如何在Python中将缺失的时间序列数据行添加到Pandas数据框中

Question

我有一个时间序列的产品数据集，如下所示：

date    product price   amount
11/17/2019  A   10  20
11/19/2019  A   15  20
11/24/2019  A   20  30
12/01/2019  C   40  50
12/05/2019  C   45  35

每个产品的起始日期和结束日期之间缺少了一些日期（“MM/dd/YYYY”）。我正在尝试使用零行填充缺失的日期，并将上一个表格转换为下面给出的表格：

date    product price   amount
11/17/2019  A   10  20
11/18/2019  A   0   0
11/19/2019  A   15  20
11/20/2019  A   0   0
11/21/2019  A   0   0
11/22/2019  A   0   0
11/23/2019  A   0   0
11/24/2019  A   20  30
12/01/2019  C   40  50
12/02/2019  C   0   0
12/03/2019  C   0   0
12/04/2019  C   0   0
12/05/2019  C   45  35

为了进行此转换，我使用了以下代码：

import pandas as pd
import numpy as np
data=pd.read_csv("test.txt", sep="\t", parse_dates=['date'])
data=data.set_index(["date", "product"])
start=data.first_valid_index()[0]
end=data.last_valid_index()[0]
df=data.set_index("date").reindex(pd.date_range(start,end, freq="1D"), fill_values=0)

然而，这段代码会报错。有没有什么方法可以高效地进行这种转换？

- user3104352

Answer 1

如果需要为每个产品单独添加缺失日期时间的0，请使用GroupBy.apply中的自定义函数，并使用DataFrame.reindex按最小和最大日期时间重新索引。

df = pd.read_csv("test.txt", sep="\t", parse_dates=['date'])

f = lambda x: x.reindex(pd.date_range(x.index.min(), 
                                      x.index.max(), name='date'), fill_value=0)
df = (df.set_index('date')
        .groupby('product')
        .apply(f)
        .drop('product', axis=1)
        .reset_index())
print (df)
   product       date  price  amount
0        A 2019-11-17     10      20
1        A 2019-11-18      0       0
2        A 2019-11-19     15      20
3        A 2019-11-20      0       0
4        A 2019-11-21      0       0
5        A 2019-11-22      0       0
6        A 2019-11-23      0       0
7        A 2019-11-24     20      30
8        C 2019-12-01     40      50
9        C 2019-12-02      0       0
10       C 2019-12-03      0       0
11       C 2019-12-04      0       0
12       C 2019-12-05     45      35

- jezrael

Answer 2

这种情况有一个更简单的方法：

#create the full date range, and then create a DataFrame with the range
#if needed, you can expand the range a bit using datetime.timedelta()
alldates=pd.DataFrame(pd.date_range(data.index.min()-timedelta(1),data.index.max()+timedelta(4), freq="1D",name="newdate"))

#make 'newdate' the index, and you no longer need it as a column
alldates.index=alldates.newdate
alldates.drop(columns="newdate", inplace=True)

#now, join the tables, missing dates in the original table will be filled with NaN
data=alldates.join(data)

- Rafael

Answer 3

一个选择是使用pyjanitor中的complete函数来显示每个组中缺失的行：

#pip install git+https://github.com/pyjanitor-devs/pyjanitor.git
import pandas as pd
import janitor

# build the dates to be applied per group
dates = dict(date = lambda df: pd.date_range(df.min(), df.max(), freq='1D'))

df.complete(dates, by='product', sort = True).fillna(0, downcast='infer')

                   date product  price  amount
0   2019-11-17 00:00:00       A     10      20
1   2019-11-18 00:00:00       A      0       0
2   2019-11-19 00:00:00       A     15      20
3   2019-11-20 00:00:00       A      0       0
4   2019-11-21 00:00:00       A      0       0
5   2019-11-22 00:00:00       A      0       0
6   2019-11-23 00:00:00       A      0       0
7   2019-11-24 00:00:00       A     20      30
8   2019-12-01 00:00:00       C     40      50
9   2019-12-02 00:00:00       C      0       0
10  2019-12-03 00:00:00       C      0       0
11  2019-12-04 00:00:00       C      0       0
12  2019-12-05 00:00:00       C     45      35

- sammywemmy