I came across this exercise of checking whether or not the simple brackets "(", ")" in a given string are matched evenly.
I have seen examples here using the stack command which I haven't encountered yet. So I attempted a different approach. Can anyone tell me where I am going wrong?
def matched(str):
ope = []
clo = []
for i in range(0,len(str)):
l = str[i]
if l == "(":
ope = ope + ["("]
else:
if l == ")":
clo = clo + [")"]
else:
return(ope, clo)
if len(ope)==len(clo):
return True
else:
return False
The idea is to pile up "(" and ")" into two separate lists and then compare the length of the lists. I also had another version where I had appended the lists ope and clo with the relevant I which held either ( or ) respectively.
A very slightly more elegant way to do this is below. It cleans up the for loop and replaces the lists with a simple counter variable. It also returns false if the counter drops below zero so that matched(")(") will return False.
def matched(str):
count = 0
for i in str:
if i == "(":
count += 1
elif i == ")":
count -= 1
if count < 0:
return False
return count == 0
This checks whether parentheses are properly matched, not just whether there is an equal number of opening and closing parentheses. We use a list as a stack and push onto it when we encounter opening parentheses and pop from it when we encounter closing parentheses.
The main problem with your solution is that it only counts the number of parentheses but does not match them. One way of keeping track of the current depth of nesting is by pushing opening parentheses onto a stack and popping them from the stack when we encounter a closing parenthesis.
def do_parentheses_match(input_string):
s = []
balanced = True
index = 0
while index < len(input_string) and balanced:
token = input_string[index]
if token == "(":
s.append(token)
elif token == ")":
if len(s) == 0:
balanced = False
else:
s.pop()
index += 1
return balanced and len(s) == 0
My solution here works for brackets, parentheses & braces
openList = ["[", "{", "("]
closeList = ["]", "}", ")"]
def balance(myStr):
stack = []
for i in myStr:
if i in openList:
stack.append(i)
elif i in closeList:
pos = closeList.index(i)
if stack and (openList[pos] == stack[-1]):
stack.pop()
else:
return "Unbalanced"
if len(stack) == 0:
return "Balanced"
print(balance("{[()](){}}"))
Most blatant error done by you is:
if l == ")":
clo = clo + [")"]
else:
return(ope, clo) # here
By using return, you exit from function when first char not equal to "(" or ")" is encountered. Also some indentation is off.
Minimal change which allows your code to run (although it won't give correct answers for all possible input strings) is:
def matched(str):
ope = []
clo = []
for i in range(0,len(str)):
l = str[i]
if l == "(":
ope = ope + ["("]
elif l == ")":
clo = clo + [")"]
if len(ope)==len(clo):
return True
else:
return False
The problem with your approach is that you don't consider the order. Following line would pass: ))) (((.
I'd suggest to keep the count of open and closed parenthesis:
counter starts from 0
every ( symbol increments counter
every ) symbol decrements counter
if at any moment counter is negative it is an error
if at the end of the line counter is 0 - string has matching parenthesis
a = "((a+b)*c)+(b*a))"
li = list(a)
result = []
for i in range(0, len(a)):
if a[i] == "(":
result.append(i)
elif a[i] == ")":
if len(result) > 0:
result.pop()
else:
li.pop(i)
for i in range(0, len(result)):
li.pop(result[i])
print("".join(li))
this code works fine
def matched(s):
p_list=[]
for i in range(0,len(s)):
if s[i] =='(':
p_list.append('(')
elif s[i] ==')' :
if not p_list:
return False
else:
p_list.pop()
if not p_list:
return True
else:
return False
You can do this in a couple of lines using accumulate (from itertools). The idea is to compute a cumulative parenthesis level going through the string with opening parentheses counting as level+1 and closing parentheses counting as level-1. If, at any point, the accumulated level falls below zero then there is an extra closing parenthesis. If the final level is not zero, then there is a missing closing parenthesis:
from itertools import accumulate
def matched(s):
levels = list(accumulate((c=="(")-(c==")") for c in s))
return all( level >= 0 for level in levels) and levels[-1] == 0
An alternative to check for balanced nested parentheses:
def is_balanced(query: str) -> bool:
# Alternative: re.sub(r"[^()]", "", query)
query = "".join(i for i in query if i in {"(", ")"})
while "()" in query:
query = query.replace("()", "")
return not query
for stmt in [
"(()()()())", # True
"(((())))", # True
"(()((())()))", # True
"((((((())", # False
"()))", # False
"(()()))(()", # False
"foo", # True
"a or (b and (c or d)", # False
"a or (b and (c or d))" # True
"a or (b and (c or (d and e)))", # True
]:
print(stmt)
print("Balanced:", is_balanced(stmt))
print()
It works by:
Removing everything but parentheses
Recursively remove innermost parentheses pairs
If you're left with anything besides the empty string, the statement is not balanced. Otherwise, it is.
if the parenthesis sequence is not an issue (strings like )( ) this code is faster :
def matched_parenthesis(s):
return s.count('(') == s.count(')')
Tested with 15KB string, it is ~20μs v.s. 1ms iterating over the whole string.
