I am creating a program that takes in a boolean expression as a string and converts the infix formula to postfix, while making sure the formula is in a valid form. What I'm having a hard time doing is figuring out a way to check if the inputed formula is valid or not. NO IMPORTING is permitted (use built-in python functions/methods), loops and recursion are allowed. If the formula is invalid, return None.
The formula can contain:
variables in 'abcdefghijklmnopqrstuvwxyz'
operators in '-+*'
where - is NOT, + is OR, * is AND
Here are some valid formulas (as Python strings).
"x"
"-y"
"(x*y)"
"((-x+y)*(-y+x))"
Here are some strings that are not formulas.
"X" variable not lower case letter
"x*y" missing parentheses
"-(x)" extraneous parentheses
"(x+(y)*z)" mismatched parentheses
Some conversion examples are:
(x+y) -> xy+
(x*y) -> xy*
-(x+y) -> xy+-
-x -> x-
((x*y)+(z*x)) -> xy*zx*+
A full working program isn't necessary, an algorithm to check if the formula is valid or not is fine.
My current implementation to convert formula from infix to postfix:
def infix_to_postfix(infix_expression):
precedence = {}
precedence["*"] = 2
precedence["+"] = 2
precedence["-"] = 2
precedence["("] = 1
storage_stack = Stack()
postfix_list = []
tokenList = list(infix_expression)
for token in tokenList:
if(token not in "-+*()"):
postfix_list.append(token)
elif(token == '-'):
storage_stack.push(token)
elif(token == '('):
storage_stack.push(token)
elif(token == ')'):
topToken = storage_stack.pop()
while(topToken != '('):
postfix_list.append(topToken)
topToken = storage_stack.pop()
else:
while(not storage_stack.is_empty() and precedence[storage_stack.peek()] >= precedence[token]):
postfix_list.append(storage_stack.pop())
storage_stack.push(token)
while(not storage_stack.is_empty()):
postfix_list.append(storage_stack.pop())
result = "".join(postfix_list)
return result
I need to find a way to check that the formula is valid while changing the positions of the operators and variables.
Edit:
I've come up with part of an algorithm to check if the formula is valid or not:
((a+b)*(c+d))
F1 = (a+b)
F2 = (c+d)
((a+b)*(c+d)) = (F1 * F2)
If F1 and F2 are valid, then the whole thing is valid.
A formula enclosed with parenthesis is valid if:
There is ( and ), there is either a + or * in between 2 sub-formulas, and both sub-formulas are valid.
I have this idea of checking, but I have no idea how to implement it. Most likely recursion.
This is an old question, however here is a potential solution to someone having the same issue.
import re
def formula(string):
if not string: return 0
string = re.sub("[^0-9+*/%-=]", "", string)
try:
string, result = string.split("=")
return eval(string) == int(result)
except:
return 0
Related
I am a beginner to Python. I have tried learning python and C++ in the past had learnt about like classes and stuff but then had to abandon it for reasons, now I am learning python from the beginning as I have forgotten it all.
So I was trying to make a calculator like the ones you have in mobile using python but without any GUI. Now, the problem I am having right now is, see in your mobile calculator you can do one operation after the other i.e. say you typed 95+8x2, that mobile calculator will have no problem in deducing operator precedence from your input and give result as 111 and I am trying to do something similar.
But the problem is, the way I know how to do it as a beginner would require a lot of code, it would not be complex but get too long and hence a lot of time wasted. Here is how I have thought of doing it right now :
Find the location of each of the operators in the input for that I am using their indexes i.e. I have named the input as 'alg_operation' and for example, I am using alg_operation.find('*') to where the multiplaction operator is, I am calling this index as location_prod using this I am able to calculate the product simply via converting the part of the string that comes before the operator into float then multiply it with the other part that comes just after (obviously converting it into float as well).
After finding the location of each of the 5 operators that I have decided to include i.e. exponentiation, multiplication, division (// not /), addition and subtraction, I would have to write code for 120 different cases of the arrangement of these operators, which may not be complex but definitely will take a lot of time.
How can I quickly deduce operator precedence from the string input ?
I will update this post if I learn anything new, since I am a beginner to programming.
You can indeed evaluate an arbitrary python expression with eval. Use of eval is a pretty big code smell, because it lets you execute anything, but for completeness it would be done like this:
expr = input("Please, please, please only write maths: ")
print(eval(exp))
note that you could type import shutil; shutil.rmtree("/home") and python would cheerfully run it. So obviously we don't want to do this.
