从Python文本文件中的字典

Posted 2021-03-27

Problem:

我有一个这种格式的txt文件：

Intestinal infectious diseases (001-003)  
001 Cholera  
002 Fever  
003 Salmonella   
Zoonotic bacterial diseases (020-022)  
020 Plague  
021 Tularemia  
022 Anthrax  
External Cause Status (E000)  
E000 External cause status  
Activity (E001-E002)  
E001 Activities involving x and y  
E002 Other activities

其中以3整数代码/ E + 3整数代码/ V + 3整数代码开头的每一行都是前一个标题的值，这是我字典的键。在我看到的其他问题中，使用列或冒号可以用来解析每一行来创建一个键/值对，但是我的txt文件的格式不允许我这样做。

是一种将这样的txt文件制作成字典的方法，其中键是组名，值是代码+疾病名称？

我还需要将代码和疾病名称解析为第二个字典，因此我最终得到一个包含组名作为键的字典，其中值是第二个字典，其中代码为键，疾病名称为值。

Script:

def process_file(filename):
    myDict={}
        f = open(filename, 'r')
        for line in f:
            if line[0] is not int:
                if line.startswith("E"):
                    if line[1] is int:
                        line = dictionary1_values
                    else:
                        break
                else:
                    line = dictionary1_key
            myDict[dictionary1_key].append[line]

期望的输出格式是： {"Intestinal infectious diseases (001-003)": {"001": "Cholera", "002": "Fever", "003": "Salmonella"}, "Zoonotic bacterial diseases (020-022)": {"020": "Plague", "021": "Tularemia", "022": "Anthrax"}, "External Cause Status (E000)": {"E000": "External cause status"}, "Activity (E001-E002)": {"E001": "Activities involving x and y", "E002": "Other activities"}}

答案

def process_file(filename):
    myDict = {}
    rootkey = None
    f = open(filename, 'r')
    for line in f:
        if line[1:3].isdigit():           # if the second and third character from the checked string (line) is the ASCII Code in range 0x30..0x39 ("0".."9"), i.e.: str.isdigit()
            subkey, data = line.rstrip().split(" ",1)     # split into two parts... the first one is the number with or without "E" at begin
            myDict[rootkey][subkey] = data
        else:
            rootkey = line.rstrip()       # str.rstrip() is used to delete newlines (or another so called "empty spaces")
            myDict[rootkey] = {}          # prepare a new empty rootkey into your myDict
    f.close()
    return myDict

在Python控制台中测试：

>>> d = process_file('/tmp/file.txt')
>>>
>>> d['Intestinal infectious diseases (001-003)']
{'003': 'Salmonella', '002': 'Fever', '001': 'Cholera'}
>>> d['Intestinal infectious diseases (001-003)']['002']
'Fever'
>>> d['Activity (E001-E002)']
{'E001': 'Activities involving x and y', 'E002': 'Other activities'}
>>> d['Activity (E001-E002)']['E001']
'Activities involving x and y'
>>>
>>> d
{'Activity (E001-E002)': {'E001': 'Activities involving x and y', 'E002': 'Other activities'}, 'External Cause Status (E000)': {'E000': 'External cause status'}, 'Intestinal infectious diseases (001-003)': {'003': 'Salmonella', '002': 'Fever', '001': 'Cholera'}, 'Zoonotic bacterial diseases (020-022)': {'021': 'Tularemia', '020': 'Plague', '022': 'Anthrax'}}

警告：文件中的第一行必须只是一个“rootkey”！不是“子项”或数据！否则原因可能是加注错误:-)

注意：也许你应该删除第一个“E”字符。或者不能这样做？你需要在某个地方留下这个“E”角色吗？

另一答案

尝试使用正则表达式来确定它是标题还是疾病

import re
mydict = {}
with open(filename, "r") as f:
    header = None
    for line in f:
        match_desease = re.match(r"(E?ddd) (.*)", line)
        if not match_desease:
            header = line
        else:
            code = match_desease.group(1)
            desease = match_desease.group(2)
            mydict[header][code] = desease

另一答案

一种解决方案是使用正则表达式来帮助您表征和解析在此文件中可能遇到的两种类型的行：

import re
header_re = re.compile(r'([ws]+) (([ws-]+))')
entry_re = re.compile(r'([EV]?d{3}) (.+)')

这使您可以非常轻松地检查您遇到的线路类型，并根据需要将其分开：

# Check if a line is a header:
header = header_re.match(line)
if header:
    header_name, header_codes = header.groups()  # e.g. ('Intestinal infectious diseases', '001-009')
    # Do whatever you need to do when you encounter a new group
    # ...
else:
    entry = entry_re.match(line)
    # If the line wasn't a header, it ought to be an entry,
    # otherwise we've encountered something we didn't expect
    assert entry is not None
    entry_number, entry_name = entry.groups()  # e.g. ('001', 'Cholera')
    # Do whatever you need to do when you encounter an entry in a group
    # ...

使用它来重新运行您的函数，我们可以编写以下内容：

import re

def process_file(filename):
    header_re = re.compile(r'([ws]+) (([ws-]+))')
    entry_re = re.compile(r'([EV]?d{3}) (.+)')

    all_groups = {}
    current_group = None

    with open(filename, 'r') as f:
        for line in f:
            # Check if a line is a header:
            header = header_re.match(line)
            if header:
                current_group = {}
                all_groups[header.group(0)] = current_group
            else:
                entry = entry_re.match(line)
                # If the line wasn't a header, it ought to be an entry,
                # otherwise we've encountered something we didn't expect
                assert entry is not None
                entry_number, entry_name = entry.groups()  # e.g. ('001', 'Cholera')

                current_group[entry_number] = entry_name

    return all_groups