entityMap|blocks|key|21cep|text|结合pandas+melt和pyjanitor的complete可以帮助转换数据：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|cvdrk|+#+pip+install+pyjanitor
+import+pandas+as+pd
+import+janitor
(df.melt(['email',+'room'],+value_name+=+'Start_Date')
.reindex([3,1,2,0])
+#+complete+is+a+wrapper+around+pandas+functions
+#+to+expose+missing+values+...+in+this+case+it+exposes+the+
+#+missing+dates+for+each+group+in+by
.complete([{'Start_Date':lambda+df:+pd.date_range(df.min(),+df.max(),freq='H')}],+
+++++++++++by=['email',+'room'])
.assign(End_Date+=+lambda+df:+df.Start_Date.add(pd.Timedelta('1+hour')))
.query('variable+!=+"End+Date"').drop(columns='variable'))

++++++++++email+++room++++++++++Start_Date++++++++++++End_Date
0++abc@corp.com++L11M2+2021-02-01+08:00:00+2021-02-01+09:00:00
1++abc@corp.com++L11M2+2021-02-01+09:00:00+2021-02-01+10:00:00
2++abc@corp.com++L11M2+2021-02-01+10:00:00+2021-02-01+11:00:00
4++xyz@corp.com++L12M4+2021-02-01+08:00:00+2021-02-01+09:00:00
5++xyz@corp.com++L12M4+2021-02-01+09:00:00+2021-02-01+10:00:00|code-block|syntax|javascript|ap13h^0|9|4|O|8|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@$A|N|B|O|C|D]|$A|P|B|Q|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|R|9|@]|E|@]|F|$J|K]]|$2|L|4|-4|6|7|8|S|9|@]|E|@]|F|$]]]]

A combination of pandas <code>melt</code>, with pyjanitor's <code>complete</code> could help transform the data:
<pre><code> # pip install pyjanitor
 import pandas as pd
 import janitor
(df.melt(['email', 'room'], value_name = 'Start_Date')
.reindex([3,1,2,0])
 # complete is a wrapper around pandas functions
 # to expose missing values ... in this case it exposes the 
 # missing dates for each group in by
.complete([{'Start_Date':lambda df: pd.date_range(df.min(), df.max(),freq='H')}], 
 by=['email', 'room'])
.assign(End_Date = lambda df: df.Start_Date.add(pd.Timedelta('1 hour')))
.query('variable != &quot;End Date&quot;').drop(columns='variable'))

 email room Start_Date End_Date
0 abc@corp.com L11M2 2021-02-01 08:00:00 2021-02-01 09:00:00
1 abc@corp.com L11M2 2021-02-01 09:00:00 2021-02-01 10:00:00
2 abc@corp.com L11M2 2021-02-01 10:00:00 2021-02-01 11:00:00
4 xyz@corp.com L12M4 2021-02-01 08:00:00 2021-02-01 09:00:00
5 xyz@corp.com L12M4 2021-02-01 09:00:00 2021-02-01 10:00:00
</code></pre>

entityMap|blocks|key|5vubi|text|让我们创建一些示例数据|type|unstyled|depth|inlineStyleRanges|entityRanges|data|d07gr|from+datetime+import+datetime,+timedelta

ref+=+now.replace(minute=0,+second=0,+microsecond=0)
def+shifted(i):+return+ref+%2B+timedelta(hour=i)

