我希望能够将所有括号配对成一个字符串，如果它们没有配对，那么它们会得到它们的索引号和False。似乎它一遍又一遍地重复一些值，即cl == pop [1]。我试图看看问题出在哪里，但无论我怎么努力，我都看不到它。所以我问是否有人帮助我找到错误，甚至可能改进我的代码;）
def check_parentheses(string):
pending = 0
brackets = []
'''Checks if parens are paired, otherwise they are bad.'''
parenstack = collections.deque()
for ch in string:
if ch in lrmap:
try:
cl = string.index(ch, pending)
pending = cl + 1
except:
cl = False
if ch in lparens:
parenstack.append([ch, cl])
print parenstack
elif ch in rparens:
try:
pop = parenstack.pop()
if lrmap[pop[0]] != ch:
print 'wrong type of parenthesis popped from stack',\
pop[0], ch, pop[1], cl
brackets.append([pop[1], False])
brackets.append([cl, False])
else:
brackets.append([pop[1], cl])
except IndexError:
print 'no opening parenthesis left in stack'
brackets.append([cl, False])
# if we are not out of opening parentheses, we have a mismatch
for p in parenstack:
brackets.append([p[1],False])
return brackets
答案 0 :(得分：20)
您可以将我的代码调整为类似的问题：
def Evaluate(str):
stack = []
pushChars, popChars = "<({[", ">)}]"
for c in str :
if c in pushChars :
stack.append(c)
elif c in popChars :
if not len(stack) :
return False
else :
stackTop = stack.pop()
balancingBracket = pushChars[popChars.index(c)]
if stackTop != balancingBracket :
return False
else :
return False
return not len(stack)
答案 1 :(得分：9)
iparens = iter('(){}[]<>')
parens = dict(zip(iparens, iparens))
closing = parens.values()
def balanced(astr):
stack = []
for c in astr:
d = parens.get(c, None)
if d:
stack.append(d)
elif c in closing:
if not stack or c != stack.pop():
return False
return not stack
示例：
>>> balanced('[1<2>(3)]')
True
>>> balanced('[1<2(>3)]')
False
答案 2 :(得分：4)
BRACES = { '(': ')', '[': ']', '{': '}' }
def group_check(s):
stack = []
for b in s:
c = BRACES.get(b)
if c:
stack.append(c)
elif not stack or stack.pop() != b:
return False
return not stack
答案 3 :(得分：0)
谢谢hughdbrown你的代码轻而易举地工作，它真的很短！你刚刚让我头疼：D
如果没问题，将其转换为pep8：）
修改
charset = dict(opening='{[(<',\
closing='}])>',\
string = ('"', "'"),\
comment=(('<!--', '-->')))
charset = dict(opening='{[(<',\
closing='}])>',\
string = ('"', "'"),\
comment=(("'''", "'''"), ('"""', '"""'), ('#', '\n')))
charset = dict(opening='{[(<',\
closing='}])>',\
string = ('"', "'"),\
comment=(('/*', '*/'), ('//', '\n')))
你明白了吗？：）
charset = dict(opening='{[(<',\
closing='}])>',\
string = ('"', "'"),\
comment=(('<!--', '-->'), ('"""', '"""'), ('#', '\n')))
allowed = ''.join([x[0][0] + x[1][0] for x in charset['comment']])
allowed += ''.join(charset['string'])
allowed += charset['opening']
allowed += charset['closing']
def brace_check(text):
o = []
c = []
notr = []
found = []
busy = False
last_pos = None
for i in xrange(len(text)):
ch = text[i]
if not busy:
cont = True
for comment in charset['comment']:
if ch == comment[0][0]:
como = text[i:len(comment[0])]
if como == comment[0]:
busy = comment[1]
if ch in charset['opening']:
last_pos = i
cont = False
break
if cont:
if ch in charset['string']:
busy = ch
elif ch in charset['opening']:
o.append((ch, i))
elif ch in charset['closing']:
c.append((ch, i))
else:
if ch == busy[0]:
if len(busy) == 1:
comc = ch
else:
comc = text[i:i + len(busy)]
if comc == busy:
if last_pos is not None:
if busy[-1] in charset['closing']:
found.append((last_pos, i))
last_pos = None
text = text[:i] + '\n' * len(comc) +\
text[i + len(comc):]
busy = not busy
elif busy in charset['string']:
if ch == '\n':
busy = not busy
for t, e in reversed(o):
try:
n = next((b, v) for b, v in c\
if b == charset['closing'][\
charset['opening'].find(t)] and v > e)
c.remove(n)
n = n[1]
if found != []:
if e < found[-1][0] and n > found[-1][0] and n < found[-1][1]\
or e < found[-1][1] and n > found[-1][1] and e > found[-1][0]:
found.append((n, False))
n = False
except StopIteration:
n = False
found.append((e, n))
for t, e in c:
found.append((e, False))
return found
答案 4 :(得分：0)
Python 3中可理解的解决方案：
object Form1: TForm1
Left = 0
Top = 0
ClientHeight = 362
ClientWidth = 619
Color = clBtnFace
Font.Charset = OEM_CHARSET
Font.Color = clWindowText
Font.Height = 14
Font.Name = 'System'
Font.Pitch = fpFixed
Font.Style = [fsBold]
OldCreateOrder = False
OnClose = FormClose
PixelsPerInch = 96
TextHeight = 14
object Memo1: TMemo
Left = 0
Top = 31
Width = 619
Height = 331
Align = alClient
ScrollBars = ssBoth
TabOrder = 0
ExplicitTop = 32
end
object Panel1: TPanel
Left = 0
Top = 0
Width = 619
Height = 31
Align = alTop
TabOrder = 1
object Wczytaj1: TSpeedButton
Left = 34
Top = 4
Width = 23
Height = 22
Glyph.Data = {
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OnClick = Wczytaj1Click
end
object Zamknij1: TSpeedButton
Left = 232
Top = 4
Width = 23
Height = 22
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OnClick = Zamknij1Click
end
object Konwertuje1: TSpeedButton
Left = 92
Top = 4
Width = 23
Height = 22
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OnClick = Konwertuje1Click
end
object Wyczysc1: TSpeedButton
Left = 150
Top = 4
Width = 23
Height = 22
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D0D4C8D0D4C8D0D40000C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8
D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4
C8D0D4C8D0D4C8D0D4C8D0D40000C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4
C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0
D4C8D0D4C8D0D4C8D0D4C8D0D4C8D0D40000}
OnClick = Wyczysc1Click
end
object Zapiszjako1: TSpeedButton
Left = 59
Top = 4
Width = 23
Height = 22
Glyph.Data = {
F6000000424DF600000000000000760000002800000010000000100000000100
04000000000080000000C40E0000C40E00001000000000000000000000000000
8000008000000080800080000000800080008080000080808000C0C0C0000000
FF0000FF000000FFFF00FF000000FF00FF00FFFF0000FFFFFF00888888888888
888888000000000000088044000FF0000408804400F99F000408804400F99F00
04088044000FF0000408804444F99F444408804400F99F004408804088F999F8
040880408F8F999F04088040F9F8F99F04088040F99F999F04088040F99999F8
000880408F999F880808800000FFF00000088888888888888888}
OnClick = Zapiszjako1Click
end
object Czcionka1: TSpeedButton
Left = 175
Top = 4
Width = 23
Height = 22
Caption = 'A'
Font.Charset = OEM_CHARSET
Font.Color = clWindowText
Font.Height = -21
Font.Name = 'System'
Font.Pitch = fpFixed
Font.Style = [fsBold]
ParentFont = False
OnClick = Czcionka1Click
end
object Info1: TSpeedButton
Left = 200
Top = 4
Width = 23
Height = 22
Caption = 'i'
Font.Charset = OEM_CHARSET
Font.Color = clWindowText
Font.Height = -21
Font.Name = 'System'
Font.Pitch = fpFixed
Font.Style = [fsBold]
ParentFont = False
OnClick = Info1Click
end
object Filtruj1: TSpeedButton
Left = 118
Top = 4
Width = 23
Height = 22
OnClick = Filtruj1Click
end
object PrzywrcTXT1: TSpeedButton
Left = 4
Top = 4
Width = 23
Height = 22
Glyph.Data = {
F6000000424DF600000000000000760000002800000010000000100000000100
04000000000080000000CE0E0000C40E00001000000000000000000000000000
80000080000000808000800000008000800080800000C0C0C000808080000000
FF0000FF000000FFFF00FF000000FF00FF00FFFF0000FFFFFF00777777777777
77777777777777777777000000000007777700333333333077770B0333333333
07770FB03333333330770BFB0333333333070FBFB000000000000BFBFBFBFB07
77770FBFBFBFBF0777770BFB0000000777777000777777770007777777777777
7007777777770777070777777777700077777777777777777777}
OnClick = PrzywrcTXT1Click
end
object Edit1: TEdit
Left = 268
Top = 4
Width = 80
Height = 22
TabOrder = 0
Text = 'Edit1'
end
object Edit2: TEdit
Left = 354
Top = 4
Width = 80
Height = 22
TabOrder = 1
Text = 'Edit2'
end
end
object OpenDialog1: TOpenDialog
Left = 440
end
object SaveDialog1: TSaveDialog
Left = 472
end
object FontDialog1: TFontDialog
Font.Charset = DEFAULT_CHARSET
Font.Color = clWindowText
Font.Height = -11
Font.Name = 'Tahoma'
Font.Style = []
Left = 504
end
end
答案 5 :(得分：0)
试试这个：
<TextBlock Text="..." Foreground="{DynamicResource AnotherBrush}" />
答案 6 :(得分：0)
以下代码将显示缺少的括号和给定字符串中缺少的次数。
from collections import Counter
def find_missing(str):
stack1 = []
stack2 = []
result = []
res_dict = {}
open_set = '<[{('
closed_set = '>]})'
a = list(str)
