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python - 如何用正则表达式匹配数据

问题描述

我有一个数组列表，如：

data = ['- TEST BEGA','R8=11K(10,15A)B','R9=1K(0,3A)B','R10_R84=13MEG(7,14K)R','R85_R84<100K(970,1000K)R',
'R85_R86=10K(9,11K)R']

我想像这样拆分数组列表

SCN: TEST BEGA

STEP R8
CHILD R8
Operator =
MEASURE_CHILD 11K(10,15A)B

STEP R9
CHILD R9
Operator =
MEASURE_CHILD 1K(0,3A)B

STEP R10_R84
CHILD R10_R84
Operator =
MEASURE_CHILD 13MEG(7,14K)R

STEP R85
CHILD R84
Operator <
MEASURE_CHILD 100K(970,1000K)R
CHILD R86
Operator =
MEASURE_CHILD 10K(9,11K)R

我使用这段代码来做这些事情，但我不知道出了什么问题：

def createTreeStandardBloc( self ):
    data = ['- TEST BEGA','R8=11K(10,15A)B','R9=1K(0,3A)B','R10_R84=13MEG(7,14K)R','R85_R84<100K(970,1000K)R','R85_R85=10K(9,11K)R']
    last_s = None
    for i, line in enumerate(data):
        if i == 0:
            print("SCN:", line.strip("- "))
        elif line.strip():
            s, c, op, mc = re.match("^\s*([^_]+)(_\w+)?([<>=])(.*)\s*$", line).groups()
            if s != last_s:
                print("STEP", s)
            print("CHILD", c or s)
            print("Operator",op)
            print("MEASURE_CHILD", mc)
            last_s = s

问题是数据 R10_R84 的步长被划分为 R10 的步长和 R84 的子级我希望当前缀像 R85 一样重复时，任何数据都会被划分。

标签： pythonpython-3.xalgorithmsplitmatch

解决方案

我相信其他人会想出一个更好的解决方案，但是就这样吧。

from collections import defaultdict

def get_operator(string):
    '''
    returns the operator found in the string
    '''
    operators = '=><'
    for i in operators:
        if i in string:
            return i

    return None

def createTreeStandardBloc(data):
    # parsed is a default dict of lists which will default
    # to an empty list if a new key is added
    parsed = defaultdict(list)

    # this loop does a few things
    for line in data[1:]:
        # it gets the operator
        oper = get_operator(line)
        # splits the line based on the operator
        split_line = line.split(oper)

        prefixes = split_line[0].split('_')

        # if there aren't 2 prefixes
        # it sets the child to the first and only prefix
        # otherwise it sets it to the second
        if len(prefixes) == 1:
            child = prefixes[0]
        else:
            child = prefixes[1]

        # then it adds it preformatted to the defaultdict
        # this means that any additional items found with
        # the same step prefix will just get added onto that step
        # as a child 
        parsed[prefixes[0]].append('CHILD ' + child)
        parsed[prefixes[0]].append('Operator ' + oper)
        parsed[prefixes[0]].append('MEASURE_CHILD ' + split_line[1])

    # here we start the final formatting of data
    formatted = []
    formatted.append('SCN: ' + data[0].strip('- '))

    for key, items in parsed.items():
        formatted.append(' ')

        # we get the first child prefix here
        child_prefix = items[0][6:]

        # if the child is different from the step 
        # and there are only 3 items
        # we should join them back together
        # I know mutating a collection were iterating over 
        # is sinful but I did it anyway ;)
        if len(items) == 3 and key != child_prefix:
            key = key + '_' + child_prefix
            items[0] = 'CHILD ' + key

        # now we can safely add our step to the formatted list
        formatted.append('STEP ' + key)

        # and the items
        for item in items:
            formatted.append(item)

    return formatted


data = ['- TEST BEGA', 
        'R8=11K(10,15A)B',
        'R9=1K(0,3A)B',
        'R10_R84=13MEG(7,14K)R',
        'R85_R84<100K(970,1000K)R',
        'R85_R86=10K(9,11K)R']

new_data = createTreeStandardBloc(data)

for line in new_data:
    print(line)

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