blocks|key|4023253|text|你已经很接近了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4023254|holidays+=+[dt.date(2018,+10,+11)]
days+=+[dt.date(2017,+1+,+1)+%2B+dt.timedelta(k)+for+k+in+range(365*3)]
def+lastt_bus_day(date,+format='%25Y-%25m-%25d'):
++++if+not+isinstance(date,+dt.date):
++++++++date+=+dt.datetime.strptime(date,+format).date()
++++return+max(
++++++++[d+for+d+in+days+if+d.weekday()+not+in+[5,+6]
++++++++++++++++++++++++++++and+d+not+in+holidays
++++++++++++++++++++++++++++and+d+<=+date
++++++++]
++++)|code-block|syntax|javascript|4023255|然后将其应用于整个数据帧：|4023256|df['business_day']+=+df['day']
df['business_day'].loc[df['category']+==+'B']+=+df.loc[df['category']+==+'B',+'day'].apply(lastt_bus_day)|4023257|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You're pretty close already:

<pre><code>holidays = [dt.date(2018, 10, 11)]
days = [dt.date(2017, 1 , 1) + dt.timedelta(k) for k in range(365*3)]
def lastt_bus_day(date, format='%Y-%m-%d'):
 if not isinstance(date, dt.date):
 date = dt.datetime.strptime(date, format).date()
 return max(
 [d for d in days if d.weekday() not in [5, 6]
 and d not in holidays
 and d &lt;= date
 ]
 )
</code></pre>

Then just apply this across the dataframe:

<pre><code>df['business_day'] = df['day']
df['business_day'].loc[df['category'] == 'B'] = df.loc[df['category'] == 'B', 'day'].apply(lastt_bus_day)
</code></pre>

blocks|key|4023286|text|通过使用pandas+BDay|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4023287|df.day.update(df.loc[(df.category=='B')&((df.day.dt.weekday.isin([5,6])%7C(df.day.isin(holidays+)))),'day']-pd.tseries.offsets.BDay(1))
df
Out[22]:+
++category++++++++day
0++++++++A+2018-10-10
1++++++++B+2018-10-10
2++++++++C+2018-10-12
3++++++++B+2018-10-12
4++++++++C+2018-10-14
5++++++++A+2018-10-15|code-block|syntax|javascript|4023288|entityMap^0|4|6|B|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

By using <code>pandas</code> <code>BDay</code> 

<pre><code>df.day.update(df.loc[(df.category=='B')&amp;((df.day.dt.weekday.isin([5,6])|(df.day.isin(holidays )))),'day']-pd.tseries.offsets.BDay(1))
df
Out[22]: 
 category day
0 A 2018-10-10
1 B 2018-10-10
2 C 2018-10-12
3 B 2018-10-12
4 C 2018-10-14
5 A 2018-10-15
</code></pre>

blocks|key|4023359|text|您可以在子集where+category+==+'B'上对所有非假日工作日使用pd.merge_asof，并为所有其他类别分配日期。设置allow_exact_matches=False以确保您不会与B的同一天匹配。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4023360|import+pandas+as+pd

mask+=+df.category+==+'B'

#+DataFrame+of+all+non-holiday+days
df_days+=+pd.DataFrame(days,+columns=['day'])
df_days+=+df_days.loc[(df_days.day.dt.weekday<5)+&+~df_days.day.isin(holidays)]

dfb+=+pd.merge_asof(
++++++++df.loc[mask],+
++++++++df_days.assign(new_day=df_days.day),+
++++++++on='day',+
++++++++direction='backward',
++++++++allow_exact_matches=False)

dfnb+=+df.assign(new_day+=+df.day)[~mask]

pd.concat([dfnb,+dfb],+ignore_index=True).sort_values('day')|code-block|syntax|javascript|4023361|输出：|4023362|+++++++++day+category++++new_day
0+2018-10-10++++++++A+2018-10-10
4+2018-10-11++++++++B+2018-10-10
1+2018-10-12++++++++C+2018-10-12
5+2018-10-13++++++++B+2018-10-12
2+2018-10-14++++++++C+2018-10-14
3+2018-10-15++++++++A+2018-10-15|4023363|entityMap^0|C|F|13|D|1W|P|2S|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

You can use <code>pd.merge_asof</code> on the subset where <code>category == 'B'</code> with all non-holiday business days and assign the date for all other categories. Set <code>allow_exact_matches=False</code> to ensure you don't match with the same day for <code>B</code>.

<pre><code>import pandas as pd

mask = df.category == 'B'

# DataFrame of all non-holiday days
df_days = pd.DataFrame(days, columns=['day'])
df_days = df_days.loc[(df_days.day.dt.weekday&lt;5) &amp; ~df_days.day.isin(holidays)]

dfb = pd.merge_asof(
 df.loc[mask], 
 df_days.assign(new_day=df_days.day), 
 on='day', 
 direction='backward',
 allow_exact_matches=False)

dfnb = df.assign(new_day = df.day)[~mask]

pd.concat([dfnb, dfb], ignore_index=True).sort_values('day')
</code></pre>

<h3>Output:</h3>

<pre><code> day category new_day
0 2018-10-10 A 2018-10-10
4 2018-10-11 B 2018-10-10
1 2018-10-12 C 2018-10-12
5 2018-10-13 B 2018-10-12
2 2018-10-14 C 2018-10-14
3 2018-10-15 A 2018-10-15
</code></pre>

I have a DataFrame with two columns, a date and a category. I want to create a new date column according to the rule: if category is <code>B</code> then the value should the business day closest to the date (only from the past or the day itself), else it's the value of the date column itself.

I define business day as any day which isn't on a weekend, nor present in the list <code>holidays</code> defined in the minimal example below. 

Please consider the following DataFrame <code>df</code>:

<pre><code>import datetime as dt
import pandas as pd
from IPython.display import display

holidays = [dt.datetime(2018, 10, 11)]
df = pd.DataFrame({"day": ["2018-10-10", "2018-10-11", "2018-10-12",
 "2018-10-13", "2018-10-14", "2018-10-15"
 ],
 "category":["A", "B", "C", "B", "C", "A"]
 }
)

df["day"] = pd.to_datetime(df.day, format="%Y-%m-%d")
display(df)

 day category
0 2018-10-10 A
1 2018-10-11 B
2 2018-10-12 C
3 2018-10-13 B
4 2018-10-14 C
5 2018-10-15 A
</code></pre>

How do I get a third column whose values are the ones listed below?

<pre><code>2018-10-10
2018-10-10
2018-10-12
2018-10-12
2018-10-14
2018-10-15
</code></pre>

I have a created a function that finds the last business day when working with lists, if that's any help.

<pre><code># creates a list whose elements are all days in the years 2017, 2018 and 2019
days = [dt.datetime(2017, 1 , 1) + dt.timedelta(k) for k in range(365*3)]


def lastt_bus_day(date):
 return max(
 [d for d in days if d.weekday() not in [5, 6]
 and d not in holidays
 and d &lt;= date
 ]
 )

for d in df.day:
 print(last_bus_day(d))
#prints
2018-10-10 00:00:00
2018-10-10 00:00:00
2018-10-12 00:00:00
2018-10-12 00:00:00
2018-10-12 00:00:00
2018-10-15 00:00:00
</code></pre>

Get previous business day in a DataFrame

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个包含两列的DataFrame，一个日期和一个类别。我想根据以下规则创建一个新的日期列:如果category是B，则值应该是最接近日期的工作日(仅来自过去或日期本身)，否则是日期列本身的值。我将工作日定义为不在周末的任何一天，也不存在于下面最小示例中定义的列表holidays中。请考虑以下DataFrame df...

问在DataFrame中获取前一个工作日
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