df+=+pd.DataFrame([
++++('A',+'B',+shifted(1),+shifted(10)),+
++++('C',+'D',+shifted(-5),+shifted(-1))],+
++++columns=['name',+'email',+'start',+'end'])|code-block|syntax|javascript|f4r9u|数据如下所示|bu2vb|++name+email+++++++++++++++start+++++++++++++++++end
0++++A+++++B+2021-08-27+12:00:00+2021-08-27+21:00:00
1++++C+++++D+2021-08-27+06:00:00+2021-08-27+05:00:00|1dcam|您可以使用apply函数拆分每一行，确保返回一个pd.Series。|offset|length|style|CODE|dgj5c|new_start+=+df.apply(lambda+row:+pd.Series(pd.date_range(row.start,+row.end,+freq='H')),+axis=`).stack()|363bf|在此之后，new_start是每小时的开始，有一个双索引，一个是原始索引，一个是特定块的顺序，也可能是有用的。|69ki1|0++0+++2021-08-27+12:00:00
+++1+++2021-08-27+13:00:00
+++2+++2021-08-27+14:00:00
+++3+++2021-08-27+15:00:00
+++4+++2021-08-27+16:00:00
+++5+++2021-08-27+17:00:00
+++6+++2021-08-27+18:00:00
+++7+++2021-08-27+19:00:00
+++8+++2021-08-27+20:00:00
+++9+++2021-08-27+21:00:00
1++0+++2021-08-27+06:00:00
+++1+++2021-08-27+07:00:00
+++2+++2021-08-27+08:00:00
+++3+++2021-08-27+09:00:00
+++4+++2021-08-27+10:00:00
dtype:+datetime64[ns]|2eccc|现在，只需将此连接到原始框架。|7lgh2|res+=+df[["name",+"email"]].join(
new_start.reset_index(1,+drop=True).rename("start"))|bplog|您可以像下面这样添加end列|1j459|res["end"]+=+res.start+%2B+timedelta(hours=1)|53r33^0|0|0|0|0|5|5|O|9|0|0|5|9|0|0|0|0|A|3|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|16|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|17|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|18|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|19|9|@]|A|@]|B|$F|G]]|$2|L|4|M|6|7|8|1A|9|@$N|1B|O|1C|P|Q]|$N|1D|O|1E|P|Q]]|A|@]|B|$]]|$2|R|4|S|6|E|8|1F|9|@]|A|@]|B|$F|G]]|$2|T|4|U|6|7|8|1G|9|@$N|1H|O|1I|P|Q]]|A|@]|B|$]]|$2|V|4|W|6|E|8|1J|9|@]|A|@]|B|$F|G]]|$2|X|4|Y|6|7|8|1K|9|@]|A|@]|B|$]]|$2|Z|4|10|6|E|8|1L|9|@]|A|@]|B|$F|G]]|$2|11|4|12|6|7|8|1M|9|@$N|1N|O|1O|P|Q]]|A|@]|B|$]]|$2|13|4|14|6|E|8|1P|9|@]|A|@]|B|$F|G]]|$2|15|4|-4|6|7|8|1Q|9|@]|A|@]|B|$]]]]

Let's create some sample data
<pre><code>from datetime import datetime, timedelta

ref = now.replace(minute=0, second=0, microsecond=0)
def shifted(i): return ref + timedelta(hour=i)

df = pd.DataFrame([
 ('A', 'B', shifted(1), shifted(10)), 
 ('C', 'D', shifted(-5), shifted(-1))], 
 columns=['name', 'email', 'start', 'end'])
</code></pre>
The data looks like this
<pre><code> name email start end
0 A B 2021-08-27 12:00:00 2021-08-27 21:00:00
1 C D 2021-08-27 06:00:00 2021-08-27 05:00:00
</code></pre>
You can split up each row with the <code>apply</code> function, making sure you return a <code>pd.Series</code>.
<pre><code>new_start = df.apply(lambda row: pd.Series(pd.date_range(row.start, row.end, freq='H')), axis=`).stack()
</code></pre>
After this, <code>new_start</code> is the start of every hour, with a double index, one is the original index and one is the order of that particular block, could be useful as well.
<pre><code>0 0 2021-08-27 12:00:00
 1 2021-08-27 13:00:00
 2 2021-08-27 14:00:00
 3 2021-08-27 15:00:00
 4 2021-08-27 16:00:00
 5 2021-08-27 17:00:00
 6 2021-08-27 18:00:00
 7 2021-08-27 19:00:00
 8 2021-08-27 20:00:00
 9 2021-08-27 21:00:00
1 0 2021-08-27 06:00:00
 1 2021-08-27 07:00:00
 2 2021-08-27 08:00:00
 3 2021-08-27 09:00:00
 4 2021-08-27 10:00:00
dtype: datetime64[ns]
</code></pre>
Now just join this to the original frame.
<pre><code>res = df[[&quot;name&quot;, &quot;email&quot;]].join(
new_start.reset_index(1, drop=True).rename(&quot;start&quot;))
</code></pre>
And you can add the <code>end</code> column like this
<pre><code>res[&quot;end&quot;] = res.start + timedelta(hours=1)
</code></pre>

entityMap|blocks|key|eiab2|text|下面是一个使用pandas.date_range和explode的简单解决方案|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|g08c|df['Start+Date']+=+df.apply(lambda+d:+pd.date_range(d['Start+Date'],
++++++++++++++++++++++++++++++++++++++++++++++++++++d['End+Date'],+
++++++++++++++++++++++++++++++++++++++++++++++++++++freq='h')[:-1],+
++++++++++++++++++++++++++++axis=1)
df+=+df.explode('Start+Date')
df['End+Date']+=+df['Start+Date']+%2B+pd.Timedelta('1h')|code-block|syntax|javascript|8m51k|输出：|fb9s2|++++++++++email+++room++++++++++Start+Date++++++++++++End+Date
0++abc@corp.com++L11M2+2021-02-01+08:00:00+2021-02-01+09:00:00
0++abc@corp.com++L11M2+2021-02-01+09:00:00+2021-02-01+10:00:00
0++abc@corp.com++L11M2+2021-02-01+10:00:00+2021-02-01+11:00:00
1++xyz@corp.com++L12M4+2021-02-01+08:00:00+2021-02-01+09:00:00
1++xyz@corp.com++L12M4+2021-02-01+09:00:00+2021-02-01+10:00:00|hvli^0|7|H|P|7|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|Q|9|@$A|R|B|S|C|D]|$A|T|B|U|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|V|9|@]|E|@]|F|$J|K]]|$2|L|4|M|6|7|8|W|9|@]|E|@]|F|$]]|$2|N|4|O|6|I|8|X|9|@]|E|@]|F|$J|K]]|$2|P|4|-4|6|7|8|Y|9|@]|E|@]|F|$]]]]

Here is a simple solution using <code>pandas.date_range</code> and <code>explode</code>:
<pre><code>df['Start Date'] = df.apply(lambda d: pd.date_range(d['Start Date'],
 d['End Date'], 
 freq='h')[:-1], 
 axis=1)
df = df.explode('Start Date')
df['End Date'] = df['Start Date'] + pd.Timedelta('1h')
</code></pre>
output:
<pre><code> email room Start Date End Date
0 abc@corp.com L11M2 2021-02-01 08:00:00 2021-02-01 09:00:00
0 abc@corp.com L11M2 2021-02-01 09:00:00 2021-02-01 10:00:00
0 abc@corp.com L11M2 2021-02-01 10:00:00 2021-02-01 11:00:00
1 xyz@corp.com L12M4 2021-02-01 08:00:00 2021-02-01 09:00:00
1 xyz@corp.com L12M4 2021-02-01 09:00:00 2021-02-01 10:00:00
</code></pre>

I have a dataset that shows who is booking which room at which timing and it looks like this.
<pre><code>email room Start Date End Date
abc@corp.com L11M2 2021-02-01 08:00:00 2021-02-01 11:00:00
xyz@corp.com L12M4 2021-02-01 08:00:00 2021-02-01 10:00:00
</code></pre>
I want to split this up into different hours such that one row only contains one hour of data.
This is the dataframe that I want to get.
<pre><code>email room Start Date End Date
abc@corp.com L11M2 2021-02-01 08:00:00 2021-02-01 09:00:00
abc@corp.com L11M2 2021-02-01 09:00:00 2021-02-01 10:00:00
abc@corp.com L11M2 2021-02-01 10:00:00 2021-02-01 11:00:00
xyz@corp.com L12M4 2021-02-01 08:00:00 2021-02-01 09:00:00
xyz@corp.com L12M4 2021-02-01 09:00:00 2021-02-01 10:00:00
</code></pre>
Is there any way that I can do this in python?

Split 1 row into multiple rows of hourly data based on datetime column

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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 我有一个数据集，它显示了谁在哪个时间预订了哪个房间，它看起来像这样。 email         room   Start Date           End Dateabc@corp.com  L11M2  2021-02-01 08:00:00  2021-02-01 11:00:00xyz@corp.com ...

问根据datetime列将一行拆分为多行每小时数据
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问根据datetime列将一行拆分为多行每小时数据EN