I’m currently trying to create a Python script that will autogenerate space-delimited arithmetic expressions which are valid. However, I get sample output that looks like this: ( 32 - 42 / 95 + 24 ( ) ( 53 ) + ) 21
While the empty parentheses are perfectly OK by me, I can’t use this autogenerated expression in calculations since there’s no operator between the 24 and the 53, and the + before the 21 at the end has no second argument.
What I want to know is, is there a way to account for/fix these errors using a Pythonic solution? (And before anyone points it out, I’ll be the first to acknowledge that the code I posted below is probably the worst code I’ve pushed and conforms to…well, very few of Python’s core tenets.)
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
Yes, you can generate random arithmetic expressions in a Pythonic way. You need to change your approach, though. Don’t try to generate a string and count parens. Instead generate a random expression tree, then output that.
By an expression tree, I mean an instance of a class called, say,
Expressionwith subclassesNumber,PlusExpression,MinusExpression, 'TimesExpression,DivideExpression, andParenthesizedExpression. Each of these, exceptNumberwill have fields of typeExpression. Give each a suitable__str__method. Generate some random expression objects and just print the “root.”Can you take it from here or would you like me to code it up?
ADDENDUM: Some sample starter code. Doesn’t generate random expressions (yet?) but this can be added….
** ADDENDUM 2 **
Getting back into Python is really fun. I couldn’t resist implementing the random expression generator. It is built on the code above. SORRY ABOUT THE HARDCODING!!
Here is the output I got:
Ain’t tooooo pretty. I think it puts out too many parents. Maybe change the probability of choosing between parenthesized expressions and binary expressions might work well….