我目前正在尝试创建一个Python脚本,它将自动生成有效的空格分隔的算术表达式.但是,我得到的示例输出如下所示:( 32 - 42 / 95 + 24 ( ) ( 53 ) + ) 21
虽然我的空括号完全没问题,但我不能在计算中使用这个自动生成的表达式,因为24和53之间没有运算符,而结束时21之前的+没有第二个参数.
我想知道的是,有没有办法使用Pythonic解决方案解释/修复这些错误？(在任何人指出它之前,我将首先承认我下面发布的代码可能是我推动的最差代码并且符合......嗯,很少有Python的核心原则.)
import random
parentheses = ['(',')']
ops = ['+','-','*','/'] + parentheses
lines = 0
while lines < 1000:
fname = open('test.txt','a')
expr = []
numExpr = lines
if (numExpr % 2 == 0):
numExpr += 1
isDiv = False # Boolean var, makes sure there's no Div by 0
# isNumber, isParentheses, isOp determine whether next element is a number, parentheses, or operator, respectively
isNumber = random.randint(0,1) == 0 # determines whether to start sequence with number or parentheses
isParentheses = not isNumber
isOp = False
# Counts parentheses to ensure parentheses are matching
numParentheses = 0
while (numExpr > 0 or numParentheses > 0):
if (numExpr < 0 and numParentheses > 0):
isDiv = False
expr.append(')')
numParentheses -= 1
elif (isOp and numParentheses > 0):
rand = random.randint(0,5)
expr.append(ops[rand])
isDiv = (rand == 3) # True if div op was just appended
# Checks to see if ')' was appended
if (rand == 5):
isNumber = False
isOp = True
numParentheses -= 1
# Checks to see if '(' was appended
elif (rand == 4):
isNumber = True
isOp = False
numParentheses += 1
# All other operations go here
else:
isNumber = True
isOp = False
# Didn't add parentheses possibility here in case expression in parentheses somehow reaches 0
elif (isNumber and isDiv):
expr.append(str(random.randint(1,100)))
isDiv = False
isNumber = False
isOp = True
# If a number's up, decides whether to append parentheses or a number
elif (isNumber):
rand = random.randint(0,1)
if (rand == 0):
expr.append(str(random.randint(0,100)))
isNumber = False
isOp = True
elif (rand == 1):
if (numParentheses == 0):
expr.append('(')
numParentheses += 1
else:
rand = random.randint(0,1)
expr.append(parentheses[rand])
if rand == 0:
numParentheses += 1
else:
numParentheses -= 1
isDiv = False
numExpr -= 1
fname.write(' '.join(expr) + '\n')
fname.close()
lines += 1
Ray*_*oal 15
是的,您可以以Pythonic方式生成随机算术表达式.但是,你需要改变你的方法.不要试图生成一个字符串并计算parens.而是生成随机表达式树,然后输出.
通过表达式目录树,我的意思是叫,比如类,实例Expression与子类Number,PlusExpression,MinusExpression , 'TimesExpression,DivideExpression和ParenthesizedExpression.除了Number将具有类型的字段之外,其中每一个都是Expression.给每个人一个合适的__str__方法.生成一些随机表达式对象,然后打印"root".
你能从这里拿走它还是要我编码？
附录:一些示例入门代码.不生成随机表达式(但？)但可以添加....
# This is just the very beginning of a script that can be used to process
# arithmetic expressions. At the moment it just defines a few classes
# and prints a couple example expressions.
# Possible additions include methods to evaluate expressions and generate
# some random expressions.
class Expression:
pass
class Number(Expression):
def __init__(self, num):
self.num = num
def __str__(self):
return str(self.num)
class BinaryExpression(Expression):
def __init__(self, left, op, right):
self.left = left
self.op = op
self.right = right
def __str__(self):
return str(self.left) + " " + self.op + " " + str(self.right)
class ParenthesizedExpression(Expression):
def __init__(self, exp):
self.exp = exp
def __str__(self):
return "(" + str(self.exp) + ")"
e1 = Number(5)
print e1
e2 = BinaryExpression(Number(8), "+", ParenthesizedExpression(BinaryExpression(Number(7), "*", e1)))
print e2
**ADDENDUM 2**
回到Python非常有趣.我无法抗拒实现随机表达式生成器.它建立在上面的代码之上.抱怨HARDCODING !!
