使用多个键将文本文件转换为字典

Question

我正在尝试使用以下格式的文本文件阅读：

student
    first name: John
    last name: Doe
    grade: 9
    gpa: 4.0
school
    name: Richard High School
    city: Kansas City

####

student
    first name: Jane
    last name: Doe
    grade: 10
    gpa: 3.0
school
    name: Richard High School
    city: Kansas City

进入Python词典。试图最终结果看起来像：

{0:{'student':{'first name': 'John',
    'last name': 'Doe',
    'grade': '9',
    'gpa': '4.0'},
    "school": {'name': 'Richard High School',
               'city': 'Kansas City'},
1:{'student':{'first name': 'Jane',
    'last name': 'Doe',
    'grade': '10',
    'gpa': '3.0'},
    'school': {'name': 'Richard High School',
               'city': 'Kansas City'}
}

到目前为止，我知道如何处理内部钥匙：

with open('<filename>') as f:
    dict = {}
    for line in f:
        x, y = line.split(": ")
        dict[x] = y
    print(dict)

但是除​​此之外，我仍然陷入困境。

Answer 1

这是一个可能的解决方案：

import re

file = open("a.txt")
dictionaryMain = {}
dictionaryElement = {}
dictionaryStudent = {}
dictionarySchool = {}


text = file.read()
elements = text.split("####")

i = 0
for element in elements:
    firstName = re.search('first name: (.+)', text).group(1)
    lastName = re.search('last name: (.+)', text).group(1)
    grade = re.search('grade: (.+)', text).group(1)
    gpa = re.search('gpa: (.+)', text).group(1)
    name = re.search('name: (.+)', text).group(1)
    city = re.search('city: (.+)', text).group(1)
    dictionaryStudent['first name'] = firstName
    dictionaryStudent['last name'] = lastName
    dictionaryStudent['grade'] = grade
    dictionaryStudent['gpa'] = gpa
    dictionarySchool['name'] = name
    dictionarySchool['city'] = city
    dictionaryElement['student'] = dictionaryStudent
    dictionaryElement['school'] = dictionarySchool
    i = i+1
    dictionaryMain[i] = dictionaryElement

print(dictionaryMain)

输入文件：

student
    first name: John
    last name: Doe
    grade: 9
    gpa: 4.0
school
    name: Richard High School
    city: Kansas City

####

student
    first name: Jane
    last name: Doe
    grade: 10
    gpa: 3.0
school
    name: Richard High School
    city: Kansas City

####

student
    first name: Jane
    last name: Doe
    grade: 10
    gpa: 3.0
school
    name: Richard High School
    city: Kansas City

输出：

{
  1: {
    'student': {
      'first name': 'John',
      'last name': 'Doe',
      'grade': '9',
      'gpa': '4.0'
    },
    'school': {
      'name': 'John',
      'city': 'Kansas City'
    }
  },
  2: {
    'student': {
      'first name': 'John',
      'last name': 'Doe',
      'grade': '9',
      'gpa': '4.0'
    },
    'school': {
      'name': 'John',
      'city': 'Kansas City'
    }
  },
  3: {
    'student': {
      'first name': 'John',
      'last name': 'Doe',
      'grade': '9',
      'gpa': '4.0'
    },
    'school': {
      'name': 'John',
      'city': 'Kansas City'
    }
  }
}

我不完全知道您的用例是什么，但是如果您具有如此严格的格式，您应该真正考虑使用数据类。

Answer 2

如果您的数据的模式完全如您所写，并且您不介意拥有平面词典，每个学生一个：

pattern = re.compile(r"""
student
    first name: (?P<first_name>.*)
    last name: (?P<last_name>.*)
    grade: (?P<grade>\d*)
    gpa: (?P<gpa>\d+.?\d*)
school
    name: (?P<school>.*)
    city: (?P<city>.*)""".strip())


with open(<filename>, "r") as f:
    data = f.read()


students = [match.groupdict() for match in pattern.finditer(data)]

输出：

[{'first_name': 'John',
  'last_name': 'Doe',
  'grade': '9',
  'gpa': '4.0',
  'school': 'Richard High School',
  'city': 'Kansas City'},
 {'first_name': 'Jane',
  'last_name': 'Doe',
  'grade': '10',
  'gpa': '3.0',
  'school': 'Richard High School',
  'city': 'Kansas City'}]

我没有看到您想要的数据结构的好处，因此我的建议以获得更有利于表格数据分析的东西。

编辑：现在我们正在谈论 Pandas，

In [4]: df = pd.DataFrame(students)

In [5]: df
Out[5]:
  first_name last_name grade  gpa               school         city
0       John       Doe     9  4.0  Richard High School  Kansas City
1       Jane       Doe    10  3.0  Richard High School  Kansas City

获取每个年级的学生数量：

In [6]: df.groupby("grade").size()
Out[6]:
grade
10    1
9     1
dtype: int64

您还可以按任意数量的列进行分组，例如按年级和学校：

In [7]: df.groupby(["grade", "school"]).size()
Out[7]:
grade  school
10     Richard High School    1
9      Richard High School    1
dtype: int64

Answer 3

import re
temp = 0
data = {temp:{}}
with open('txt.txt') as f:
    for line in f:
        if len(line.strip()) == 0:
            continue
        if re.match("^[^:]*:.*$", line):
            key, value = line.split(':', 1)
            data[temp][main_key][key.strip()] = value.strip()
        elif re.match("^[^\#]*$", line):
            main_key = line.strip()
            if main_key in (data[temp].keys()):
                temp += 1
                data[temp] = {}
            data[temp][main_key] = {}

如果我正确地理解了你的目标，这就是答案。但要小心，它基于正则表达式，您现在可以在 regex101.com 中了解更多信息，

如果我看到像“”这样的行并且充满了空！ （断线）
第二，我检查行格式是否类似于“key：value”，如果不是，那么它是主键，我将其添加到主字典中，否则，我将其添加到主字典中的最后一个字典中

Answer 4

您可以这样做，但请记住，此方法非常特定于原始问题中定义的输入和输出：

d = dict()
k = 0
with open('foo.txt') as infile:
    for line in map(str.strip, infile):
        if len(line) > 0:
            match line:
                case 'student':
                    td = dict()
                    d[k] = {line: td}
                    k += 1
                case 'school':
                    td[line] = dict()
                    td = td[line]
                case _:
                    k_, *v = line.split(':')
                    if v:
                        td[k_] = v[0].strip()

print(d)

输出：

{0: {'student': {'first name': 'John', 'last name': 'Doe', 'grade': '9', 'gpa': '4.0', 'school': {'name': 'Richard High School', 'city': 'Kansas City'}}}, 1: {'student': {'first name': 'Jane', 'last name': 'Doe', 'grade': '10', 'gpa': '3.0', 'school': {'name': 'Richard High School', 'city': 'Kansas City'}}}}