将txt解析为块 [英] Parse txt to blocks

问题描述

我有一个txt文件，该文件具有以下结构

i have a txt file, which has the following structure

start
id=1
date=21.05.2018
summ=500
end

start
id=7
date=23.05.2018
summ=500
owner=guest
end

并且我需要在字典列表中解析它(str:str(即使它是int类型或日期:将其转换为字符串)).即使用start end将其拆分为块，然后将其拆分为=符号. start end之间的行数可以不同. d 但是无法实现.我尝试过这样的事情:

and i need to parse it in a list of dictionaries (str : str (even it is int type or date: convert it to string)). i.e. split it on block with start end, and after that split it on = symbol. The amount of lines between start end can be different. D But a can't realize it. I tried something like this:

d ={}
arr = []
ind = 0
for line in plines:
    ind = ind + 1
    if 'startpayment' in line:
        print('ind = ' + str(ind))
        for i in range(ind, len(plines)):
            print(i)
            key, value = plines[i].strip().split('=')
            if type(value) == 'str':
                d[key] = str(value)
            elif type(value) == 'int':
                 d[key] = int(value)
            arr.append(d)
            if 'endpayment' in line:
                break

有人可以帮我吗?谢谢

推荐答案

您还可以尝试以下方法:

You could also try something like this:

from itertools import takewhile

with open('data.txt') as in_file:
    items = [line.strip() for line in in_file.read().split()]
    # ['start', 'id=1', 'date=21.05.2018', 'summ=500', 'end', 'start', 'id=7', 'date=23.05.2018', 'summ=500', 'owner=guest']

    pos = [i for i, item in enumerate(items) if item == 'start']
    # [0, 5]

    blocks = [list(takewhile(lambda x: x != 'end', items[i+1:])) for i in pos]
    # [['id=1', 'date=21.05.2018', 'summ=500'], ['id=7', 'date=23.05.2018', 'summ=500', 'owner=guest']]

    print([dict(x.split('=') for x in block) for block in blocks])

哪些输出:

[{'id': '1', 'date': '21.05.2018', 'summ': '500'}, {'id': '7', 'date': '23.05.2018', 'summ': '500', 'owner': 'guest'}]

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