我有一个这样的数据框：
[['1','3086703274'], ['1','3086703274'], ['1','3086703274',]]
我想将end-start> 1 year（请参阅end = 2023 and start = 2020的最后一行）的行拆分，将A列的值保持相同，同时按比例拆分B列中的值：
Start date end date A B
01.01.2020 30.06.2020 2 3
01.01.2020 31.12.2020 3 1
01.04.2020 30.04.2020 6 2
01.01.2021 31.12.2021 2 3
01.07.2020 31.12.2020 8 2
01.01.2020 31.12.2023 1 2
.......
有什么主意吗？
答案 0 :(得分：1)
这是我的解决方案。请参阅以下评论：
import io
# TEST DATA:
text=""" start end A B
01.01.2020 30.06.2020 2 3
01.01.2020 31.12.2020 3 1
01.04.2020 30.04.2020 6 2
01.01.2021 31.12.2021 2 3
01.07.2020 31.12.2020 8 2
31.12.2020 20.01.2021 12 12
31.12.2020 01.01.2021 22 22
30.12.2020 01.01.2021 32 32
10.05.2020 28.09.2023 44 44
27.11.2020 31.12.2023 88 88
31.12.2020 31.12.2023 100 100
01.01.2020 31.12.2021 200 200
"""
df= pd.read_csv(io.StringIO(text), sep=r"\s+", engine="python", parse_dates=[0,1])
#print("\n----\n df:",df)
#----------------------------------------
# SOLUTION:
def split_years(r):
"""
Split row 'r' where "end"-"start" greater than 0.
The new rows have repeated values of 'A', and 'B' divided by the number of years.
Return: a DataFrame with rows per year.
"""
t1,t2 = r["start"], r["end"]
ys= t2.year - t1.year
kk= 0 if t1.is_year_end else 1
if ys>0:
l1=[t1] + [ t1+pd.offsets.YearBegin(i) for i in range(1,ys+1) ]
l2=[ t1+pd.offsets.YearEnd(i) for i in range(kk,ys+kk) ] + [t2]
return pd.DataFrame({"start":l1, "end":l2, "A":r.A,"B": r.B/len(l1)})
print("year difference <= 0!")
return None
# Create two groups, one for rows where the 'start' and 'end' is in the same year, and one for the others:
grps= df.groupby(lambda idx: (df.loc[idx,"start"].year-df.loc[idx,"end"].year)!=0 ).groups
print("\n---- grps:\n",grps)
# Extract the "one year" rows in a data frame:
df1= df.loc[grps[False]]
#print("\n---- df1:\n",df1)
# Extract the rows to be splitted:
df2= df.loc[grps[True]]
print("\n---- df2:\n",df2)
# Split the rows and put the resulting data frames into a list:
ldfs=[ split_years(df2.loc[row]) for row in df2.index ]
print("\n---- ldfs:")
for fr in ldfs:
print(fr,"\n")
# Insert the "one year" data frame to the list, and concatenate them:
ldfs.insert(0,df1)
df_rslt= pd.concat(ldfs,sort=False)
#print("\n---- df_rslt:\n",df_rslt)
# Housekeeping:
df_rslt= df_rslt.sort_values("start").reset_index(drop=True)
print("\n---- df_rslt:\n",df_rslt)
输出：
---- grps:
{False: Int64Index([0, 1, 2, 3, 4], dtype='int64'), True: Int64Index([5, 6, 7, 8, 9, 10, 11], dtype='int64')}
---- df2:
start end A B
5 2020-12-31 2021-01-20 12 12
6 2020-12-31 2021-01-01 22 22
7 2020-12-30 2021-01-01 32 32
8 2020-10-05 2023-09-28 44 44
9 2020-11-27 2023-12-31 88 88
10 2020-12-31 2023-12-31 100 100
11 2020-01-01 2021-12-31 200 200
---- ldfs:
start end A B
0 2020-12-31 2020-12-31 12 6.0
1 2021-01-01 2021-01-20 12 6.0
start end A B
0 2020-12-31 2020-12-31 22 11.0
1 2021-01-01 2021-01-01 22 11.0
start end A B
0 2020-12-30 2020-12-31 32 16.0
1 2021-01-01 2021-01-01 32 16.0
start end A B
0 2020-10-05 2020-12-31 44 11.0
1 2021-01-01 2021-12-31 44 11.0
2 2022-01-01 2022-12-31 44 11.0
3 2023-01-01 2023-09-28 44 11.0
start end A B
0 2020-11-27 2020-12-31 88 22.0
1 2021-01-01 2021-12-31 88 22.0
2 2022-01-01 2022-12-31 88 22.0
3 2023-01-01 2023-12-31 88 22.0
start end A B
0 2020-12-31 2020-12-31 100 25.0
1 2021-01-01 2021-12-31 100 25.0
2 2022-01-01 2022-12-31 100 25.0
3 2023-01-01 2023-12-31 100 25.0
start end A B
0 2020-01-01 2020-12-31 200 100.0
1 2021-01-01 2021-12-31 200 100.0
---- df_rslt:
start end A B
0 2020-01-01 2020-06-30 2 3.0
1 2020-01-01 2020-12-31 3 1.0
2 2020-01-01 2020-12-31 200 100.0
3 2020-01-04 2020-04-30 6 2.0
4 2020-01-07 2020-12-31 8 2.0
5 2020-10-05 2020-12-31 44 11.0
6 2020-11-27 2020-12-31 88 22.0
7 2020-12-30 2020-12-31 32 16.0
8 2020-12-31 2020-12-31 12 6.0
9 2020-12-31 2020-12-31 100 25.0
10 2020-12-31 2020-12-31 22 11.0
11 2021-01-01 2021-12-31 100 25.0
12 2021-01-01 2021-12-31 88 22.0
13 2021-01-01 2021-12-31 44 11.0
14 2021-01-01 2021-01-01 32 16.0
15 2021-01-01 2021-01-01 22 11.0
16 2021-01-01 2021-01-20 12 6.0
17 2021-01-01 2021-12-31 2 3.0
18 2021-01-01 2021-12-31 200 100.0
19 2022-01-01 2022-12-31 88 22.0
20 2022-01-01 2022-12-31 100 25.0
21 2022-01-01 2022-12-31 44 11.0
22 2023-01-01 2023-09-28 44 11.0
23 2023-01-01 2023-12-31 88 22.0
24 2023-01-01 2023-12-31 100 25.0
答案 1 :(得分：0)
另一种方法，添加新列而不是新行。但是我认为这完成了您想要做的事情。
df["years_apart"] = (
(df["end_date"] - df["start_date"]).dt.days / 365
).astype(int)
for years in range(1, df["years_apart"].max().astype(int)):
df[f"{years}_end_date"] = pd.NaT
df.loc[
df["years_apart"] == years, f"{years}_end_date"
] = df.loc[
df["years_apart"] == years, "start_date"
] + dt.timedelta(days=365*years)
df["B_bis"] = df["B"] / df["years_apart"]
输出
start_date end_date years_apart 1_end_date 2_end_date ...
2018-01-01 2018-01-02 0 NaT NaT
2018-01-02 2019-01-02 1 2019-01-02 NaT
2018-01-03 2020-01-03 2 NaT 2020-01-03
答案 2 :(得分：0)
我解决了它造成的日期差和一个计数器，该计数器为重复的行增加了几年：
#calculate difference between start and end year
table['diff'] = (table['end'] - table['start'])//timedelta(days=365)
table['diff'] = table['diff']+1
#replicate rows depending on number of years
table = table.reindex(table.index.repeat(table['diff']))
#counter that increase for diff>1, assign increasing years to the replicated rows
table['count'] = table['diff'].groupby(table['diff']).cumsum()//table['diff']
table['start'] = np.where(table['diff']>1, table['start']+table['count']-1, table['start'])
table['end'] = table['start']
#split B among years
table['B'] = table['B']//table['diff']