And for me the order is not an issue as the underlying protocol guaranties that the string is well-formed.
In case u also need to find the position of the first mismatching bracket from left u can use the below code which also cover certain edge cases:
def isBalanced(expr):
opening=set('([{')
new=set(')]}{[(')
match=set([ ('(',')'), ('[',']'), ('{','}') ])
stack=[]
stackcount=[]
for i,char in enumerate(expr,1):
if char not in new:
continue
elif char in opening:
stack.append(char)
stackcount.append(i)
else:
if len(stack)==0:
print(i)
return False
lastOpen=stack.pop()
lastindex=stackcount.pop()
if (lastOpen, char) not in match:
print (i)
return False
length=len(stack)
if length!=0:
elem=stackcount[0]
print (elem)
return length==0
string =input()
ans=isBalanced(string)
if ans==True:
print("Success")
if "(" ,")" these two characters are not present then we don't want to return true or false just return no matching found. if matching found i just checking the count of both characters are same then return true, else return false
def matched(str):
count1=0
count2=1
for i in str:
if i =="(":
count1+=1:
elif i==")":
count2+=1:
else:
print "no matching found for (,)"
if count1==count2:
return True
else:
return False
Simplest of all , though all of you guys have done good:
def wellbracketed(s):
left=[]
right=[]
for i in range(0,len(s)):``
if s[i]=='(':
left=left+['(']
elif s[i]==')':
if len(left)!=0:
right=right+[')']
else:
return False
return(len(left)==len(right))
here's another way to solve it by having a counter that tracks how many open parentheses that are difference at this very moment.
this should take care all of the cases.
def matched(str):
diffCounter = 0
length = len(str)
for i in range(length):
if str[i] == '(':
diffCounter += 1
elif str[i] == ')':
diffCounter -= 1
if diffCounter == 0:
return True
else:
return False
input_str = "{[()](){}}"
strblance=""
for i in input_str:
if not strblance:
strblance = strblance+i
elif (i is '}' and strblance[len(strblance)-1] is '{') \
or ( i is']'and strblance[len(strblance)-1] is '[') \
or ( i is ')'and strblance[len(strblance)-1] is '('):
strblance = strblance[:len(strblance)-1]
else:
strblance = strblance+i
if not strblance:
print ("balanced")
else:
print ("Not balanced")
More advanced example in which you additionally need to check a matching of square brackets '[]' and braces '{}' pars.
string = '([]{})'
def group_match(string):
d = {
')':'(',
']':'[',
'}':'{'
}
list_ = []
for index, item in enumerate(string):
if item in d.values():
list_.append(item)
elif (item in d.keys()) and (d.get(item) in list_):
list_.pop()
return len(list_) == 0
The simplest code ever!!
def checkpar(x):
while len(''.join([e for e in x if e in "()"]).split('()'))>1: x=''.join(x.split('()'))
return not x
you can check this code.
This code don't use stack operations.
def matched(s):
count = 0
for i in s:
if i is "(":
count += 1
elif i is ")":
if count != 0:
count -= 1
else:
return (False)
if count == 0:
return (True)
else:
return (False)
#function to check if number of closing brackets is equal to the number of opening brackets
#this function also checks if the closing bracket appears after the opening bracket
def matched(str1):
if str1.count(")")== str1.count("("):
p1=str1.find("(")
p2=str1.find(")")
if p2 >= p1:
str1=str1[p1+1:p2]+ str1[p2+1:]
if str1.count(")")>0 and str1.count("(")>0:
matched(str1)
return True
else:
return False
else:
return False
matched(str1)
parenthesis_String = input("Enter your parenthesis string")
parenthesis_List = []
for p in parenthesis_String:
parenthesis_List.append(p)
print(parenthesis_List)
if len(parenthesis_List)%2 != 0:
print("Not Balanced Wrong number of input")
for p1 in parenthesis_List:
last_parenthesis = parenthesis_List.pop()
print(last_parenthesis)
if (p1 == '{' and last_parenthesis == '}' or p1 == '[' and last_parenthesis == ']' or p1 == '(' and last_parenthesis == ')'):
print("Balanced")
else:
print("Not balanced")
A little different one.
expression = '{(){({)}}'
brackets = '[](){}'
stack = []
balanced = False
for e in expression:
if e in brackets and stack: # Popping from the stack if it is closing bracket
if stack [-1] == brackets[brackets.index(e)-1]:
stack.pop()
balanced = True
continue # it will go to the new iteration skipping the next if below
if e in brackets: # Push to stack if new bracket in the expression
stack .append(e)
balanced = False
balanced = 'Balanced' if balanced and not stack else 'Unbalanced'
print(balanced, stack)
just modified Henry Prickett-Morgan's code a little bit to handle it more sensibly, namely taking into account that the number of "(" matches that of ")" but string starts with ")" or ends with "(" which are apparently not right.
def ValidParenthesis(s):
count = 0
if s[0] == ')' or s[-1] == '(':
return False
else:
for c in s:
if c == '(':
count += 1
elif c == ')':
count -= 1
else:
continue
return count == 0
The best way to understand this snippet is to follow along with all kind of scenarios.
in_data = ['{','[','(']
out_data = ['}',']',')']
def check_match(statements):
stack = []
for ch in statements:
if ch in in_data:
stack.append(ch)
if ch in out_data:
last = None
if stack:
last = stack.pop()
if last is '{' and ch is '}':
continue
elif last is '[' and ch is ']':
continue
elif last is '(' and ch is ')':
continue
else:
return False
if len(stack) > 0:
return False
else:
return True
print(check_match("{www[eee}ee)eee"))
print(check_match("(ee)(eee[eeew]www)"))
print(check_match("(ss(ss[{ss}]zs)zss)"))
print(check_match("([{[[]]}])"))
def matched(str):
braces = {"{": "}", "(": ")", "[": "]"}
stack = []
for c in str:
if c in braces.keys():
stack.append(c)
elif c in braces.values():
if not stack:
return False
last_brace = stack.pop()
if braces[last_brace] != c:
return False
if stack:
return False
return True
print(matched("()"))
>> True
print(matched("(}"))
>> False
print(matched("}{"))
>> False
print(matched("}"))
>> False
print(matched("{"))
>> False
print(matched("(ff{fgg} [gg]h)"))
>> True
Given a string s containing just the characters '(', ')', '{', '}', '[' and ']',
determine if the input string is valid.
def isValid(s):
stack = []
for i in s:
if i in open_list:
stack.append(i)
elif i in close_list:
pos = close_list.index(i)
if open_list[pos] == stack[len(stack)-1]:
stack.pop()
else:
return False
if len(stack) == 0:
return True
else:
return False
print(isValid("{[(){}]}"))
s='{[]{()}}}{'
t=list(s)
cntc=0
cnts=0
cntp=0
cntc=min(t.count("{"),t.count("}"))
cnts=min(t.count("["),t.count("]"))
cntp=min(t.count("("),t.count(")"))
print(cntc+cnts+cntp)
for a balanced string, we can find an opening brace followed by it closing brace. if you do this basic check you could remove the checked substring and check the remaining string. At the end, if the string is not empty then it is not balanced.
def is_balanced(s: str) -> bool:
while any([x in s for x in ["", "", ""]]):
s=s.replace("{}", "").replace("[]","").replace("()","")
return s==""
def parenthesis_check(parenthesis):
chars = []
matches = {')':'(',']':'[','}':'{'}
for i in parenthesis:
if i in matches:
if chars.pop() != matches[i]:
return False
else:
chars.append(i)
return chars == []
foo1="()()())("
def bracket(foo1):
count = 0
for i in foo1:
if i == "(":
count += 1
else:
if count==0 and i ==")":
return False
count -= 1
if count == 0:
return True
else:
return False
bracket(foo1)
Although I'm not proposing a fix to your implementation, I suggest a cleaner and more pythonic version of the #kreld solution:
def check_parentheses(expr):
s = []
for c in expr:
if c in '(':
s.append(c)
elif c in ')':
if not len(s):
break
else:
s.pop()
else:
return not len(s)
return False
# test -----------------------------------------------------------------
test_expr = [')(', '(()', '())', '(', ')', '((', '))', '(()())', '(())',
'()', '()(())']
for i, t in enumerate(test_expr, 1):
print '%i\t%s\t%s' % (i, t, check_parentheses(t))
# output ---------------------------------------------------------------
1 )( False
2 (() False
3 ()) False
4 ( False
5 ) False
6 (( False
7 )) False
8 (()()) True
9 (()) True
10 () True
11 ()(()) True
I am trying to split some text. Basically I want to separate level-1 brackets, like "('1','a',NULL),(2,'b')" => ["('1','a',NULL)", "(2,'b')]", but I need to be aware of possible quoted strings inside. It needs to at least satisfy the following py.tests:
from splitter import split_text
def test_normal():
assert split_text("('1'),('2')") == ["('1')", "('2')"]
assert split_text("(1),(2),(3)") == ["(1)", "(2)", "(3)"]
def test_complex():
assert split_text("('1','a'),('2','b')") == ["('1','a')", "('2','b')"]
assert split_text("('1','a',NULL),(2,'b')") == ["('1','a',NULL)", "(2,'b')"]
def test_apostrophe():
assert split_text("('\\'1','a'),('2','b')") == ["('\\'1','a')", "('2','b')"]
def test_coma_in_string():
assert split_text("('1','a,c'),('2','b')") == ["('1','a,c')", "('2','b')"]
def test_bracket_in_string():
assert split_text("('1','a)c'),('2','b')") == ["('1','a)c')", "('2','b')"]
def test_bracket_and_coma_in_string():
assert split_text("('1','a),(c'),('2','b')") == ["('1','a),(c')", "('2','b')"]
def test_bracket_and_coma_in_string_apostrophe():
assert split_text("('1','a\\'),(c'),('2','b')") == ["('1','a\\'),(c')", "('2','b')"]
I have tried the following:
1) Regular expressions
This looks like the best solution, but unfortunately I did not come up with anything satisfying all tests.
My best try is:
def split_text(text):
return re.split('(?<=\)),(?=\()', text)
But obviously, that is rather simplistic and fails test_bracket_and_coma_in_string and test_bracket_and_coma_in_string_apostrophe.
2) Finite-state-machine-like solution
I tried to code the FSM myself:
OUTSIDE, IN_BRACKETS, IN_STRING, AFTER_BACKSLASH = range(4)
def split_text(text):
state = OUTSIDE
read = []
result = []
for character in text:
if state == OUTSIDE:
if character == ',':
result.append(''.join(read))
read = []
elif character == '(':
read.append(character)
state = IN_BRACKETS
else:
read.append(character)
elif state == IN_BRACKETS:
read.append(character)
if character == ')':
state = OUTSIDE
elif character == "'":
state = IN_STRING
elif state == IN_STRING:
read.append(character)
if character == "'":
state = IN_BRACKETS
elif character == '\\':
state = AFTER_BACKSLASH
elif state == AFTER_BACKSLASH:
read.append(character)
state = IN_STRING
result.append(''.join(read)) # The rest of string
return result
It works, passes all tests, but is very slow.
3) pyparsing
from pyparsing import QuotedString, ZeroOrMore, Literal, Group, Suppress, Word, nums
null_value = Literal('NULL')
number_value = Word(nums)
string_value = QuotedString("'", escChar='\\', unquoteResults=False)
value = null_value | number_value | string_value
one_bracket = Group(Literal('(') + value + ZeroOrMore(Literal(',') + value) + Literal(')'))
all_brackets = one_bracket + ZeroOrMore(Suppress(',') + one_bracket)
def split_text(text):
parse_result = all_brackets.parseString(text)
return [''.join(a) for a in parse_result]
Also passes all tests, but surprisingly it is even slower than solution #2.
Any ideas how to make the solution fast and robust? I have this feeling that I am missing something obvious.
One way would be to use the newer regex module which supports the (*SKIP)(*FAIL) functionality:
import regex as re
def split_text(text):
rx = r"""'.*?(?<!\\)'(*SKIP)(*FAIL)|(?<=\)),(?=\()"""
return re.split(rx, text)
Broken down it says:
'.*?(?<!\\)' # look for a single quote up to a new single quote
# that MUST NOT be escaped (thus the neg. lookbehind)
(*SKIP)(*FAIL)| # these parts shall fail
(?<=\)),(?=\() # your initial pattern with a positive lookbehind/ahead
This succeeds on all your examples.