We could try to protect ourselves by sanitising the input. In this case this might actually work, with something like:
safe_chars = (".", "*", "+", "/", "-"," ", *range(10))
if not all(x in safe_chars for x in expr):
raise ValueError("You tried to enter dangerous data!")
I can't immediately think of any way to do anything dangerous with input consisting only of those chars, but doubtless someone will point it out immediately in the comments [in which case I'll add it here]. More generally, sanitising data like this is hard, because in order to know what's safe you really need to understand the input, by which point you've just written a parser.
Please do note that eval is inherently dangerous. It can be a useful hack for once-off code, although even then... and it is of course useful when you actually want to evaluate python code.
Converting it to reverse polish notation will solve your problem
def readNumber(string: str,index: int):
number = ""
while index < len(string) and isNumber(string[index]):
number += string[index]
index += 1
return float(number), index
def isOperator(token):
operators = ['+', '-', '*', '/', '%']
return token in operators
def isNumber(token):
return (token >= '0' and token <= '9') or token == "."
def toRpn(string: str):
"""
Converts infix notation to reverse polish notation
"""
precedence = {
'(': 0,
'-': 1,
'+': 1,
'*': 2,
'/': 2,
'%': 2,
}
i = 0
fin = []
ops = []
while i < len(string):
token = string[i]
if isNumber(token):
number, i = readNumber(string,i)
fin.append(number)
continue
if isOperator(token):
top = ops[-1] if ops else None
if top is not None and precedence[top] >= precedence[token]:
fin.append(ops.pop())
ops.append(token)
i += 1
continue
if token == '(':
ops.append(token)
i += 1
continue
if token == ')':
while True:
operator = ops.pop()
if operator == '(':
break
fin.append(operator)
if not ops:
break
i += 1
continue
i += 1
while ops:
fin.append(ops.pop())
return fin
def calculate_rpn(rpn: list):
"""
Calculates the result of an expression in reverse polish notation
"""
stack = []
for token in rpn:
if isOperator(token):
a = stack.pop()
b = stack.pop()
if token == '+':
stack.append(b + a)
elif token == '-':
stack.append(b - a)
elif token == '*':
stack.append(b * a)
elif token == '/':
stack.append(b / a)
elif token == '%':
stack.append(b % a)
else:
stack.append(token)
return stack[0]
print ("90165: ", calculate_rpn(toRpn("27+38+81+48*33*53+91*53+82*14+96")))
print ("616222: ", calculate_rpn(toRpn("22*26*53+66*8+7*76*25*44+78+100")))
print (calculate_rpn(toRpn("(22+4)*4")))
My Github
You can easily add more operators and their precedence if you want.
You have to modify the precedence array and the isOperator function. Also you should modify the function of the respective operator in the calculate_rpn function.
I want to use recursion to reverse a string in python so it displays the characters backwards (i.e "Hello" will become "olleh"/"o l l e h".
I wrote one that does it iteratively:
def Reverse( s ):
result = ""
n = 0
start = 0
while ( s[n:] != "" ):
while ( s[n:] != "" and s[n] != ' ' ):
n = n + 1
result = s[ start: n ] + " " + result
start = n
return result
But how exactly do I do this recursively? I am confused on this part, especially because I don't work with python and recursion much.
Any help would be appreciated.
def rreverse(s):
if s == "":
return s
else:
return rreverse(s[1:]) + s[0]
(Very few people do heavy recursive processing in Python, the language wasn't designed for it.)
To solve a problem recursively, find a trivial case that is easy to solve, and figure out how to get to that trivial case by breaking the problem down into simpler and simpler versions of itself.
What is the first thing you do in reversing a string? Literally the first thing? You get the last character of the string, right?
So the reverse of a string is the last character, followed by the reverse of everything but the last character, which is where the recursion comes in. The last character of a string can be written as x[-1] while everything but the last character is x[:-1].
Now, how do you "bottom out"? That is, what is the trivial case you can solve without recursion? One answer is the one-character string, which is the same forward and reversed. So if you get a one-character string, you are done.
But the empty string is even more trivial, and someone might actually pass that in to your function, so we should probably use that instead. A one-character string can, after all, also be broken down into the last character and everything but the last character; it's just that everything but the last character is the empty string. So if we handle the empty string by just returning it, we're set.
Put it all together and you get:
def backward(text):
if text == "":
return text
else:
return text[-1] + backward(text[:-1])
Or in one line:
backward = lambda t: t[-1] + backward(t[:-1]) if t else t
As others have pointed out, this is not the way you would usually do this in Python. An iterative solution is going to be faster, and using slicing to do it is going to be faster still.
