我有一个打印字符串的递归函数。 我希望函数返回当前打印的字符串。
代码基本上打印出不同级别的递归所需的字符串的不同部分。
我正在考虑使用全局变量，但并不认为这听起来非常pythonic。
作为更大程序的一部分，我将多次调用此函数。
如果您想查看当前的代码：
def check_if_multiple_sections(the_string):
level = 0
counter = 0
for char in the_string:
if char == '(':
level = level + 1
if level == 0:
counter = counter + 1
if char == ')':
level = level - 1
if level == 0:
counter = counter + 1
return counter
def break_it_up(the_string):
level = 0
counter = 0
break_points = []
for char in the_string:
if char == '(':
level = level + 1
if char == ')':
level = level - 1
if level == 0:
break_points.append(counter + 1)
counter = counter + 1
return break_points
def recur_markov_change(the_string, key):
#print("the_string: " + str(the_string) + " : key: " +str(key))
if key is not None:
#print(the_string)
#print("YEA")
print(str(the_string).split(' ')[0] +'^'+str(key) + ' ', end='')
else:
print('(TOP ', end ='')
key = ((the_string.split(' '))[0])[1:]
if len(the_string) < 2:
return
remaining_string = the_string.split(' ', 1)[1][:-1]
results = []
results.append(((remaining_string.split(' '))[0])[1:])
level = 0
counter = 0
if the_string.count('(') == 1:
items = the_string[1:-1]
both = items.split(' ')
print(str(the_string).split(' ',1)[1], end="") #This prints leaves
return
for char in remaining_string:
if char == '(':
level = level + 1
if char == ')':
level = level - 1
if level == 0 and char == ' ':
results.append(((remaining_string[counter+1:].split(' '))[0])[1:])
counter = counter + 1
answer = (key, results, remaining_string)
#print(answer)
if check_if_multiple_sections(remaining_string) > 1:
break_points = break_it_up(remaining_string)
sublines = []
prev_spot = 0
for breaks in break_points:
sublines.append(remaining_string[prev_spot:breaks].strip())
prev_spot = breaks
sublines.append(remaining_string[breaks:].strip())
for line in sublines:
if len(line) > 2:
print(' ', end='')
recur_markov_change(line, key)
print(')', end='')
given_string = '(TOP (SBARQ (WHNP_WDT Which) (SQ_VP (VBZ is) (ADJP_JJ last))) (PUNC ?))'
string_that_I_want_function_to_return = '(TOP (SBARQ^TOP (WHNP_WDT^SBARQ Which) (SQ_VP^SBARQ (VBZ^SQ_VP is) (ADJP_JJ^SQ_VP last))) (PUNC^TOP ?))'
recur_markov_change(given_string, None)
print("\ndesired string:")
print(string_that_I_want_function_to_return)
答案 0 :(得分：2)
首先，我添加了一个函数递归添加的基本字符串。我调用了此initial，然后我用inititial+=替换了所有打印件，最后返回initial。
def check_if_multiple_sections(the_string):
level = 0
counter = 0
for char in the_string:
if char == '(':
level = level + 1
if level == 0:
counter = counter + 1
if char == ')':
level = level - 1
if level == 0:
counter = counter + 1
return counter
def break_it_up(the_string):
level = 0
counter = 0
break_points = []
for char in the_string:
if char == '(':
level = level + 1
if char == ')':
level = level - 1
if level == 0:
break_points.append(counter + 1)
counter = counter + 1
return break_points
def recur_markov_change(the_string, key, initial):
#initial+=("the_string: " + str(the_string) + " : key: " +str(key))
if key is not None:
#initial+=(the_string)
#initial+=("YEA")
initial+=(str(the_string).split(' ')[0] +'^'+str(key) + ' ')
else:
initial+=('(TOP ')
key = ((the_string.split(' '))[0])[1:]
if len(the_string) < 2:
return
remaining_string = the_string.split(' ', 1)[1][:-1]
results = []
results.append(((remaining_string.split(' '))[0])[1:])
level = 0
counter = 0
if the_string.count('(') == 1:
items = the_string[1:-1]
both = items.split(' ')
initial+=(str(the_string).split(' ',1)[1]) #This initial+=s leaves
return initial
for char in remaining_string:
if char == '(':
level = level + 1
if char == ')':
level = level - 1
if level == 0 and char == ' ':
results.append(((remaining_string[counter+1:].split(' '))[0])[1:])
counter = counter + 1
answer = (key, results, remaining_string)
#initial+=(answer)
if check_if_multiple_sections(remaining_string) > 1:
break_points = break_it_up(remaining_string)
sublines = []
prev_spot = 0
for breaks in break_points:
sublines.append(remaining_string[prev_spot:breaks].strip())
prev_spot = breaks
sublines.append(remaining_string[breaks:].strip())
for line in sublines:
if len(line) > 2:
initial+=(' ')
initial = recur_markov_change(line, key, initial)
initial+=(')')
return initial
given_string = '(TOP (SBARQ (WHNP_WDT Which) (SQ_VP (VBZ is) (ADJP_JJ last))) (PUNC ?))'
