Lor*_*ish 5 python recursion generator
有没有办法使下面的代码工作?
add = lambda n: (yield n) or add(n+1)
Run Code Online (Sandbox Code Playgroud)
(答案不需要是功能样式)
def add(n):
yield n
for m in add(n+1):
yield m
Run Code Online (Sandbox Code Playgroud)
使用递归生成器,可以轻松构建复杂的回溯器:
def resolve(db, goals, cut_parent=0):
try:
head, tail = goals[0], goals[1:]
except IndexError:
yield {}
return
try:
predicate = (
deepcopy(clause)
for clause in db[head.name]
if len(clause) == len(head)
)
except KeyError:
return
trail = []
for clause in predicate:
try:
unify(head, clause, trail)
for each in resolve(db, clause.body, cut_parent + 1):
for each in resolve(db, tail, cut_parent):
yield head.subst
except UnificationFailed:
continue
except Cut, cut:
if cut.parent == cut_parent:
raise
break
finally:
restore(trail)
else:
if is_cut(head):
raise Cut(cut_parent)
...
for substitutions in resolve(db, query):
print substitutions
Run Code Online (Sandbox Code Playgroud)
这是一个由递归生成器实现的 Prolog 引擎。db 是一个字典,代表事实和规则的 Prolog 数据库。unify() 是统一函数,它创建当前目标的所有替换并将更改附加到跟踪中,以便以后可以撤消它们。Restore() 执行撤消操作,is_cut() 测试当前目标是否为“!”,以便我们可以进行分支修剪。
| 归档时间: |
|
| 查看次数: |
1832 次 |
| 最近记录: |