from random import random, randint, choice
def randomExpression(prob):
p = random()
if p > prob:
return Number(randint(1, 100))
elif randint(0, 1) == 0:
return ParenthesizedExpression(randomExpression(prob / 1.2))
else:
left = randomExpression(prob / 1.2)
op = choice(["+", "-", "*", "/"])
right = randomExpression(prob / 1.2)
return BinaryExpression(left, op, right)
for i in range(10):
print(randomExpression(1))
这是我得到的输出:
(23)
86 + 84 + 87 / (96 - 46) / 59
((((49)))) + ((46))
76 + 18 + 4 - (98) - 7 / 15
(((73)))
(55) - (54) * 55 + 92 - 13 - ((36))
(78) - (7 / 56 * 33)
(81) - 18 * (((8)) * 59 - 14)
(((89)))
(59)
不是太漂亮了.我认为它让太多的父母.也许改变括号表达式和二进制表达式之间选择的概率可能会很好....
我发现这个线程有类似的任务，即生成用于符号计算的单元测试的随机表达式。在我的版本中，我包含一元函数并允许符号是任意字符串，即您可以使用数字或变量名称。
from random import random, choice
UNARIES = ["sqrt(%s)", "exp(%s)", "log(%s)", "sin(%s)", "cos(%s)", "tan(%s)",
"sinh(%s)", "cosh(%s)", "tanh(%s)", "asin(%s)", "acos(%s)",
"atan(%s)", "-%s"]
BINARIES = ["%s + %s", "%s - %s", "%s * %s", "%s / %s", "%s ** %s"]
PROP_PARANTHESIS = 0.3
PROP_BINARY = 0.7
def generate_expressions(scope, num_exp, num_ops):
scope = list(scope) # make a copy first, append as we go
for _ in xrange(num_ops):
if random() < PROP_BINARY: # decide unary or binary operator
ex = choice(BINARIES) % (choice(scope), choice(scope))
if random() < PROP_PARANTHESIS:
ex = "(%s)" % ex
scope.append(ex)
else:
scope.append(choice(UNARIES) % choice(scope))
return scope[-num_exp:] # return most recent expressions
正如从以前的答案中复制的那样，我只是在二元运算符周围加上一些括号PROP_PARANTHESIS（这有点作弊）。二元运算符比一元运算符更常见，因此我也将其留作配置 ( PROP_BINARY)。示例代码是：
scope = [c for c in "abcde"]
for expression in generate_expressions(scope, 10, 50):
print expression
这将生成类似以下内容的内容：
e / acos(tan(a)) / a * acos(tan(a)) ** (acos(tan(a)) / a + a) + (d ** b + a)
(a + (a ** sqrt(e)))
acos((b / acos(tan(a)) / a + d) / (a ** sqrt(e)) * (a ** sinh(b) / b))
sin(atan(acos(tan(a)) ** (acos(tan(a)) / a + a) + (d ** b + a)))
sin((b / acos(tan(a)) / a + d)) / (a ** sinh(b) / b)
exp(acos(tan(a)) / a + acos(e))
tan((b / acos(tan(a)) / a + d))
acos(tan(a)) / a * acos(tan(a)) ** (acos(tan(a)) / a + a) + (d ** b + a) + cos(sqrt(e))
(acos(tan(a)) / a + acos(e) * a + e)
((b / acos(tan(a)) / a + d) - cos(sqrt(e))) + sinh(b)
放置PROP_BINARY = 1.0和应用
scope = range(100)
让我们回到输出
43 * (50 * 83)
34 / (29 / 24)
66 / 47 - 52
((88 ** 38) ** 40)
34 / (29 / 24) - 27
(16 + 36 ** 29)
55 ** 95
70 + 28
6 * 32
(52 * 2 ** 37)