将名称和数据的非结构化列表转换为嵌套字典

Question

我有一个看起来像这样的“非结构化”列表：

info = [
    'Joe Schmoe',
    'W / M / 64',
    'Richard Johnson',
    'OFFICER',
    'W / M /48',
    'Adrian Stevens',
    '? / ? / 27'
    ]

非结构化，因为列表由以下集合组成：

• （姓名、军官身份、人口统计信息）三胞胎，或
• （姓名、人口统计信息）对。

在后一种情况下，Officer=False，在前一种情况下，Officer=True。人口统计信息字符串表示Race / Gender / Age，NaNs 由文字问号表示。这是我想去的地方：

res = {
    'Joe Schmoe': {
        'race': 'W',
        'gender': 'M',
        'age': 64,
        'officer': False
        },
    'Richard Johnson': {
        'race': 'W',
        'gender': 'M',
        'age': 48,
        'officer': True
        },
    'Adrian Stevens': {
        'race': 'NaN',
        'gender': 'NaN',
        'age': 27,
        'officer': False
        }
    }

现在我已经构建了两个函数来执行此操作。第一个在下面并处理 Demographic Info 字符串。 （这个我没问题；只是放在这里供参考。）

import re

def fix_demographic(info):
    # W / M / ?? --> W / M / NaN
    # ?/M/?  --> NaN / M / NaN
    # Keep as str NaN rather than np.nan for now
    race, gender, age = re.split('\s*/\s*', re.sub('\?+', 'NaN', info))
    return race, gender, age

第二个函数解构列表并将其值扔到字典结果中的不同位置：

demographic = re.compile(r'(\w+|\?+)\s*\/\s*(\w+|\?+)\s*\/\s*(\w+|\?+)')


def parse_victim_info(info: list):
    res = defaultdict(dict)
    for i in info:
        if not demographic.fullmatch(i) and i.lower() != 'officer':
            # We have a name
            previous = 'name'
            name = i
        if i.lower() == 'officer':
            res[name]['officer'] = True
            previous = 'officer'
        if demographic.fullmatch(i):
            # We have demographic info; did "OFFICER" come before it?
            if previous == 'name':
                res[name]['officer'] = False
            race, gender, age = fix_demographic(i)
            res[name]['race'] = race
            res[name]['gender'] = gender
            res[name]['age'] = int(age) if age.isnumeric() else age
            previous = None
    return res

>>> parse_victim_info(info)
defaultdict(dict,
            {'Adrian Stevens': {'age': 27,
              'gender': 'NaN',
              'officer': False,
              'race': 'NaN'},
             'Richard Johnson': {'age': 48,
              'gender': 'M',
              'officer': True,
              # ... ...

第二个函数感觉太冗长乏味了。

有没有更好的方法可以更智能地记住迭代中看到的最后一个值的分类？

Answer 1

这种东西非常适合generator：

代码：

def find_triplets(data):
    data = iter(data)
    while True:
        name = next(data)
        demo = next(data)
        officer = demo == 'OFFICER'
        if officer:
            demo = next(data)
        yield name, officer, demo

测试代码：

info = [
    'Joe Schmoe',
    'W / M / 64',
    'Lillian Schmoe',
    'W / F / 60',
    'Richard Johnson',
    'OFFICER',
    'W / M /48',
    'Adrian Stevens',
    '? / ? / 27'
]

for x in find_triplets(info):
    print(x)

结果：

('Joe Schmoe', False, 'W / M / 64')
('Lillian Schmoe', False, 'W / F / 60')
('Richard Johnson', True, 'W / M /48')
('Adrian Stevens', False, '? / ? / 27')

将元组三元组转换为dict：

import re

def fix_demographic(info):
    # W / M / ?? --> W / M / NaN
    # ?/M/?  --> NaN / M / NaN
    # Keep as str NaN rather than np.nan for now
    race, gender, age = re.split('\s*/\s*', re.sub('\?+', 'NaN', info))
    return dict(race=race, gender=gender, age=age)


data_dict = {name: dict(officer=officer, **fix_demographic(demo))
             for name, officer, demo in find_triplets(info)}

print(data_dict)

结果：

{
    'Joe Schmoe': {'officer': False, 'race': 'W', 'gender': 'M', 'age': '64'}, 
    'Lillian Schmoe': {'officer': False, 'race': 'W', 'gender': 'F', 'age': '60'}, 
    'Richard Johnson': {'officer': True, 'race': 'W', 'gender': 'M', 'age': '48'}, 
    'Adrian Stevens': {'officer': False, 'race': 'NaN', 'gender': 'NaN', 'age': '27'}
}

Answer 2

您可以在 Python3 中使用itertools.groupby：

import itertools
import re
info = [
'Joe Schmoe',
'W / M / 64',
'Lillian Schmoe',
'W / F / 60',
'Richard Johnson',
'OFFICER',
'W / M /48',
'Adrian Stevens',
'? / ? / 27'
]
data = [list(b) for a, b in itertools.groupby(info, key=lambda x:x.count('/') > 0 or x == 'OFFICER')]

final_data = {data[i][0]:{**{a:'NaN' if b == '?' else (int(b) if b.isdigit() else b) for a, b in zip(['race', 'gender', 'age'], filter(None, re.split('\s+|/', [h for h in data[i+1] if h.count('/') > 0][0])))}, **{"officer":"OFFICER" in data[i+1]}} for i in range(0, len(data), 2)} 

输出：

{'Joe Schmoe': {'race': 'W', 'gender': 'M', 'age': 64, 'officer': False}, 'Lillian Schmoe': {'race': 'W', 'gender': 'F', 'age': 60, 'officer': False}, 'Richard Johnson': {'race': 'W', 'gender': 'M', 'age': 48, 'officer': True}, 'Adrian Stevens': {'race': 'NaN', 'gender': 'NaN', 'age': 27, 'officer': False}}