来自 Python 文本文件的字典

Question

问题：

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-integer code/E+3-integer code/V+3-integer code 开头的每一行都是前面标题的值，它们是我字典的键。在我见过的其他问题中，可以使用列或冒号来解析每一行以生成键/值对，但是我的 txt 文件的格式不允许我这样做。

有没有办法将这样的txt文件制作成字典，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到第二个字典中，所以我最终得到一个包含组名作为键的字典，值是第二个字典，代码作为键，疾病名称作为值。

脚本：

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]

所需的输出格式是：

{"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"}}

Answer 1

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是“rootkey”！不是“子密钥”或数据！否则原因可能是引发错误:-)注意：也许您应该删除第一个“E”字符。还是做不到？你需要把这个“E”字符留在某个地方吗？

Answer 2

一种解决方案是使用正则表达式来帮助您表征和解析您可能在此文件中遇到的两种类型的行：import reheader_re = re.compile(r'([\w\s]+) \(([\w\s\-]+)\)')entry_re = re.compile(r'([EV]?\d{3}) (.+)')这使您可以非常轻松地检查遇到的线路类型，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    # ...使用它来重新工作您的功能，我们可以编写以下内容：import redef process_file(filename):    header_re = re.compile(r'([\w\s]+) \(([\w\s\-]+)\)')    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

Answer 3

尝试使用正则表达式来确定它是标题还是疾病import remydict = {}with open(filename, "r") as f:    header = None    for line in f:        match_desease = re.match(r"(E?\d\d\d) (.*)", line)        if not match_desease:            header = line        else:            code = match_desease.group(1)            desease = match_desease.group(2)            mydict[header][code] = desease