调用该函数并立即返回 null。您可能不希望它用于 end_time 函数。

于 2013-10-10T18:39:00.837 回答