I cooked this and it works on given tests.
tests = ["('1'),('2')",
"(1),(2),(3)",
"('1','a'),('2','b')",
"('1','a',NULL),(2,'b')",
"('\\'1','a'),('2','b')",
"('1','a,c'),('2','b')",
"('1','a)c'),('2','b')",
"('1','a),(c'),('2','b')",
"('1','a\\'),(c'),('2','b')"]
for text in tests:
tmp = ''
res = []
bracket = 0
quote = False
for idx,i in enumerate(text):
if i=="'":
if text[idx-1]!='\\':
quote = not quote
tmp += i
elif quote:
tmp += i
elif i==',':
if bracket: tmp += i
else: pass
else:
if i=='(': bracket += 1
elif i==')': bracket -= 1
if bracket: tmp += i
else:
tmp += i
res.append(tmp)
tmp = ''
print res
Output:
["('1')", "('2')"]
['(1)', '(2)', '(3)']
["('1','a')", "('2','b')"]
["('1','a',NULL)", "(2,'b')"]
["('\\'1','a')", "('2','b')"]
["('1','a,c')", "('2','b')"]
["('1','a)c')", "('2','b')"]
["('1','a),(c')", "('2','b')"]
["('1','a\\'),(c')", "('2','b')"]
The code has room for improvement, and edits are welcome. :)
This is the regular expression which seems to work and passes all the tests. Running it on real data it is about 6x faster than finite state machine implemented in Python.
PATTERN = re.compile(
r"""
\( # Opening bracket
(?:
# String
(?:'(?:
(?:\\')|[^'] # Either escaped apostrophe, or other character
)*'
)
|
# or other literal not containing right bracket
[^')]
)
(?:, # Zero or more of them separated with comma following the first one
# String
(?:'(?:
(?:\\')|[^'] # Either escaped apostrophe, or other character
)*'
)
|
# or other literal
[^')]
)*
\) # Closing bracket
""",
re.VERBOSE)
def split_text(text):
return PATTERN.findall(text)
I'm trying to make a program that checks to make sure that the text is balanced in terms of brackets (so (),[],{} and not (),[,{}). I can get it to work when it is balanced, and when it is not balanced when it is missing a closing bracket (like the previous example). What I can't get it to do is come back as unbalanced if I'm missing a bracket on the left ((),],{}). I know it's trying to pop from an empty stack but can't figure out how to counter act that. My teacher has it in her Stack class that if it's trying to pop to an empty stack, then an exception is raised automatically, and I can't change her class, which is the problem, otherwise I just would have made that as false anyways and not be in this mess. So does anyone have any ideas of how to do it before that error is raised?
Here's the code:
from ListNode import *
from Stack import Stack
ch = 0
s = 0
check = True
def parbalance():
stack = Stack()
user = input("Enter a file name: ")
file = open(user)
lines = file.readlines()
for char in lines:
for ch in char:
#print(ch)
if ch in "([{":
stack.push(ch)
if ch in ")]}":
popStack = stack.pop()
if ch == "(" and popStack != ")":
check = False
elif ch == "[" and popStack != "]":
check = False
elif ch == "{" and popStack != "}":
check = False
if stack.is_empty():
check = True
print("true")
else:
check = False
print("false")
parbalance()
In case it helps, here's her Stack class:
from ListNode import *
class Stack:
def __init__(self):
self.top = None
def push(self, item):
temp = ListNode(item)
temp.set_link(self.top)
self.top = temp
#self.top = ListNode(item, self.top)
def pop(self):
if self.top == None:
raise Exception("Trying to pop from an empty stack")
temp = self.top
self.top = temp.get_link()
return temp.get_item()
def destroy(self):
self.top = None
def is_full(self):
return False
def is_empty(self):
return self.top == None
Use try to capture an error:
try:
popStack = stack.pop()
except:
# Stack is empty, set failure and bail from the function.
check = False
return
Also, note that your tests are backwards:
if ch == "(" and popStack != ")":
ch is the closing bracket and popStack is the opening bracket, so this should be:
if ch == ")" and popStack != "(":
Without this change, your code will recognize the string "(}" as balanced.
As a side note, consider returning True or False from the function instead of setting a global variable. Using global variables to return values from functions is not a good idea.
You can put the code into a try except block. Once you catch an exception you know the stack underflows. Consequently there must be an unbalanced paranthesis.
By the way: I would not use the lengthy if else chain. Instead I would work it along the following lines:
pars = {'(':')', '[':']', '{':'}'}
....
try:
...
if ch in pars.keys():
stack.push(ch)
if ch in pars.values():
if ch != pars[stack.pop()]:
return False;
except:
return False;
return stack.is_empty()
That way you could easily add other bracket symbols if required.