我在熊貓中有以下數(shù)據(jù)框 code tank date time no_operation_flag 123 1 01-01-2019 00:00:00 1 123 1 01-01-2019 00:30:00 1 123 1 01-01-2019 01:00:00 0 123 1 01-01-2019 01:30:00 1 123 1 01-01-2019 02:00:00 1 123 1 01-01-2019 02:30:00 1 123 1 01-01-2019 03:00:00 1 123 1 01-01-2019 03:30:00 1 123 1 01-01-2019 04:00:00 1 123 1 01-01-2019 05:00:00 1 123 1 01-01-2019 14:00:00 1 123 1 01-01-2019 14:30:00 1 123 1 01-01-2019 15:00:00 1 123 1 01-01-2019 15:30:00 1 123 1 01-01-2019 16:00:00 1 123 1 01-01-2019 16:30:00 1 123 2 02-01-2019 00:00:00 1 123 2 02-01-2019 00:30:00 0 123 2 02-01-2019 01:00:00 0 123 2 02-01-2019 01:30:00 0 123 2 02-01-2019 02:00:00 1 123 2 02-01-2019 02:30:00 1 123 2 02-01-2019 03:00:00 1 123 2 03-01-2019 03:30:00 1 123 2 03-01-2019 04:00:00 1 123 1 03-01-2019 14:00:00 1 123 2 03-01-2019 15:00:00 1 123 2 03-01-2019 00:30:00 1 123 2 04-01-2019 11:00:00 1 123 2 04-01-2019 11:30:00 0 123 2 04-01-2019 12:00:00 1 123 2 04-01-2019 13:30:00 1 123 2 05-01-2019 03:00:00 1 123 2 05-01-2019 03:30:00 1 123 2 05-01-2019 04:00:00 1我想要做的是no_operation_flag在坦克級別和日級別標記連續(xù) 1 超過 5 次，但時間應(yīng)該是連續(xù)的（時間是半小時級別）。Dataframe 已經(jīng)在容器、日期和時間級別進行了排序。
3 回答
守著星空守著你
TA貢獻1799條經(jīng)驗 獲得超8個贊
我認為這是一種非常具有前瞻性且有些骯臟的方式，但很容易理解。
對于行循環(huán)，4 行后的檢查時間是 2 小時遠。
（如果 1 為真）檢查所有對應(yīng)的五個值
df['no_operation_flag']都是 1。（如果 2 為真）將 1 放入對應(yīng)的 5 個值中
df['final_flag']。
# make col with zero
df['final_flag'] = 0
for i in range(1, len(df)-4):
j = i + 4
dt1 = df['date'].iloc[i]+' '+df['time'].iloc[i]
ts1 = pd.to_datetime(dt1)
dt2 = df['date'].iloc[j]+' '+df['time'].iloc[j]
ts2 = pd.to_datetime(dt2)
# timedelta is 2 hours?
if ts2 - ts1 == datetime.timedelta(hours=2, minutes=0):
# all of no_operation_flag == 1?
if (df['no_operation_flag'].iloc[i:j+1] == 1).all():
df['final_flag'].iloc[i:j+1] = 1
慕尼黑5688855
TA貢獻1848條經(jīng)驗 獲得超2個贊
您可以使用這樣的解決方案，僅使用新助手過濾每個組的連續(xù)日期時間，DataFrame并添加所有缺少的日期merge時間，最后添加新列：
df['datetimes'] = pd.to_datetime(df['date'].astype(str) + ' ' + df['time'].astype(str))
df1 = (df.set_index('datetimes')
.groupby(['code','tank', 'date'])['no_operation_flag']
.resample('30T')
.first()
.reset_index())
shifted1 = df1.groupby(['code','tank', 'date'])['no_operation_flag'].shift()
g1 = df1['no_operation_flag'].ne(shifted1).cumsum()
mask1 = g1.map(g1.value_counts()).gt(5) & df1['no_operation_flag'].eq(1)
df1['final_flag'] = mask1.astype(int)
#print (df1.head(40))
df = df.merge(df1[['code','tank','datetimes','final_flag']]).drop('datetimes', axis=1)
print (df)
code tank date time no_operation_flag final_flag
0 123 1 01-01-2019 00:00:00 1 0
1 123 1 01-01-2019 00:30:00 1 0
2 123 1 01-01-2019 01:00:00 0 0
3 123 1 01-01-2019 01:30:00 1 1
4 123 1 01-01-2019 02:00:00 1 1
5 123 1 01-01-2019 02:30:00 1 1
6 123 1 01-01-2019 03:00:00 1 1
7 123 1 01-01-2019 03:30:00 1 1
8 123 1 01-01-2019 04:00:00 1 1
9 123 1 01-01-2019 05:00:00 1 0
10 123 1 01-01-2019 14:00:00 1 1
11 123 1 01-01-2019 14:30:00 1 1
12 123 1 01-01-2019 15:00:00 1 1
13 123 1 01-01-2019 15:30:00 1 1
14 123 1 01-01-2019 16:00:00 1 1
15 123 1 01-01-2019 16:30:00 1 1
16 123 2 02-01-2019 00:00:00 1 0
17 123 2 02-01-2019 00:30:00 0 0
18 123 2 02-01-2019 01:00:00 0 0
19 123 2 02-01-2019 01:30:00 0 0
20 123 2 02-01-2019 02:00:00 1 0
21 123 2 02-01-2019 02:30:00 1 0
22 123 2 02-01-2019 03:00:00 1 0
23 123 2 03-01-2019 03:30:00 1 0
24 123 2 03-01-2019 04:00:00 1 0
25 123 1 03-01-2019 14:00:00 1 0
26 123 2 03-01-2019 15:00:00 1 0
27 123 2 03-01-2019 00:30:00 1 0
28 123 2 04-01-2019 11:00:00 1 0
29 123 2 04-01-2019 11:30:00 0 0
30 123 2 04-01-2019 12:00:00 1 0
31 123 2 04-01-2019 13:30:00 1 0
32 123 2 05-01-2019 03:00:00 1 0
33 123 2 05-01-2019 03:30:00 1 0
34 123 2 05-01-2019 04:00:00 1 0
江戶川亂折騰
TA貢獻1851條經(jīng)驗 獲得超5個贊
利用：
df['final_flag'] = ( df.groupby([df['no_operation_flag'].ne(1).cumsum(),
'tank',
'date',
pd.to_datetime(df['time'].astype(str))
.diff()
.ne(pd.Timedelta(minutes = 30))
.cumsum(),
'no_operation_flag'])['no_operation_flag']
.transform('size')
.gt(5)
.view('uint8') )
print(df)
輸出
code tank date time no_operation_flag final_flag
0 123 1 01-01-2019 00:00:00 1 0
1 123 1 01-01-2019 00:30:00 1 0
2 123 1 01-01-2019 01:00:00 0 0
3 123 1 01-01-2019 01:30:00 1 1
4 123 1 01-01-2019 02:00:00 1 1
5 123 1 01-01-2019 02:30:00 1 1
6 123 1 01-01-2019 03:00:00 1 1
7 123 1 01-01-2019 03:30:00 1 1
8 123 1 01-01-2019 04:00:00 1 1
9 123 1 01-01-2019 05:00:00 1 0
10 123 1 01-01-2019 14:00:00 1 1
11 123 1 01-01-2019 14:30:00 1 1
12 123 1 01-01-2019 15:00:00 1 1
13 123 1 01-01-2019 15:30:00 1 1
14 123 1 01-01-2019 16:00:00 1 1
15 123 1 01-01-2019 16:30:00 1 1
16 123 2 02-01-2019 00:00:00 1 0
17 123 2 02-01-2019 00:30:00 0 0
18 123 2 02-01-2019 01:00:00 0 0
19 123 2 02-01-2019 01:30:00 0 0
20 123 2 02-01-2019 02:00:00 1 0
21 123 2 02-01-2019 02:30:00 1 0
22 123 2 02-01-2019 03:00:00 1 0
23 123 2 03-01-2019 03:30:00 1 0
24 123 2 03-01-2019 04:00:00 1 0
25 123 1 03-01-2019 14:00:00 1 0
26 123 2 03-01-2019 15:00:00 1 0
27 123 2 03-01-2019 00:30:00 1 0
28 123 2 04-01-2019 11:00:00 1 0
29 123 2 04-01-2019 11:30:00 0 0
30 123 2 04-01-2019 12:00:00 1 0
31 123 2 04-01-2019 13:30:00 1 0
32 123 2 05-01-2019 03:00:00 1 0
33 123 2 05-01-2019 03:30:00 1 0
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