Question

我有一个像这样的文本文件，我想在python中处理它

info.txt

    firstname1
    surname1
    email@email.com1
    student1
 -------------------
    firstname2
    surname2
    email@email.com2
    student2
-----------------

我想编写一个python代码，迭代并存储每个索引示例中的每一行：[firstname,surname,email@email.com,student]并忽略"-----"

python代码

 with open('log.txt') as f:
        lines = f.read().splitlines()
        x = x + 1

    for i in lines:
        print i

但我相信这是错误的我对python非常新，有些人请指出我正确的方向我希望输出给我这样的东西

输出

index 1 :first name: firstname1
         Surname: surname1
         Email: email@email.com1
         Student student1

index 2 :first name: firstname2
         Surname: surname2
         Email: email@email.com2
         student: student2

Answer 1

我知道解释如何做这样的事情的一般指导方针是更好的形式，但对于像这样的简单任务，代码本身就说明了，真的......

我会像这样实现它。

from pprint import pprint  # For nicer formatting of the output.

# For the sake of a self-contained example,
# the data is inlined here.
#
# `f` could be replaced with `open('log.txt').

f = """
    firstname1
    surname1
    email@email.com1
    student1
-------------------
    firstname2
    surname2
    email@email.com2
    student2
-----------------
""".splitlines()

data = []
current = None
for line in f:
    line = line.strip()  # Remove leading and trailing spaces
    if not line:  # Ignore empty lines
        continue  # Skip the rest of this iteration.
    if line.startswith('-----'):  # New record.
        current = None  # Clear the `current` variable
        continue  # Skip the rest of the iteration
    if current is None:  # No current entry?
        # This can happen either after a ----- line, or
        # when we're dealing with the very first line of the file.

        current = []  # Create an empty list,
        data.append(current)  # and push it to the list of data.
    current.append(line)

pprint(data)

输出是一个列表列表：

[['firstname1', 'surname1', 'email@email.com1', 'student1'],
 ['firstname2', 'surname2', 'email@email.com2', 'student2']]

Answer 2

这是一个可能更优雅的解决方案。（只要您的文件严格保留您示例中的格式，即四行数据后跟一条虚线。）

from itertools import izip # skip this line if you are using Python 3

with open('info.txt') as f:
    result = [{'first name': first.strip(), 'Surname': sur.strip(),
               'Email': mail.strip(), 'student': stud.strip()}
              for first, sur, mail, stud, _ in izip(*[f]*5)]

这为您提供了一个字典列表，如下所示：

[{'first name': 'firstname1', 'Surname': 'surname1', 'Email': 'email@email.com1', 'student': 'student1'}, {'first name': 'firstname2', 'Surname': 'surname2', 'Email': 'email@email.com2', 'student': 'student2'}]

如果您的“索引1”对应于列表的第一个元素（即result[0]），则“索引2”对应于列表的第二个元素，依此类推。

例如，您可以使用

获取index == 2的姓氏

index = 2
result[index - 1]['Surname']

如果您真的担心索引会被移位，您可以根据结果构建一个字典。演示：

>>> result = dict(enumerate(result, 1))
>>> result
{1: {'first name': 'firstname1', 'Surname': 'surname1', 'Email': 'email@email.com1', 'student': 'student1'}, 2: {'first name': 'firstname2', 'Surname': 'surname2', 'Email': 'email@email.com2', 'student': 'student2'}}
>>>
>>> result[2]['Surname']
'surname2'
>>>
>>> for index, info in result.items():
...     print index, info['first name'], info['Surname'], info['Email'], info['student']
... 
1 firstname1 surname1 email@email.com1 student1
2 firstname2 surname2 email@email.com2 student2

Python在文本文件列表中逐行存储

2 个答案: