代码之家 › 专栏 › 技术社区 › Brad Solomon

将非结构化名称和数据列表转换为嵌套字典

defaultdict nested dictionary python

Brad Solomon · 技术社区 · 8 年前

我有一个“非结构化”列表,如下所示:

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

非结构化 其中,列表由以下几组组成:

( 姓名、官员状态、人口统计信息 )三胞胎,或
( 姓名、人口统计信息

在后一种情况下, Officer=False 在前者中, Officer=True . 人口统计信息字符串表示 Race / Gender / Age 具有 NaN 用文字问号表示。以下是我想了解的内容:

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实现这一点。第一个在下面,处理人口信息字符串。(我对这个没意见,把它放在这里作为参考。)

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放入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了&对于它正在做的事情来说很乏味。

有没有更好的方法可以更智能地记住迭代中最后一个值的分类?

2 回复 | 直到 8 年前

Brad Solomon 8 年前

这种东西很适合 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'}
}

Brad Solomon 8 年前

您可以使用 itertools.groupby 在Python3中:

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}}

将非结构化名称和数据列表转换为嵌套字典

代码:

测试代码:

结果:

将元组三元组转换为 dict :

结果:

将元组三元组转换为 `dict` :