Additionally, Python imposes a limit on stack size, and there's no tail call optimization, so a recursive solution would be limited to reversing strings of only about a thousand characters. You can increase Python's stack size, but there would still be a fixed limit, while other solutions can always handle a string of any length.
I just want to add some explanations based on Fred Foo's answer.
Let's say we have a string called 'abc', and we want to return its reverse which should be 'cba'.
def reverse(s):
if s == "":
return s
else:
return reverse(s[1:]) + s[0]
s = "abc"
print (reverse(s))
How this code works is that:
when we call the function
reverse('abc') #s = abc
=reverse('bc') + 'a' #s[1:] = bc s[0] = a
=reverse('c') + 'b' + 'a' #s[1:] = c s[0] = a
=reverse('') + 'c' + 'b' + 'a'
='cba'
If this isn't just a homework question and you're actually trying to reverse a string for some greater goal, just do s[::-1].
def reverse_string(s):
if s: return s[-1] + reverse_string(s[0:-1])
else: return s
or
def reverse_string(s):
return s[-1] + reverse_string(s[0:-1]) if s else s
I know it's too late to answer original question and there are multiple better ways which are answered here already. My answer is for documentation purpose in case someone is trying to implement tail recursion for string reversal.
def tail_rev(in_string,rev_string):
if in_string=='':
return rev_string
else:
rev_string+=in_string[-1]
return tail_rev(in_string[:-1],rev_string)
in_string=input("Enter String: ")
rev_string=tail_rev(in_string,'')
print(f"Reverse of {in_string} is {rev_string}")
s = input("Enter your string: ")
def rev(s):
if len(s) == 1:
print(s[0])
exit()
else:
#print the last char in string
#end="" prints all chars in string on same line
print(s[-1], end="")
"""Next line replaces whole string with same
string, but with 1 char less"""
return rev(s.replace(s, s[:-1]))
rev(s)
if you do not want to return response than you can use this solution. This question is part of LeetCode.
class Solution:
i = 0
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
if self.i >= (len(s)//2):
return
s[self.i], s[len(s)-self.i-1] = s[len(s)-self.i-1], s[self.i]
self.i += 1
self.reverseString(s)
Recently, I got an interview question which says to convert string expressions like "1+2-3" and "-2+4" to 0 and 2 respectively. Assuming the inputs are single digits numbers followed by signs and no NULL input. I tried this output but the interviewer said I am close but not perfect solution. Please help me here. Thanks.
def ans(input):
result, j = 0, 0
for i in input:
if i == '+' or i == '-':
j = i
else:
i = int(i)
result = result j i
return result
ans("1+2-3")
ans("-2+4")
I am making some silly mistake but I am learning. Thanks in advance.
Two things need fixing to work at all:
You need to handle the initial value properly; when the initial value is non-negative, this fails. Before the loop, set j = '+' so a non-sign prefixed value is added (also, for style points, j is a terrible name, could you use op or something?).
You can't use variables as operators.
Replace:
result = result j i
with:
if j == '+':
result += i
else:
result -= i
Note: If modules are allowed, a generalization can be used to handle operators the "nice" way (though more work would be needed to obey operator precedence). You'd define:
import operator
ops = {'+': operator.add, '-': operator.sub, ...}
then make the initial value of op operator.add and change the test for operators to:
if i in ops:
op = ops[i]
else:
result = op(result, int(i))
which scales to many more operators, dynamically selecting the operation to perform without cascading if/elif checks.
Side-note: While violating the spirit of the challenge, ast.literal_eval (at least as of Python 3.5, and this may change, see bug #22525) will actually safely parse strings like this (eval is unsafe, since it can execute arbitrary code, but ast.literal_eval can only parse Python literals and apparently some basic compile-time math). So you could just do:
import ast
ans = ast.literal_eval
Sure, it handles many other literals too, but we never defined the failure case behavior anyway. :-)
Using eval() is the simplest solution. Like
eval("1+2-3")
The following code give another solution without using built-in eval
import operator
class Parse(object):
def __init__(self, input):
self.input = input
self.pos = 0
self.end = len(input)
def eval(self):
result = self.match_digits()
while self.pos < self.end:
op = self.match_operator()
operand = self.match_digits()
result = op(result, operand)
return result
def match_operator(self):
look_ahead = self.input[self.pos]
self.advance()
return operator.add if look_ahead == '+' else operator.sub
def match_digits(self):
look_ahead = self.input[self.pos]
positive = 1
if look_ahead == '-':
positive = -1
self.advance()
digits, s = 0, self.pos
while s < self.end and self.input[s].isdigit():
digits = digits * 10 + int(self.input[s])
s += 1
self.advance(s-self.pos)
return digits * positive
def advance(self, offset=1):
self.pos += offset
For testing
p = Parse(input='2+1+0-3')
print p.eval()
p = Parse(input='-2+-13+3')
print p.eval()
I think the most flexible solution (not using eval and able to handle any operations) is to parse the string into a binary (red-black) tree, where leafs are numbers and branches operators (+,-,/,*,etc).
For example, "1+(5*12)/17" would be parsed into following structure:
"+"
/ \
1 "/"
/ \
"()" 17
/
"*"
/ \
5 12
Once you've parsed a string into this structure, it's easy to compute by traversing branches depth-first, right to left.
If you need to handle variables, then you'd have to get locals() and replace accordingly, either as you parse the string, or as you traverse the tree.
EDIT:
I created a working example to illustrate this, you can find the source on github: https://github.com/MJWunderlich/py-math-expression-evaluator
what about:
def f(s):
s = s.strip()
for i, c in enumerate(s):
if c == '+':
return f(s[:i]) + f(s[i+1:])
if c == '-':
return f(s[:i]) - f(s[i+1:])
for i, c in enumerate(s):
if c == '*':
return f(s[:i]) * f(s[i+1:])
if c == '/':
return f(s[:i]) / f(s[i+1:])
return 0 if s == '' else int(s)
? Doesn't work with parenthesis
This question already has answers here:
Evaluating a mathematical expression in a string
(14 answers)
Closed 8 years ago.
I've got a python formula that randomly places operands in between numbers. The list could, for example, look like this:
['9-8+7', '7-8-6']
What I want to do is get the value of each string, so that looping through the strings, an array would see 9-8+7 and would append 8 and 7-8-6 would append -7. I can't convert a string with operands to int, so is this possible at all? Or should I change the algorithm so that instead of creating a string with each random output it calculates the value of it immediately?
Thank you in advance.
You can do eval on the list items, but that's a potential security hole and should only be used if you fully trust the source.
>>> map(eval, ['9-8+7', '7-8-6'])
[8, -7]
If you control the code producing the string, computing the values directly sounds like a better approach (safer and probably faster).
As Fredrik pointed out, you can do eval in Python. I thought I'd add a more general approach that would work in any language, and might shed some light on simple parsers for those that haven't seen them in action.
You're describing a language whose formal definition looks something like this:
expr := sum
sum := prod [("+" | "-") prod]...
prod := digit [("*" | "/") digit]...
digit := '0'..'9'
This grammar (which I'm not bothering to make correct EBNF) accepts these strings: "3", "4*5/2", and "8*3+9", and so on.
This gives us a clue how to parse it, and evaluation is no more work than accumulating results as we go. The following is working Python 2 code. Notice how closely the code follows the grammar.
class ParseFail(Exception):
pass
def eval_expr(str):
value, pos = eval_sum(str, 0)
return value
def eval_sum(str, pos):
value, pos = eval_product(str, pos)
accum = value
while pos != len(str):
op = str[pos]
if not str[pos] in ['+', '-']:
raise ParseFail("Unexpected symbol at position "
"{pos} of {str}".format(str=str, pos=pos))
value, pos = eval_product(str, pos + 1)
if op == '+':
accum += value
else:
accum -= value
return accum, pos
def eval_product(str, pos):
value, pos = eval_digit(str, pos)
accum = value
while pos != len(str):
op = str[pos]
if not str[pos] in ['*', '/']:
return accum, pos
value, pos = eval_digit(str, pos + 1)
if op == '*':
accum *= value
else:
accum /= value
return accum, pos
def eval_digit(str, pos):
if not str[pos].isdigit():
raise ParseFail("Unexpected symbol at position "
"{pos} of {str}".format(str=str, pos=pos))
return int(str[pos]), pos + 1
try:
print "3 ->", eval_expr("3")
print "3*4 ->", eval_expr("3*4")
print "2+3*4-5 ->", eval_expr("2+3*4-5")
# Should raise ParseFail
print "2+3*4^2-5 ->", eval_expr("2+3*4^2-5")
except ParseFail as err:
print
print err.args[0]
Here's a sample run:
$ python simple_expr.py
3 -> 3
3*4 -> 12
2+3*4-5 -> 9
2+3*4^2-5 ->
Unexpected symbol at position 5 of 2+3*4^2-5
It would be pretty easy to extend this to a full string calculator with more operators, such as the exponent operator '^' and multi-digit integers. Parentheses, floats and functions might be a bit of work, but not that hard either. Every programmer should try it once in their lives, in my opinion.
This of course depends on how well-behaved and restricted your expressions are.
Since subtraction is addition with a negative number, you can write the subtractions as additions with a negative number. Spit on + to find the terms. Then parse the terms of the sum to integers, and sum them. Do so for each expression.
[sum(map(int,l.replace('-', '+-').split('+'))) for l in ['9-8+7','7-8-6']]
I want to use recursion to reverse a string in python so it displays the characters backwards (i.e "Hello" will become "olleh"/"o l l e h".
I wrote one that does it iteratively:
def Reverse( s ):
result = ""
n = 0
start = 0
while ( s[n:] != "" ):
while ( s[n:] != "" and s[n] != ' ' ):
n = n + 1
result = s[ start: n ] + " " + result
start = n
return result
But how exactly do I do this recursively? I am confused on this part, especially because I don't work with python and recursion much.
Any help would be appreciated.
def rreverse(s):
if s == "":
return s
else:
return rreverse(s[1:]) + s[0]
(Very few people do heavy recursive processing in Python, the language wasn't designed for it.)
To solve a problem recursively, find a trivial case that is easy to solve, and figure out how to get to that trivial case by breaking the problem down into simpler and simpler versions of itself.
What is the first thing you do in reversing a string? Literally the first thing? You get the last character of the string, right?
So the reverse of a string is the last character, followed by the reverse of everything but the last character, which is where the recursion comes in. The last character of a string can be written as x[-1] while everything but the last character is x[:-1].
Now, how do you "bottom out"? That is, what is the trivial case you can solve without recursion? One answer is the one-character string, which is the same forward and reversed. So if you get a one-character string, you are done.
But the empty string is even more trivial, and someone might actually pass that in to your function, so we should probably use that instead. A one-character string can, after all, also be broken down into the last character and everything but the last character; it's just that everything but the last character is the empty string. So if we handle the empty string by just returning it, we're set.
Put it all together and you get:
def backward(text):
if text == "":
return text
else:
return text[-1] + backward(text[:-1])
Or in one line:
backward = lambda t: t[-1] + backward(t[:-1]) if t else t
As others have pointed out, this is not the way you would usually do this in Python. An iterative solution is going to be faster, and using slicing to do it is going to be faster still.
Additionally, Python imposes a limit on stack size, and there's no tail call optimization, so a recursive solution would be limited to reversing strings of only about a thousand characters. You can increase Python's stack size, but there would still be a fixed limit, while other solutions can always handle a string of any length.
I just want to add some explanations based on Fred Foo's answer.
Let's say we have a string called 'abc', and we want to return its reverse which should be 'cba'.
def reverse(s):
if s == "":
return s
else:
return reverse(s[1:]) + s[0]
s = "abc"
print (reverse(s))
How this code works is that:
when we call the function
reverse('abc') #s = abc
=reverse('bc') + 'a' #s[1:] = bc s[0] = a
=reverse('c') + 'b' + 'a' #s[1:] = c s[0] = a
=reverse('') + 'c' + 'b' + 'a'
='cba'
If this isn't just a homework question and you're actually trying to reverse a string for some greater goal, just do s[::-1].
def reverse_string(s):
if s: return s[-1] + reverse_string(s[0:-1])
else: return s
or
def reverse_string(s):
return s[-1] + reverse_string(s[0:-1]) if s else s
I know it's too late to answer original question and there are multiple better ways which are answered here already. My answer is for documentation purpose in case someone is trying to implement tail recursion for string reversal.
def tail_rev(in_string,rev_string):
if in_string=='':
return rev_string
else:
rev_string+=in_string[-1]
return tail_rev(in_string[:-1],rev_string)
in_string=input("Enter String: ")
rev_string=tail_rev(in_string,'')
print(f"Reverse of {in_string} is {rev_string}")
s = input("Enter your string: ")
def rev(s):
if len(s) == 1:
print(s[0])
exit()
else:
#print the last char in string
#end="" prints all chars in string on same line
print(s[-1], end="")
"""Next line replaces whole string with same
string, but with 1 char less"""
return rev(s.replace(s, s[:-1]))
rev(s)
if you do not want to return response than you can use this solution. This question is part of LeetCode.
class Solution:
i = 0
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
if self.i >= (len(s)//2):
return
s[self.i], s[len(s)-self.i-1] = s[len(s)-self.i-1], s[self.i]
self.i += 1
self.reverseString(s)