聚合函数是一种特殊类型的函数【与数组类元素相关】。我们首先创建说明所需的数据。
from pyDatalog import pyDatalog
pyDatalog.create_terms('X,Y,manager, count_of_direct_reports')
# the manager of Mary is John
+(manager['Mary'] == 'John')
+(manager['Sam'] == 'Mary')
+(manager['Tom'] == 'Mary')
最基本的聚合函数是len_()，它计算数组的规模
(count_of_direct_reports[X]==len_(Y)) <= (manager[Y]==X)
print(count_of_direct_reports['Mary']==Y)
pyDatalog 寻找 manager['Mary']==Y的所有可能解Y, 然后计算Y的数目。 .
聚合函数包括：
len_ (P[X]==len_(Y)) <= body ：P[X]是Y的值的计数（通过子句的主体与X关联）
sum_ (P[X]==sum_(Y, for_each=Z)) <= body ：P[X]是每个Z中的Y的总和。（Z用于区分可能相同的Y值）
min_， max_ (P[X]==min_(Y, order_by=Z)) <= body ：P[X]是按Z排序的Y的最小值（或最大值）。
tuple_ (P[X]==tuple_(Y, order_by=Z)) <= body ：P[X]是一个元组，包含按Z排序的Y的所有值。
concat_ (P[X]==concat_(Y, orde\r_by=Z, sep=',')) <= body ：与'sum'相同，但是用于字符串。字符串按Z排序，并用'，'分隔。
rank_ (P[X]==rank_(grou\p_by=Y, order_by=Z)) <= body ：P[X]是列表按Z排序时的Y值列表中的X的序列号。
running_sum_ (P[X]==running_sum_(N, group_by=Y, order_by=Z)) <= body ：P[X]是当Y按Z排序时每个在之前X或等于X的N值的总和 。 mean_和 linear_regression：请参阅我们的参考（ https://sites.google.com/site/pydatalog/reference ）
（对这块不是很理解，官网例子也不多，可能写得不是很清楚。）
from pyDatalog import pyDatalog
pyDatalog.create_terms('X,Y,Z, works_in, department_size, manager, indirect_manager, count_of_indirect_reports')
向集合中添加事实的方法：
# Mary works in Production
+ works_in('Mary', 'Production')
+ works_in('Sam', 'Marketing')
+ works_in('John', 'Production')
+ works_in('John', 'Marketing')
同样，字面值也可以按值查询,比在原生python中的操作简洁【隐藏了一个循环】。
# give me all the X that work in Marketing
print(works_in(X, 'Marketing'))
# procedural equivalent in Python
# for i in _works_in:
# if i[1]=='Marketing':
# print i[0]
# one of the indirect manager of X is Y, if the (direct) manager of X is Y
indirect_manager(X,Y) <= (manager[X] == Y)
# another indirect manager of X is Y, if there is a Z so that the manager of X is Z,
# and an indirect manager of Z is Y
indirect_manager(X,Y) <= (manager[X] == Z) & indirect_manager(Z,Y)
print(indirect_manager('Sam',X))
请注意，这里使用了2个独立的子句实现了隐式的“或”。
【自己总结了“字面值”和上一篇中的“函数”的区别与联系： 1.前者使用圆括号，没有值（也就不能用==）；后者使用方括号，对于括号中的元素有一个取值。 2.在用法上，manager[X] == Y 与 manager(X,Y)也是相似的， 但是按关键字查询中,manager['Mary'] == X 与 manager('Mary',X)相比会更高效。前者是一个哈希操作，而后者依然需要循环。】
当解析查询时，pyDatalog可以记得中间结果，通过这个过程被称为记忆化。这使查询更快，而且它也有助于处理无限循环！
# the manager of John is Mary (whose manager is John !)
manager['John'] = 'Mary'
manager['Mary'] = 'John'
print(indirect_manager('John',X)) # no infinite loop
这使pyDatalog成为在复杂数据结构上实现递归算法的一个很好的工具，例如表示网络。
也可以删除事实：
# John does not work in Production anymore
- works_in('John', 'Production')
# 补充：
# 【也可以用增减事实(fact)同样的方式增减定理，但注意加上括号，加定理没有加号，但是减定理需要减号】
# - (indirect_manager(X,Y) <= (manager[X] == Z) & indirect_manager(Z,Y))
# print(indirect_manager('John',X))
聚合函数也可以在字面值上定义：
(count_of_indirect_reports[X]==len_(Y)) <= indirect_manager(Y,X)
print(count_of_indirect_reports['John']==Y)
# 自己写了一个小推理规则
pyDatalog.create_terms('X,Y,Z,father,fatherOf,grandfatherOf')
(grandfatherOf[X] == Z) <= ((fatherOf[X]==Y) & (fatherOf[Y]==Z))
fatherOf["乾隆"] = "雍正"
fatherOf["雍正"] = "康熙"
print(grandfatherOf["乾隆"] == X)
树和图可以用它们的结点之间的连接定义：
pyDatalog.create_terms('link, can_reach')
# there is a link between node 1 and node 2
+link(1,2)
+link(2,3)
+link(2,4)
+link(2,5)
+link(5,6)
+link(6,7)
+link(7,2)
# 无向图，边双向连接
link(X,Y) <= link(Y,X)
