【发布时间】：2010-01-25 17:07:29
【问题描述】：
我编写了一个程序，它生成随机表达式，然后使用遗传技术来选择适合度。
程序的以下部分生成随机表达式并将其存储在树结构中。
由于这在运行期间可能会被调用数十亿次，我认为它应该针对时间进行优化。
我是编程新手，我自己工作（玩耍），就像我在互联网上搜索一样
想法，我想要一些意见，因为我觉得我正在孤立地做这件事。
瓶颈似乎是 Node.init ()，（占总时间的 22%）和 random.choice()，（占总时间的 14%）
import random
def printTreeIndented(data, level=0):
'''utility to view the tree
'''
if data == None:
return
printTreeIndented(data.right, level+1)
print ' '*level + ' '+ str(data.cargo)#+ ' '+ str(data.seq)+ ' '+ str(data.branch)
printTreeIndented(data.left, level+1)
#These are the global constants used in the Tree.build_nodes() method.
Depth = 5
Ratio = .6 #probability of terminating the current branch.
Atoms = ['1.0','2.0','3.0','4.0','5.0','6.0','7.0','8.0','9.0','x','x','x','x']
#dict of operators. the structure is: operator: number of arguements
Operators = {'+': 2, '-': 2, '*': 2, '/': 2, '**': 2}
class KeySeq:
'''Iterator to produce sequential
integers for keys in Tree.thedict
'''
def __init__(self, data = 0):
self.data = data
def __iter__(self):
return self
def next(self):
self.data = self.data + 1
return self.data
KS = KeySeq()
class Node(object):
'''
'''
def __init__(self, cargo, left=None, right=None):
object.__init__(self)
self.isRoot = False
self.cargo = cargo
self.left = left
self.right = right
self.parent = None
self.branch = None
self.seq = 0
class Tree(object):
def __init__(self):
self.thedict = {} #provides access to the nodes for further mutation and
# crossbreeding.
#When the Tree is instantiated, it comes filled with data.
self.data = self.build_nodes()
# Uncomment the following lines to see the data and a crude graphic of the tree.
# print 'data: '
# for v in self.thedict.itervalues():
# print v.cargo,
# print
# print
# printTreeIndented(self.data)
def build_nodes (self, depth = Depth, entry = 1, pparent = None,
bbranch = None):
'''
'''
r = float()
r = random.random()
#If r > Ratio, it forces a terminal node regardless of
#the value of depth.
#If entry = 1, then it's the root node and we don't want
# a tree with just a value in the root node.
if (depth <= 0) or ((r > Ratio) and (not (entry))):
'''
Add a terminal node.
'''
this_atom = (random.choice(Atoms))
this_atom = str(this_atom)
this_node = Node(this_atom)
this_node.parent = pparent
this_node.branch = bbranch
this_node.seq = KS.next()
self.thedict[this_node.seq] = this_node
return this_node
else:
'''
Add a node that has branches.
'''
this_operator = (random.choice(Operators.keys()))
this_node = Node(this_operator)
if entry:
this_node.isRoot = True
this_node.parent = pparent
this_node.branch = bbranch
this_node.seq = KS.next()
self.thedict[this_node.seq] = this_node
#branch as many times as 'number of arguements'
# it's only set up for 2 arguements now.
for i in range(Operators[this_operator]):
depth =(depth - 1)
if i == 0:
this_node.left = (self.build_nodes(entry = 0, depth =(depth),
pparent = this_node, bbranch = 'left'))
else:
this_node.right = (self.build_nodes(entry = 0, depth =(depth),
pparent = this_node, bbranch = 'right'))
return this_node
def Main():
for i in range(100000):
t = Tree()
return t
if __name__ == '__main__':
rresult = Main()
【问题讨论】：
标签： python optimization expression-trees