for i in a:
if i in open_set:
stack1.append(i)
elif i in closed_set:
stack2.append(i)
dict1 = Counter(stack1)
dict2 = Counter(stack2)
print(dict1)
print(dict2)
for i in open_set:
if dict1[i] > dict2[closed_set[open_set.index(i)]]:
res_dict[closed_set[open_set.index(i)]] = dict1[i] - dict2[closed_set[open_set.index(i)]]
result.append(closed_set[open_set.index(i)])
for i in closed_set:
if dict2[i] > dict1[open_set[closed_set.index(i)]]:
res_dict[open_set[closed_set.index(i)]] = dict2[i] - dict1[open_set[closed_set.index(i)]]
result.append(open_set[closed_set.index(i)])
return res_dict
# return result
if __name__ == '__main__':
str1 = '{This ((()bracket {[function]} <<going> crazy}'
x = find_missing(str1)
if len(x) > 0:
print("Imbalanced")
print(x)
else:
print("Balanced")
答案 7 :(得分：0)
首先我们将从左到右扫描字符串，每次我们看到一个左括号时我们将它推到一个堆栈，因为我们希望先关闭最后一个左括号。 （还记得堆栈的FILO结构！） 然后，当我们看到一个右括号时，我们通过从堆栈中弹出一个元素来检查最后打开的一个是否是相应的结束匹配。如果它是有效的匹配，那么我们继续前进，如果不是返回false。 码： https://gist.github.com/i143code/51962bfb1bd5925f75007d4dcbcf7f55
答案 8 :(得分：0)
我需要一些最近的项目，并认为我可以在OP的解决方案上建立一些。它允许检查注释模式，引号和括号，同时忽略周围的文本。我有目的地使它比它需要的更通用，以便其他人可以采取他们想要的东西并削减他们不想要的东西。
"""
This module is for testing bracket pairings within a given string
Tested with Python 3.5.4
>>> regexp = getRegexFromList(opening + closing)
>>> print(regexp)
(\\<\\-\\-|\\-\\-\\>|\\/\\*|\\/\\/|\\*\\/|\\#|\\"|\\'|\\(|\\[|\\{|\\<|\\\n|\\\n|\\"|\\'|\\)|\\]|\\}|\\>)
>>> test_string = 'l<--([0])-->1/*{<2>}*/3//<--4 &-->\\n5#"6"\\n7"/*(8)*/"9\'"10"\'11({12\ta})13[<14>]'
>>> patterns = re.findall(regexp, test_string)
>>> print(patterns)
['<--', '(', '[', ']', ')', '-->', '/*', '{', '<', '>', '}', '*/', '//', '<--', '-->', '\\n', '#', '"', '"', '\\n', '"', '/*', '(', ')', '*/', '"', '(', '{', '}', ')', '[', '<', '>', ']']
>>> doBracketsMatch(patterns)
True
>>> doBracketsMatch(['"', ')', '"', '[', ']', '\\''])
False
"""
# Dependencies
import re
# Global Variables
# Provide opening and closing patterns, along with their priorities & whether a priority is nestable
opening = ['<--', '/*', '//', '#', '"', '\'', '(', '[', '{', '<']
closing = ['-->', '*/', '\n', '\n', '"', '\'', ')', ']', '}', '>']
priority = [ 1, 1, 1, 1, 1, 1, 0, 0, 0, 0]
nestable = {0: True, 1: False}
bracket_pairs = dict(zip(opening + closing, \
[[(closing + opening)[i], (priority + priority)[i]] \
for i in range(0, opening.__len__() * 2)]))
def getRegexFromList(listOfPatterns):
"""
Generate the search term for the regular expression
:param listOfPatterns:
:return:
>>> getRegexFromList(['"', '<--', '##', 'test'])
'(\\\\t\\\\e\\\\s\\\\t|\\\\<\\\\-\\\\-|\\\\#\\\\#|\\\\")'
"""
# Longer patterns first to prevent false negatives
search_terms = sorted(listOfPatterns, key=len, reverse=True)
regex = ""
for term in search_terms:
for char in str(term):
regex = regex + '\\' + char # Search for all characters literally
regex = regex + '|' # Search pattern = (a|b|c)
return '(' + regex[:-1] + ')' # Remove excess '|' and add brackets
def doBracketsMatch(list_of_brackets):
"""
Determine if brackets match up
:param list_of_brackets:
:return:
"""
stack = []
for bracket in list_of_brackets:
# Check empty stack conditions
if stack.__len__() is 0:
# Check for openings first to catch quotes
if bracket in opening:
stack.append(bracket)
elif bracket in closing:
return False
else:
continue
# Check for a matching bracket
elif bracket == bracket_pairs[stack[-1]][0]:
stack.pop()
# Ignore cases:
# - False positives
# - Lower priority brackets
# - Equal priority brackets if nesting is not allowed
elif bracket not in bracket_pairs or \
bracket_pairs[bracket][1] < bracket_pairs[stack[-1]][1] or \
(bracket_pairs[bracket][1] == bracket_pairs[stack[-1]][1] and \
not nestable[bracket_pairs[bracket][1]]):
continue
# New open bracket
elif bracket in opening:
stack.append(bracket)
# Otherwise, unpaired close bracket
else:
return False
# If stack isn't empty, then there is an unpaired open bracket
return not bool(stack)
if __name__ == '__main__':
import doctest
doctest.testmod()