Python 将名称和数据的非结构化列表转换为嵌套字典
我有一个“非结构化”列表,如下所示:Python 将名称和数据的非结构化列表转换为嵌套字典,python,dictionary,nested,defaultdict,Python,Dictionary,Nested,Defaultdict,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' ] 非结构化的,因为列表由以下几组组成: (姓名、官员身份、人口统计信息)三胞胎,或 (姓名、人口统计信息)成对 在后一种情况下,Officer=False,在前一种情况
info = [
'Joe Schmoe',
'W / M / 64',
'Richard Johnson',
'OFFICER',
'W / M /48',
'Adrian Stevens',
'? / ? / 27'
]
非结构化的,因为列表由以下几组组成:
- (姓名、官员身份、人口统计信息)三胞胎,或
- (姓名、人口统计信息)成对李>
Officer=False
,在前一种情况下,Officer=True
。人口统计信息字符串表示种族/性别/年龄
,其中NaN
s由文字问号表示。以下是我想去的地方:
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实现这一点。第一个在下面,处理人口统计信息字符串。(我对这个没意见,只是放在这里作为参考。)
第二个函数解构列表并将其值抛出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,
# ... ...
第二个函数对于它所做的事情来说感觉太冗长和乏味了
有没有更好的方法可以更巧妙地记住迭代中最后一个值的分类?这种方法非常适合: 代码: 测试代码: 结果: 将元组三元组转换为dict: 结果:
这类事情非常适合: 代码: 测试代码: 结果: 将元组三元组转换为dict: 结果:
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}}
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}}
这是一种问题,你别无选择,只能卷起袖子,让自己因为糟糕的设计而写糟糕的代码:(这是一种问题,你别无选择,只能卷起袖子,让自己因为糟糕的设计而写糟糕的代码:(
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)
{
'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'}
}
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)}
{'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}}