3 回答
![?](http://img1.sycdn.imooc.com/533e4bd900011a1d02000200-100-100.jpg)
TA贡献1876条经验 获得超5个赞
使用是pd.Timestamp因为我们的字典键类型pd.Timestamp不是str.
import pandas as pd
d[pd.Timestamp('2020-06-30 00:00:00')]["intangibleAssets"]
295
![?](http://img1.sycdn.imooc.com/54dc06a60001ef0401000100-100-100.jpg)
TA贡献1835条经验 获得超7个赞
下面的函数是通用的,可以应用于任意数量的字典。
获取钥匙:
def matchingKeys(dictionary, searchString): return [key for key,val in dictionary if searchString in val]
要获取值:
def matchingValues(dictionary, searchString): return [val for key,val in dictionary if searchString in val]
要同时获得:
def matchingElements(dictionary, searchString): return {key:val for key,val in dictionary if searchString in val}
![?](http://img1.sycdn.imooc.com/5333a207000118af02200220-100-100.jpg)
TA贡献2021条经验 获得超8个赞
由于您使用的是 pandas,因此您可以转换为数据框并使用loc或.iloc传递您感兴趣的行和列:
与.loc:
d = {'2020-06-30 00:00:00': {'intangibleAssets': 295, 'capitalSurplus': 415},'2020-03-30 00:00:00': {'intangibleAssets': 400, 'capitalSurplus': 50}}
df = pd.DataFrame(d).loc['intangibleAssets','2020-06-30 00:00:00']
df
295
或iloc:
d = {'2020-06-30 00:00:00': {'intangibleAssets': 295, 'capitalSurplus': 415},'2020-03-30 00:00:00': {'intangibleAssets': 400, 'capitalSurplus': 50}}
df = pd.DataFrame(d).iloc[0,0]
df
295
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