在Python中打印字典列表

Question

I have the following list of lists of dicts:

l = [[{'close': 'TRUE'}], [{'error': 'FALSE'}], [{'close': 'TRUE', 'error': 'TRUE'}, {'close': 'TRUE', 'error': 'FALSE'}]]

and I would like to print it this way:

(close = TRUE) & (error = FALSE) & ((close = TRUE & error = TRUE) | (close = TRUE & error = FALSE))

For the moment, I have the following function which nearly do the job:

def pretty_string(l):
    print ' & '.join('({0})'
                        .format(' | '
                                .join('({0})'
                                      .format(' & '
                                              .join('{0} = {1}'
                                                    .format(key, value)
                                                    for key, value
                                                    in disjuncts.items()))
                                      for disjuncts
                                      in conjuncts))
                        for conjuncts
                        in l)

But it gives me:

((close = TRUE)) & ((error = FALSE)) & ((close = TRUE & error = TRUE) | (close = TRUE & error = FALSE))

Notice the extra parentheses around "(close = TRUE)" and "(error = FALSE)".

How can these be removed?

Answer 1

Just use if statement to change format string ( ('({0})' if len(disjuncts) > 1 else '{0}') ) depending on the length of your internal list. Like this:

def pretty_string(l):
    print ' & '.join(
        '({0})'.format(
            ' | '.join(
                ('({0})' if len(disjuncts) > 1 else '{0}').format(
                    ' & '.join(
                        '{0} = {1}'.format(key, value) for key, value in disjuncts.items()
                    )
                ) for disjuncts in conjuncts
            )
        ) for conjuncts in l
    )

Answer 2

def conv(item):
    if(isinstance(item, dict)):
        yield '(' + ' & '.join("{} = {}".format(*i) for i in item.items()) + ')'
    elif(isinstance(item, list)):
        for i in item:
            for s in conv(i):
                yield s

def pretty_string(l):                                                                                                       
    return ' | '.join(conv(l))