string_that_I_want_function_to_return = '(TOP (SBARQ^TOP (WHNP_WDT^SBARQ Which) (SQ_VP^SBARQ (VBZ^SQ_VP is) (ADJP_JJ^SQ_VP last))) (PUNC^TOP ?))'
print(recur_markov_change(given_string, None, ""))
print("\ndesired string:")
print(string_that_I_want_function_to_return)
经过测试和工作，就像我的魅力一样
答案 1 :(得分：0)
...This would be impossible without recursion.., - 没有递归就有可能。
import collectons, operator
s = '(TOP (SBARQ (WHNP_WDT Which) (SQ_VP (VBZ is) (ADJP_JJ last))) (PUNC ?))'
result = '(TOP (SBARQ^TOP (WHNP_WDT^SBARQ Which) (SQ_VP^SBARQ (VBZ^SQ_VP is) (ADJP_JJ^SQ_VP last))) (PUNC^TOP ?))'
我将使用自定义Node方法的类___str___。节点的有效负载将是其他节点或最内层节点的字符串。
class Node:
def __init__(self, name, index):
self.name = name
self.index = index
self.parent = None
self._payload = []
@property
def payload(self):
return self._payload
@payload.setter
def payload(self, thing):
self._payload.append(thing)
def __str__(self):
if self.parent is not None:
name = '{}^{}'.format(self.name, self.parent.name)
else:
name = self.name
#account for the extra space in a non-Node paylode
if not isinstance(self.payload[0], Node):
payload = ' ' + self.payload[0]
else:
payload = ' '.join(str(thing) for thing in self.payload)
return '({!s:} {!s:})'.format(name, payload)
该函数首先找到所有左右parens的索引。如果左边的paren后跟左边的paren，则会创建未完成的节点。 当找到左/右paren对时，节点已被完全解析（完成）。它返回最外层的节点。
def parse(s):
# Where are the parenthesis?
unfinished_nodes = collections.deque()
left, right = collections.deque(), collections.deque()
for i, c in enumerate(s):
if c == '(':
left.append(i)
if c == ')':
right.append(i)
if len(left) != len(right):
raise ValueError('Unbalanced parenthesis in s')
# helpers
paren = operator.itemgetter(0)
next_paren = operator.itemgetter(1)
while left and right:
while paren(left) > paren(right):
# back-up to previous Node
left.rotate()
#print('left:{}, right:{}'.format(paren(left), paren(right)))
# are we at the bottom? No more Nodes?
try:
if next_paren(left) < paren(left):
finished = True
else:
finished = paren(left) < paren(right) < next_paren(left)
except IndexError:
# only one more left paren - outermost Node
finished = True
if finished:
token = s[paren(left)+1:paren(right)]
#print('\ttoken:{}'.format(token))
if '(' not in token:
# this is an inner Node
name, payload = token.split()
node = Node(name, paren(left))
node.payload = payload
try:
node.parent = unfinished_nodes[-1]
node.parent.payload = node
except IndexError:
# this inner Node is the first Node in the sequence
pass
unfinished_nodes.append(node)
# finished is the outermost completely parsed node
finished = unfinished_nodes.pop()
#print('\tfinished:{!s:}'.format(finished.name))
right.popleft()
left.popleft()
#print('\t\tL:{}\t\tR:{}'.format(left, right))
else:
name = s[paren(left)+1:next_paren(left)]
name = name.strip()
node = Node(name, paren(left))
try:
node.parent = unfinished_nodes[-1]
node.parent.payload = node
except IndexError:
# this is the outermost Node
pass
#print('\tnew Unfinished:{}'.format(node.name))
unfinished_nodes.append(node)
# proceed to the next Node
left.rotate(-1)
return finished
#or if you prefer
#return str(finished)
print(parse(s))
print(result)
这里简化了，没有任何恶作剧，非递归，一次通过输入字符串 - 快速，干净，高效：
import collections
def parse2(s):
parents = collections.deque()
final = collections.deque()
t = ''
new_parent = False
new = ''
for c in s:
if c == '(':
new_parent = True
t += c
elif c == ')':
t += c
final.append(t)
t = ''
parents.pop()
else:
if c == ' ' and new_parent:
if parents:
t = t + '^' + parents[-1]
parents.append(new)
new_parent = False
new = ''
elif new_parent:
new += c
t += c
return ''.join(final)
不确定是什么让我使用Node类来提出第一个解决方案，并将其视为树/图形/其他。