Question

附件表示一个树结构，我想将其转换为java源代码，将树表示为嵌套的if语句

STOP_WORDS > 0
| NEXT_TYPE > 5: X
| NEXT_TYPE <= 5: Y
STOP_WORDS <= 0
…


If (STOP_WORDS > 0)
{
                If (NEXT_TYPE > 5)
                {
                                Return X
                }
                If ( NEXT_TYPE <= 5)
                {
                                Return Y
                }
}
If (STOP_WORDS <= 0)
{
….
}

如您所见，使用管道（|）符号的缩进级别表示父子关系你可以用Java编写这个程序，或者你喜欢用你认为合适的任何其他语言编写。

下面是我需要转换为If语句的文本文件：

STOP_WORDS > 0
| NEXT_TYPE > 5: 5
| NEXT_TYPE <= 5: 1
STOP_WORDS <= 0
| STREET_TYE > 0
| | PREVIOUS_TYPE > 5: 2
| | PREVIOUS_TYPE <= 5
| | | NEXT_TYPE > 5: 2
| | | NEXT_TYPE <= 5: 5
| STREET_TYE <= 0
| | PERSON_TITLE > 0
| | | NEXT_TYPE <= 5: 4
| | | NEXT_TYPE > 5: 5
| | PERSON_TITLE <= 0
| | | SURNAME > 0
| | | | PREVIOUS_TYPE <= 4: 3
| | | | PREVIOUS_TYPE > 4: 5
| | | SURNAME <= 0
| | | | FIRST_NAME > 0
| | | | | NEXT_TYPE <= 5: 0
| | | | | NEXT_TYPE > 5: 5
| | | | FIRST_NAME <= 0
| | | | | TOKEN_LENGTH <= 1
| | | | | | NEXT_TYPE <= 4: 0
| | | | | | NEXT_TYPE > 4
| | | | | | | NEXT_TYPE <= 5: 1
| | | | | | | NEXT_TYPE > 5: 5
| | | | | TOKEN_LENGTH >1
| | | | | | NEXT_TYPE > 4: 5
| | | | | | NEXT_TYPE <= 4: 0
| | | | | | | PREVIOUS_TYPE <= 1: 4
| | | | | | | PREVIOUS_TYPE >1
| | | | | | | | PREVIOUS_TYPE >3: 4
| | | | | | | | PREVIOUS_TYPE <= 3: 5

我的问题是我找不到跟踪以前的If语句以及何时关闭它们的逻辑。正如您所看到的，每条线都有很多管道（|），有时候管道更少，有时管道更多。我找不到它的逻辑。

任何帮助都将不胜感激。

Answer 1

你的问题非常不清楚，但这是一个开始

barometer = 0
with open('path/to/input') as infile:
    for line in infile:
        barometer += delta
        line = ''.join(itertools.dropwhile(lambda c: c in ' |', line))
        if ":" in line:
            cond, act = line.split(":",1)
            print "If (%s) { Return %s }" %(cond, act)
        else:
            delta = barometer - sum(1 for i in itertools.takewhile(lambda c: c in ' |', line))
            if delta < 0:
                print " {}"[delta/abs(delta)]*abs(delta)
            print "If (%s) {" %cond

我现在必须去睡觉了，但这应该让你开始，至少

Answer 2

的Python

with open('input.txt') as f:
    indent = 8
    prev_depth = -1
    closes = []
    for line in f:
        line = line.strip()
        if not line: continue

        depth = line.count('|')
        while prev_depth >= depth:
            prev_depth -= 1
            print(closes.pop())
        pad = ' ' * (depth*indent)
        print(pad + 'If ({})'.format(line.lstrip('| ').split(':', 1)[0]))
        print(pad + '{')
        closes.append(pad + '}')
        if ':' in line:
            pad2 = ' ' * ((depth+1)*indent)
            print(pad2 + 'Return {}'.format(line[line.find(':')+1:].strip()))
        prev_depth = depth
    while closes:
        print(closes.pop())

将树结构解析为If语句

2 个答案:

的Python