我有一个数据框df可以用这个创建：
data={'id':[1,1,1,1,2,2,2,2],
'date1':[datetime.date(2016,1,1),datetime.date(2016,1,2),datetime.date(2016,1,3),datetime.date(2016,1,4),
datetime.date(2016,1,2),datetime.date(2016,1,4),datetime.date(2016,1,3),datetime.date(2016,1,1)],
'date2':[datetime.date(2016,1,5),datetime.date(2016,1,3),datetime.date(2016,1,5),datetime.date(2016,1,5),
datetime.date(2016,1,4),datetime.date(2016,1,5),datetime.date(2016,1,4),datetime.date(2016,1,1)],
'score1':[5,7,3,2,9,3,8,3],
'score2':[1,3,0,5,2,20,7,7]}
df=pd.DataFrame.from_dict(data)
And looks like this:
id date1 date2 score1 score2
0 1 2016-01-01 2016-01-05 5 1
1 1 2016-01-02 2016-01-03 7 3
2 1 2016-01-03 2016-01-05 3 0
3 1 2016-01-04 2016-01-05 2 5
4 2 2016-01-02 2016-01-04 9 2
5 2 2016-01-04 2016-01-05 3 20
6 2 2016-01-03 2016-01-04 8 7
7 2 2016-01-01 2016-01-01 3 7
我需要做的是为score1和score2中的每一个创建一个列，该列创建两列，分别对score1和score2的值进行求和关于usedate是否介于date1和date2之间。通过获取usedate最小值和date1最大值之间的所有日期来创建date2。我用它来创建日期范围：
drange=pd.date_range(df.date1.min(),df.date2.max())
结果数据框newdf应如下所示：
usedate score1sum score2sum
0 2016-01-01 8 8
1 2016-01-02 21 6
2 2016-01-03 32 13
3 2016-01-04 30 35
4 2016-01-05 13 26
为了澄清，在usedate 2016-01-01，score1sum是8，这是通过查看2016-01-01所在的df中的行来计算的date1和date2，它们对row0（5）和row8（3）求和。在usedate 2016-01-04，score2sum是35，这是通过查看df中2016-01-04所在行（包括date1和date2，其中对row0（1），row3（0），row4（5），row5（2），row6（20），row7（7）求和。
也许某种groupby或melt然后groupby？
答案 0 :(得分：2)
您可以将apply与lambda函数一起使用：
df['date1'] = pd.to_datetime(df['date1'])
df['date2'] = pd.to_datetime(df['date2'])
df1 = pd.DataFrame(index=pd.date_range(df.date1.min(), df.date2.max()), columns = ['score1sum', 'score2sum'])
df1[['score1sum','score2sum']] = df1.apply(lambda x: df.loc[(df.date1 <= x.name) &
(x.name <= df.date2),
['score1','score2']].sum(), axis=1)
df1.rename_axis('usedate').reset_index()
输出：
usedate score1sum score2sum
0 2016-01-01 8 8
1 2016-01-02 21 6
2 2016-01-03 32 13
3 2016-01-04 30 35
4 2016-01-05 13 26
答案 1 :(得分：1)
这是不优雅的，但是嘿，它有效！ （编辑：在下面添加了第二种方法。）
# Convert datetime.date to pandas timestamps for easier comparisons
df['date1'] = pd.to_datetime(df['date1'])
df['date2'] = pd.to_datetime(df['date2'])
# solution
newdf = pd.DataFrame(data=drange, columns=['usedate'])
# for each usedate ud, get all df rows whose dates contain ud,
# then sum the scores of these rows
newdf['score1sum'] = [df[(df['date1'] <= ud) & (df['date2'] >= ud)]['score1'].sum() for ud in drange]
newdf['score2sum'] = [df[(df['date1'] <= ud) & (df['date2'] >= ud)]['score2'].sum() for ud in drange]
# output
newdf
usedate score1sum score2sum
2016-01-01 8 8
2016-01-02 21 6
2016-01-03 32 13
2016-01-04 30 35
2016-01-05 13 26
transform（或apply）newdf = pd.DataFrame(data=drange, columns=['usedate'])
def sum_scores(d):
return df[(df['date1'] <= d) & (df['date2'] >= d)][['score1', 'score2']].sum()
# apply works here too, and is about equally fast in my testing
newdf[['score1sum', 'score2sum']] = newdf['usedate'].transform(sum_scores)
# newdf is same to above
# Jupyter timeit cell magic
%%timeit
newdf['score1sum'] = [df[(df['date1'] <= d) & (df['date2'] >= d)]['score1'].sum() for d in drange]
newdf['score1sum'] = [df[(df['date1'] <= d) & (df['date2'] >= d)]['score2'].sum() for d in drange]
100 loops, best of 3: 10.4 ms per loop
# Jupyter timeit line magic
%timeit newdf[['score1sum', 'score2sum']] = newdf['usedate'].transform(sum_scores)
100 loops, best of 3: 8.51 ms per loop