列表的内存中完全外部连接

Question

如何着手在具有以下签名的列表上编写类似 SQL 的完全连接操作？

fullJoin :: [a] -> [b] -> (a -> b -> Bool) -> [(Maybe a, Maybe b)]
fullJoin xs [] _ = map (\x -> (Just x, Nothing)) xs
fullJoin [] ys _ = map (\y -> (Nothing, Just y)) ys
fullJoin xs@(xh:xt) ys@(yh:yt) on = 
  if on xh yh
    then (Just xh, Just yh) : fullJoin xt yt
    else ???

所写条件由a -> b -> Bool提供。在示例中，它被设置为 (\n i -> length n == i)，这意味着来自 names 的记录与 nums 中的数字具有相同的长度。

示例：

names = ["Foo","Bar", "John" , "Emily", "Connor"]
nums = [1,2,3,4,5]

fullJoin names nums (\n i -> length n == i)
  == [ (Just "Foo", Just 3)
     , (Just "Bar", Just 3)
     , (Just "John", Just 4)
     , (Just "Emily", Just 5)
     , (Just "Connor", Nothing)
     , (Nothing, Just 1)
     , (Nothing, Just 2)
     ]

为了阐明上述函数的确切 SQL 等价物，下面是它在 PostgreSQL 中的写法：

create table names as
select * from (values 
               ('Foo'),
               ('Bar'),
               ('John'),
               ('Emily'),
               ('Connor')
              ) 
              as z(name);

create table nums as
select * from (values 
               (1),
               (2),
               (3),
               (4),
               (5)
              ) 
              as z(num);

select * from names
full join nums
on char_length(name) = num

并且运行这将产生：

name    num
(null)  1
(null)  2
Foo     3
Bar     3
John    4
Emily   5
Connor  (null)

Fiddle link : sqlfiddle

Answer 1

只是迟钝地实现你想要它做的事情。超级低效：

fullJoin :: [a] -> [b] -> (a -> b -> Bool) 
         -> [(Maybe a, Maybe b)]
fullJoin xs ys p = concatMap (map (Just *** Just) . snd) a
                    ++ [(Just x, Nothing) | (x, []) <- a]
                    ++ [(Nothing, Just y) | (y, []) <- b]
  where
    a = [(x, [(x, y) | y <- ys, p x y]) | x <- xs]
    b = [(y, [(x, y) | x <- xs, p x y]) | y <- ys]

如果允许我们在 a 和 b 类型上添加 Eq 约束，我们可以使用 Data.List.\ 来查找不匹配的元素，而不是使第二扫.

测试：

> mapM_ print $ fullJoin names nums (\n i -> length n == i)
(Just "Foo",Just 3)
(Just "Bar",Just 3)
(Just "John",Just 4)
(Just "Emily",Just 5)
(Just "Connor",Nothing)
(Nothing,Just 1)
(Nothing,Just 2)

Answer 2

条件 rel(x, y) 下表 X 和 Y 之间的完全外部联接可以被认为是三个（不相交的）部分的并集：

对的集合 (x, y) 其中 rel(x,y)
对 (x0, null) 的集合，其中对于所有 y in Y、not (rel(x0, y))
对 (null, y0) 的集合，其中对于所有 x in X、not (rel(x, y0))

我们可以用类似的方式构造我们的 Haskell 程序：

fullJoin :: [a] -> [b] -> (a -> b -> Bool) -> [(Maybe a, Maybe b)]
fullJoin xs ys rel = concat $
  [[(Just x, Just y) | y <- ys, x `rel` y] | x <- xs] -- for each x, find all the related ys
  <> [[(Just x, Nothing)] | x <- xs, all (\y -> not (x `rel` y)) ys] -- find all xs that have no related ys
  <> [[(Nothing, Just y)] | y <- ys, all (\x -> not (x `rel` y)) xs] -- find all ys that have no related xs

问题的提出方式，你不能比 O(n^2) 更快。也就是说，我的解决方案虽然是 O(n^2)，但并不是最优的：它遍历了 xs 两次。你可以想想你必须做什么才能只遍历 xs 一次。它与找到一种方法来跟踪您已经看过的xs有关。

Answer 3

这是一个天真的 Python 实现。是O(n^2).

#! /usr/bin/env python

def full_join_cond (list1, list2, condition_fn):
    answer = []
    for x in list1:
        matches = []
        for y in list2:
            if condition_fn(x, y):
                matches.append(y)
        if len(matches):
            answer.extend([[x, y] for y in matches])
        else:
            answer.append([x, None])

    for y in list2:
        matched = False
        for x in list1:
            if condition_fn(x, y):
                matched = True
                break
        if not matched:
            answer.append([None, y])

    return answer


names = ["Foo","Bar", "John" , "Emily", "Connor"]
nums = [1,2,3,4,5]
cond = lambda x, y: len(x) == y

print(full_join_cond(names, nums, cond))

这是一个更接近于数据库执行此连接的方式的实现。 O(size_of_output) 经常是 O(n).

def list_map_to_dict (l, m):
    d = {}
    for x in l:
        mx = m(x)
        if mx in d:
            d[mx].append(x)
        else:
            d[mx] = [x]
    return d

def full_join_map (list1, map1, list2, map2):
    dict1 = list_map_to_dict(list1, map1)
    dict2 = list_map_to_dict(list2, map2)

    answer = []
    for mx, xs in dict1.iteritems():
        if mx in dict2:
            for y in dict2[mx]:
                answer.extend([[x, y] for x in xs])
            dict2.pop(mx)
        else:
            answer.extend([[x, None] for x in xs])

    for my, ys in dict2.iteritems():
        answer.extend([[None, y] for y in ys])
    return answer

print(full_join_map(names, lambda x: len(x), nums, lambda y: y))

对于咯咯笑和咧嘴笑，这两者可以结合在一个既通用又可以运行快速的函数中。（我没有尝试使函数签名合理 - 只是想展示这个想法。）

def full_join (list1, map1, list2, map2, cond=None):
    if map1 is None:
        map1 = lambda x: None

    if map2 is None:
        map2 = lambda y: None

    if cond is None:
        cond = lambda x, y: True

    dict1 = list_map_to_dict(list1, map1)
    dict2 = list_map_to_dict(list2, map2)

    answer = []
    for mx, xs in dict1.iteritems():
        if mx in dict2:
            answer.extend(full_join_cond(xs, dict2[mx], cond))
            dict2.pop(mx)
        else:
            answer.extend([[x, None] for x in xs])

    for my, ys in dict2.iteritems():
        answer.extend([[None, y] for y in ys])
    return answer

列表的内存中完全外部连接

In-memory full outer join for lists

sql

algorithm

haskell

full-outer-join