Question

我有df数据。我需要能够根据星期几返回某些数据行。

如果是星期一，我需要打印包含过去3天数据的行。如果是星期三或星期五，我需要打印包含过去两天数据的行。

合并两个df以创建一个：

<sequence name="seTestChangeNamespace" trace="disable" xmlns="http://ws.apache.org/ns/synapse">
(...)
<enrich>
<source type="body"/>
<target type="property" property="INPUT_MESSAGE"/>
</enrich>
<enrich>
<source type="inline">
<myns:Envelope xmlns:myns="http://schemas.xmlsoap.org/soap/envelope/">
<myns:Body/>
</myns:Envelope>
</source>
<target type="envelope"/>
</enrich>
<enrich>
<source type="property" property="INPUT_MESSAGE"/>
<target type="body"/>
</enrich>
(...)
</sequence>

根据今天的日期获取正确的日期：

df_new = pd.concat([outcomes_df, specialists_df], ignore_index=True)

df_new['Published Date'] = pd.to_datetime(df_new['Published Date'])

按日期范围过滤

N=0
if datetime.today().weekday() == 0:
    N = 3
elif datetime.today().weekday() == 2 or datetime.today().weekday() == 4:
    N = 2
else:
    pass

mydate = datetime.now() - timedelta(days=N)
print(mydate)

df_new = df_new[(df_new['Published Date'] >= mydate) & (df_new['Published Date'] <= datetime.today())]得出正确的日期

print(mydate)导致错误：

空数据框列：[col1，col2，col3，发布日期] 索引：[]

Answer 1

这里不需要if，因为这也需要循环，在这里我正在使用np.select

import numpy as np
s=df_new['Published Date'].dt.weekday
d=np.select([s==0,s.isin([2,4])],[3,2],0)


df_new['mydate']=(pd.to_datetime('today')-pd.to_timedelta(d,unit='D'))

df_new = df_new[(df_new['Published Date'] >= df_new['mydate']) & (df_new['Published Date'] <= datetime.today())]

如何按特定日期过滤df？

1 个答案: