数据库设计中有时会使用带主键的日期表。
| date_id | Date | Record_timestamp | Day | Week | Month | Quarter | Year_half | Year |
|---------+----------------+---------------------+-----------+-------+--------+-------------+-------------+----------|
| 0 | 2000-01-01 | NaN | Saturday | 52 | 1 | 1 | 1 | 2000 |
| 1 | 2000-01-02 | NaN | Sunday | 52 | 1 | 1 | 1 | 2000 |
| 2 | 2000-01-03 | NaN | Monday | 1 | 1 | 1 | 1 | 2000 |
如何在熊猫中做到这一点?
答案 0 :(得分:7)
使用dt
accessor:
In [11]: def create_date_table2(start='2000-01-01', end='2050-12-31'):
...: df = pd.DataFrame({"Date": pd.date_range(start, end)})
...: df["Day"] = df.Date.dt.weekday_name
...: df["Week"] = df.Date.dt.weekofyear
...: df["Quarter"] = df.Date.dt.quarter
...: df["Year"] = df.Date.dt.year
...: df["Year_half"] = (df.Quarter + 1) // 2
...: return df
In [12]: create_date_table2().head()
Out[12]:
Date Day Week Quarter Year Year_half
0 2000-01-01 Saturday 52 1 2000 1
1 2000-01-02 Sunday 52 1 2000 1
2 2000-01-03 Monday 1 1 2000 1
3 2000-01-04 Tuesday 1 1 2000 1
4 2000-01-05 Wednesday 1 1 2000 1
In [13]: create_date_table2().tail()
Out[13]:
Date Day Week Quarter Year Year_half
18623 2050-12-27 Tuesday 52 4 2050 2
18624 2050-12-28 Wednesday 52 4 2050 2
18625 2050-12-29 Thursday 52 4 2050 2
18626 2050-12-30 Friday 52 4 2050 2
18627 2050-12-31 Saturday 52 4 2050 2
注意:您可能想要动态计算这些而不是将它们存储为列!
答案 1 :(得分:0)
使用此功能
def create_date_table(start='2000-01-01', end='2050-12-31'):
start_ts = pd.to_datetime(start).date()
end_ts = pd.to_datetime(end).date()
# record timetsamp is empty for now
dates = pd.DataFrame(columns=['Record_timestamp'],
index=pd.date_range(start_ts, end_ts))
dates.index.name = 'Date'
days_names = {
i: name
for i, name
in enumerate(['Monday', 'Tuesday', 'Wednesday',
'Thursday', 'Friday', 'Saturday',
'Sunday'])
}
dates['Day'] = dates.index.dayofweek.map(days_names.get)
dates['Week'] = dates.index.week
dates['Month'] = dates.index.month
dates['Quarter'] = dates.index.quarter
dates['Year_half'] = dates.index.month.map(lambda mth: 1 if mth <7 else 2)
dates['Year'] = dates.index.year
dates.reset_index(inplace=True)
dates.index.name = 'date_id'
return dates
答案 2 :(得分:0)
我喜欢Andy和Robin的方法并根据我的需要稍微修改他们的create_date_table
,以防你有兴趣确定date_id
。我觉得这很有帮助,所以在其他未来的ETL过程中,给定一个日期,不需要担心额外的查找步骤。
def create_date_table3(start='1990-01-01', end='2080-12-31'):
df = pd.DataFrame({"date": pd.date_range(start, end)})
df["week_day"] = df.date.dt.weekday_name
df["day"] = df.date.dt.day
df["month"] = df.date.dt.day
df["week"] = df.date.dt.weekofyear
df["quarter"] = df.date.dt.quarter
df["year"] = df.date.dt.year
df.insert(0, 'date_id', (df.year.astype(str) + df.month.astype(str).str.zfill(2) + df.day.astype(str).str.zfill(2)).astype(int))
return df