下面两个子句解释了如何确定两个结点X,Y之间的可达关系：
# can Y be reached from X ?
can_reach(X,Y) <= link(X,Y) # direct link
# via Z
can_reach(X,Y) <= link(X,Z) & can_reach(Z,Y) & (X!=Y)
print (can_reach(1,Y))
请注意，尽管图中有循环，但pyDatalog足够聪明以解决查询问题。
这个例子（ https://github.com/pcarbonn/pyDatalog/blob/master/pyDatalog/examples/graph.py ） 中有更多的图算法的例子。
8皇后问题
通过结合我们迄今为止所学的，我们可以用声明式编程处理复杂问题，并让计算机找到解决它们的过程。 作为一个例子，让我们为8皇后问题编程找到一个有效的解决方案 。任何N皇后问题的解决方案可以在这里找到( https://github.com/pcarbonn/pyDatalog/blob/master/pyDatalog/examples/queens_N.py ) 。
from pyDatalog import pyDatalog
pyDatalog.create_terms('N,X0,X1,X2,X3,X4,X5,X6,X7')
pyDatalog.create_terms('ok,queens,next_queen')
# the queen in the first column can be in any row
queens(X0) <= (X0._in(range(8)))
# to find the queens in the first 2 columns, find the first one first, then find a second one
queens(X0,X1) <= queens(X0) & next_queen(X0,X1)
# repeat for the following queens
queens(X0,X1,X2) <= queens(X0,X1) & next_queen(X0,X1,X2)
queens(X0,X1,X2,X3) <= queens(X0,X1,X2) & next_queen(X0,X1,X2,X3)
queens(X0,X1,X2,X3,X4) <= queens(X0,X1,X2,X3) & next_queen(X0,X1,X2,X3,X4)
queens(X0,X1,X2,X3,X4,X5) <= queens(X0,X1,X2,X3,X4) & next_queen(X0,X1,X2,X3,X4,X5)
queens(X0,X1,X2,X3,X4,X5,X6) <= queens(X0,X1,X2,X3,X4,X5) & next_queen(X0,X1,X2,X3,X4,X5,X6)
queens(X0,X1,X2,X3,X4,X5,X6,X7) <= queens(X0,X1,X2,X3,X4,X5,X6) & next_queen(X0,X1,X2,X3,X4,X5,X6,X7)
# the second queen can be in any row, provided it is compatible with the first one
next_queen(X0,X1) <= queens(X1) & ok(X0,1,X1)
# to find the third queen, first find a queen compatible with the second one, then with the first
# re-use the previous clause for maximum speed, thanks to memoization
next_queen(X0,X1,X2) <= next_queen(X1,X2) & ok(X0,2,X2)
# repeat for all queens
next_queen(X0,X1,X2,X3) <= next_queen(X1,X2,X3) & ok(X0,3,X3)
next_queen(X0,X1,X2,X3,X4) <= next_queen(X1,X2,X3,X4) & ok(X0,4,X4)
next_queen(X0,X1,X2,X3,X4,X5) <= next_queen(X1,X2,X3,X4,X5) & ok(X0,5,X5)
next_queen(X0,X1,X2,X3,X4,X5,X6) <= next_queen(X1,X2,X3,X4,X5,X6) & ok(X0,6,X6)
next_queen(X0,X1,X2,X3,X4,X5,X6,X7) <= next_queen(X1,X2,X3,X4,X5,X6,X7) & ok(X0,7,X7)
# it's ok to have one queen in row X1 and another in row X2 if they are separated by N columns
ok(X1, N, X2) <= (X1 != X2) & (X1 != X2+N) & (X1 != X2-N)
# give me one solution to the 8-queen puzzle
print(queens(X0,X1,X2,X3,X4,X5,X6,X7).data[0])
系列链接：
pyDatalog: python的逻辑编程引擎(用于推理、查询等)【一：序言】
pyDatalog: python的逻辑编程引擎【二：基础教程（上）】
pyDatalog: python的逻辑编程引擎【三：基础教程（下）】
pyDatalog: python的逻辑编程引擎【四：从文件中加载和执行程序】
pyDatalog: python的逻辑编程引擎【五：与“知识图谱”的交互】