我使用to_dict()来转换数据帧,如下所示: it is very simple 我得到了
{'point_dis': {'025ec525073fea4f6433b691c65b45cd': 12365.093765441219,
'02b697dc2e9214a46aa40e7fb0609310': 2315.3360224791882}
如果我只想要一个没有{column - >的dict,我该怎么办? ,只是非嵌套的字典'
{'025ec525073fea4f6433b691c65b45cd': 12365.093765441219,
'02b697dc2e9214a46aa40e7fb0609310': 2315.3360224791882}
答案 0 :(得分:0)
首先选择列,然后使用Series.to_dict
:
d = df['point_dis'].to_dict()
print (d)
{'025ec525073fea4f6433b691c65b45cd': 12365.093765441219,
'02b697dc2e9214a46aa40e7fb0609310': 2315.3360224791882}
或使用DataFrame.to_dict
,然后按dict
选择嵌套key
:
d = df.to_dict()['point_dis']
print (d)
{'025ec525073fea4f6433b691c65b45cd': 12365.093765441219,
'02b697dc2e9214a46aa40e7fb0609310': 2315.3360224791882}