Question

我有两个数据帧，test1和test2。对于ID中的每个test2值，我想检查date中的test2并将其与{{ 1}}。如果ID中的test1中的任何一个在date中的日期范围内，则对test2列求和，并将该总和分配为{{1}中的其他列}。

输出：

因此，新的test1 df将具有一列amount，该列是test1中所有金额的总和，其中test1在{{1 }}-为此amount_sum

test2

Answer 1

使用：

#outer join both df by ID columns
df = test1.merge(test2, on='ID', how='outer')
#filter by range
df = df[(df.date > df.date1) & (df.date < df.date2)]
#thank you @Abhi for alternative
#df = df[df.date.between(df.date1, df.date2, inclusive=False)]
#aggregate sum
s = df.groupby(['ID','date1','date2'])['amount'].sum()
#add new column to test1
test = test1.join(s, on=['ID','date1','date2'])

示例：

#https://stackoverflow.com/q/21494489
np.random.seed(123)

#https://stackoverflow.com/a/50559321/2901002
def gen(start, end, n):
    start_u = start.value//10**9
    end_u = end.value//10**9

    return pd.to_datetime(np.random.randint(start_u, end_u, n), unit='s')

n = 10
test1 = pd.DataFrame({
    'ID':np.random.choice(list('abc'), n),
    'date1': gen(pd.to_datetime('2010-01-01'),pd.to_datetime('2010-03-01'), n).floor('d'),
    'date2':gen(pd.to_datetime('2010-03-01'),pd.to_datetime('2010-06-01'), n).floor('d')
})

m = 5
test2 = pd.DataFrame({
    'ID': np.random.choice(list('abc'), m),
    'amount':np.random.randint(10, size=m),
    'date':gen(pd.to_datetime('2010-01-01'), pd.to_datetime('2010-06-01'), m).floor('d')
})

print (test1)
  ID      date1      date2
0  c 2010-01-15 2010-05-22
1  b 2010-02-08 2010-04-16
2  c 2010-01-24 2010-04-12
3  c 2010-02-01 2010-04-09
4  a 2010-01-19 2010-05-20
5  c 2010-01-27 2010-05-24
6  c 2010-02-23 2010-03-15
7  b 2010-01-31 2010-05-09
8  c 2010-02-23 2010-03-29
9  b 2010-01-08 2010-03-07

print (test2)
  ID  amount       date
0  a       4 2010-05-15
1  b       6 2010-03-26
2  a       1 2010-01-07
3  b       5 2010-02-07
4  a       6 2010-04-13

#outer join both df by ID columns
df = test1.merge(test2, on='ID', how='outer')
#filter by range
df = df[(df.date > df.date1) & (df.date < df.date2)]
print (df)
   ID      date1      date2  amount       date
6   b 2010-02-08 2010-04-16     6.0 2010-03-26
8   b 2010-01-31 2010-05-09     6.0 2010-03-26
9   b 2010-01-31 2010-05-09     5.0 2010-02-07
11  b 2010-01-08 2010-03-07     5.0 2010-02-07
12  a 2010-01-19 2010-05-20     4.0 2010-05-15
14  a 2010-01-19 2010-05-20     6.0 2010-04-13

#thank you @Abhi for alternative
#df = df[df.date.between(df.date1, df.date2, inclusive=False)]
#aggregate sum
s = df.groupby(['ID','date1','date2'])['amount'].sum()
#add new column to test1
test = test1.join(s, on=['ID','date1','date2'])
print (test)
  ID      date1      date2  amount
0  c 2010-01-15 2010-05-22     NaN
1  b 2010-02-08 2010-04-16     6.0
2  c 2010-01-24 2010-04-12     NaN
3  c 2010-02-01 2010-04-09     NaN
4  a 2010-01-19 2010-05-20    10.0
5  c 2010-01-27 2010-05-24     NaN
6  c 2010-02-23 2010-03-15     NaN
7  b 2010-01-31 2010-05-09    11.0
8  c 2010-02-23 2010-03-29     NaN
9  b 2010-01-08 2010-03-07     5.0

熊猫将一个df中的日期与另一个时间范围中的时间进行匹配，然后按分组总和

1 个答案: