Question

考虑嵌套字典：

 mapSel={'lassoPoints': {'mapbox': [[-9.51, 38.96],    [-9.28, 38.78], 
[-9.24, 38.78],    [-9.22, 38.70],    [-9.29, 38.68],    [-9.25,
38.70],    [-9.32, 38.69],    [-9.38, 38.60]]},  'points': [{'curveNumber': 0,   
'location': 'Cascais',    'pointIndex': 152,    'pointNumber': 152,   
'z': 187.769},   {'curveNumber': 0,    'location': 'Oeiras',   
'pointIndex': 158,    'pointNumber': 158,    'z': 186.113},  
{'curveNumber': 0,    'location': 'Sintra',    'pointIndex': 159,   
'pointNumber': 159,    'z': 221.223}]}

第一个键“套索点”并不重要。我只想考虑 Key="points" 有一个“位置”列表或一个数据框，如：

 location
0 Cascais
1 Oeiras
2 Sintra

我尝试使用本尼迪克特：

安装：pip install python-benedict

from benedict import benedict
mapSel= benedict(mapSel, keypath_separator='.')
val = mapSel.get('points.location')
val

什么都没有

Answer 1

为什么不只使用 pandas？

>>> import pandas as pd
>>> df = pd.DataFrame(mapSel["points"])
>>> df
   curveNumber location  pointIndex  pointNumber        z
0            0  Cascais         152          152  187.769
1            0   Oeiras         158          158  186.113
2            0   Sintra         159          159  221.223

如果您只需要 location，那么您只需访问该特定列即可。

>>> df["location"]
0    Cascais
1     Oeiras
2     Sintra
Name: location, dtype: object

Answer 2

这是你想要的吗？

mapSel={'lassoPoints': {'mapbox': [[-9.51, 38.96],    [-9.28, 38.78], 
[-9.24, 38.78],    [-9.22, 38.70],    [-9.29, 38.68],    [-9.25,
38.70],    [-9.32, 38.69],    [-9.38, 38.60]]},  
        'points': [{'curveNumber': 0,   
'location': 'Cascais',    'pointIndex': 152,    'pointNumber': 152,   
'z': 187.769},   {'curveNumber': 0,    'location': 'Oeiras',   
'pointIndex': 158,    'pointNumber': 158,    'z': 186.113},  
{'curveNumber': 0,    'location': 'Sintra',    'pointIndex': 159,   
'pointNumber': 159,    'z': 221.223}]}

print([elt['location'] for elt in mapSel['points']])
# >>> ['Cascais', 'Oeiras', 'Sintra']

如何从嵌套字典中获取键？

2 个答案: