Pandas中如何根据数据范围在差异表中进行平均。

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英文:

How to average based on data range in a difference table in Pandas

问题

使用两个表,Valuesdates,我想要计算日期范围内的平均值。
Values 表如下所示:

日期和时间
2023-01-01 10:00 1
2023-01-01 11:00 2
2023-01-02 10:00 4
2023-01-04 10:00 4
2023-01-07 10:00 4

dates 表如下所示:

分组 起始日期 结束日期
1 2023-01-01 2023-01-05
2 2023-01-03 2023-01-10

正如您所见,日期范围可以重叠。

我试图计算这些范围内的平均值,所以在这个示例中,输出应该是类似以下的内容:

分组 起始日期 结束日期 平均值
1 2023-01-01 2023-01-05 2.75
2 2023-01-03 2023-01-10 4

目前我的代码如下(一行代码):

Values.groupby(np.where(Values['Date'].between(Dates['StartDay'], Dates['EndDay']), 'pre', 'post'))['Value'].mean()

然而,这会导致以下错误:

ValueError: Can only compare identically-labeled Series objects

这是基于其他类似的问题编写的,但似乎不适用于此情况,因为它涉及两个表和日期范围。

英文:

With two tables, Values and dates, I would like to get the average value between the date ranges.
Values looks like:

Date Value
2023-01-01 10:00 1
2023-01-01 11:00 2
2023-01-02 10:00 4
2023-01-04 10:00 4
2023-01-07 10:00 4

and dates looks like

Group StartDay EndDay
1 2023-01-01 2023-01-05
2 2023-01-03 2023-01-10

As you can see, the date ranges can overlap.

I am trying to calculate the averages over these ranges, so in this example the output should be something along the lines of

Group StartDay EndDay Mean
1 2023-01-01 2023-01-05 2.75
2 2023-01-03 2023-01-10 4

Currently my code looks like (all one line):

Values.groupby(np.where(Values['Date'].between(Dates['StartDay'],Dates['EndDay']),'pre','post'))['value'].mean()

however this results in
ValueError: Can only compare identically-labeled Series objects

This was based on other similar questions, however does not appear to apply here due to it being over two tables / using ranges.

答案1

得分: 3

尝试:

# 将值转换为日期时间(如果尚未转换):
df1['Date'] = pd.to_datetime(df1['Date'])
df2['StartDay'] = pd.to_datetime(df2['StartDay'])
df2['EndDay'] = pd.to_datetime(df2['EndDay'])

df1 = df1.sort_values(by='Date').set_index('Date')

df2['Mean'] = df2.apply(lambda x: df1[x['StartDay']:x['EndDay']].mean(), axis=1)
print(df2)

打印结果:

   Group   StartDay     EndDay  Mean
0      1 2023-01-01 2023-01-05  2.75
1      2 2023-01-03 2023-01-10  4.00
英文:

Try:

# convert the values to datetime (if not already):
df1['Date'] = pd.to_datetime(df1['Date'])
df2['StartDay'] = pd.to_datetime(df2['StartDay'])
df2['EndDay'] = pd.to_datetime(df2['EndDay'])

df1 = df1.sort_values(by='Date').set_index('Date')

df2['Mean'] = df2.apply(lambda x: df1[x['StartDay']:x['EndDay']].mean(), axis=1)
print(df2)

Prints:

   Group   StartDay     EndDay  Mean
0      1 2023-01-01 2023-01-05  2.75
1      2 2023-01-03 2023-01-10  4.00

答案2

得分: 0

这似乎是某种形式的不等式连接 - 如果是这种情况,您可以使用conditional_join来从pyjanitor获取结果,然后进行分组 - 应该比apply快,具体取决于数据大小:

# pip install pyjanitor
import pandas as pd
import janitor

df1 = pd.read_clipboard(sep="\t", engine='python', parse_dates=['Date'])
df2 = pd.read_clipboard(sep="\t", engine='python', parse_dates=['StartDay', 'EndDay'])

(df1
.conditional_join(
    df2, 
    # column from left, column from right, comparator
    ('Date', 'StartDay', '>='),  
    ('Date', 'EndDay', '<='), 
    # depending on the data size,
    # you might get more performance with numba
    use_numba = False,
    df_columns = 'Value')
.groupby(df2.columns.tolist())
.Value
.mean()
)

Group  StartDay    EndDay    
1      2023-01-01  2023-01-05    2.75
2      2023-01-03  2023-01-10    4.00
Name: Value, dtype: float64
英文:

Looks like some form of inequality join - if that is the case, you can use conditional_join from pyjanitor to get your results, before grouping - should be faster than apply, depending on the data size:

# pip install pyjanitor
import pandas as pd
import janitor

df1 = pd.read_clipboard(sep=&quot;\t&quot;, engine=&#39;python&#39;, parse_dates = [&#39;Date&#39;])
df2 = pd.read_clipboard(sep=&quot;\t&quot;, engine=&#39;python&#39;, parse_dates = [&#39;StartDay&#39;, &#39;EndDay&#39;])

(df1
.conditional_join(
    df2, 
    # column from left, column from right, comparator
    (&#39;Date&#39;, &#39;StartDay&#39;, &#39;&gt;=&#39;),  
    (&#39;Date&#39;, &#39;EndDay&#39;, &#39;&lt;=&#39;), 
    # depending on the data size,
    # you might get more performance with numba
    use_numba = False,
    df_columns = &#39;Value&#39;)
.groupby(df2.columns.tolist())
.Value
.mean()
)

Group  StartDay    EndDay    
1      2023-01-01  2023-01-05    2.75
2      2023-01-03  2023-01-10    4.00
Name: Value, dtype: float64

huangapple
  • 本文由 发表于 2023年2月23日 22:59:02
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