Python: How can I drop the first 5 minutes of each day in my time serie?
3
I have a dataframe with columns: Date of a transaction , Time of the transaction and Price. I want to drop the first and last 5 minutes in each day.
Here is an example:
----------------------------------------
Date | Time | Price
----------------------------------------
03/03/2014 | 09:36:36.814 | 43.90
---------------------------------------
03/03/2014 | 09:37:02.381 | 43.40
---------------------------------------
03/03/2014 | 09:41:02.381 | 43.40
---------------------------------------
03/03/2014 | 09:50:02.381 | 43.40
---------------------------------------
I want to get this output:
----------------------------------------
Date | Time | Price
---------------------------------------
03/03/2014 | 09:50:02.381 | 43.40
---------------------------------------
I need to do this for each day of time serie.
I tried this code:
trades14081.insert(2,'DateTime',pd.to_datetime(trades14081['Date']+trades14081['Time'], format = "%d/%m/%Y%H:%M:%S.%f" ))
delta=datetime.timedelta(minutes=5)
i=0
j=0
start=
end=
while trades14081['Date'][i]==trades14081['Date'][j] and j<len(trades14081):
if trades14081['DateTime'][j]-trades14081['DateTime'][i]<delta:
j=j+1
else:
start.append(i)
end.append(j)
j=j+1
while trades14081['Date'][i]==trades14081['Date'][j] and j<len(trades14081):
j=j+1
i=j
for i in range(len(start)):
trades14081=trades14081.drop(trades14081.index[start[i]:end[i]])
But I keep on getting this error:
KeyError: 19996
> 12 l.append(j)
> 13 j=j+1
> ---> 14 while trades14081['Date'][i]==trades14081['Date'][j]:
> 15 j=j+1
> 16 i=j
19996 is the length of my dataframe trades14081.
Any ideas?
python python-3.x pandas pandas-groupby timedelta
edited Nov 11 at 12:05
jpp
90.4k2052101
asked Nov 11 at 10:34
Nada Baili
193
add a comment |
3
I have a dataframe with columns: Date of a transaction , Time of the transaction and Price. I want to drop the first and last 5 minutes in each day.
Here is an example:
----------------------------------------
Date | Time | Price
----------------------------------------
03/03/2014 | 09:36:36.814 | 43.90
---------------------------------------
03/03/2014 | 09:37:02.381 | 43.40
---------------------------------------
03/03/2014 | 09:41:02.381 | 43.40
---------------------------------------
03/03/2014 | 09:50:02.381 | 43.40
---------------------------------------
I want to get this output:
----------------------------------------
Date | Time | Price
---------------------------------------
03/03/2014 | 09:50:02.381 | 43.40
---------------------------------------
I need to do this for each day of time serie.
I tried this code:
trades14081.insert(2,'DateTime',pd.to_datetime(trades14081['Date']+trades14081['Time'], format = "%d/%m/%Y%H:%M:%S.%f" ))
delta=datetime.timedelta(minutes=5)
i=0
j=0
start=
end=
while trades14081['Date'][i]==trades14081['Date'][j] and j<len(trades14081):
if trades14081['DateTime'][j]-trades14081['DateTime'][i]<delta:
j=j+1
else:
start.append(i)
end.append(j)
j=j+1
while trades14081['Date'][i]==trades14081['Date'][j] and j<len(trades14081):
j=j+1
i=j
for i in range(len(start)):
trades14081=trades14081.drop(trades14081.index[start[i]:end[i]])
But I keep on getting this error:
KeyError: 19996
> 12 l.append(j)
> 13 j=j+1
> ---> 14 while trades14081['Date'][i]==trades14081['Date'][j]:
> 15 j=j+1
> 16 i=j
19996 is the length of my dataframe trades14081.
Any ideas?
python python-3.x pandas pandas-groupby timedelta
edited Nov 11 at 12:05
jpp
90.4k2052101
asked Nov 11 at 10:34
Nada Baili
193
Can you share sample input and expected output? That'll make it clearer for us.
– Mayank Porwal
Nov 11 at 11:00
This problem domain looks like it's more suited to something like Observables. RxJS/Marble diagrams rxmarbles.com - egghead.io and angular university have good course around this if you were coming at it from a JS perspective. Perhaps github.com/ReactiveX/RxPY would work - have never used it..
– JGFMK
Nov 11 at 11:02
add a comment |
3
3
3
I have a dataframe with columns: Date of a transaction , Time of the transaction and Price. I want to drop the first and last 5 minutes in each day.
Here is an example:
----------------------------------------
Date | Time | Price
----------------------------------------
03/03/2014 | 09:36:36.814 | 43.90
---------------------------------------
03/03/2014 | 09:37:02.381 | 43.40
---------------------------------------
03/03/2014 | 09:41:02.381 | 43.40
---------------------------------------
03/03/2014 | 09:50:02.381 | 43.40
---------------------------------------
I want to get this output:
----------------------------------------
Date | Time | Price
---------------------------------------
03/03/2014 | 09:50:02.381 | 43.40
---------------------------------------
I need to do this for each day of time serie.
I tried this code:
trades14081.insert(2,'DateTime',pd.to_datetime(trades14081['Date']+trades14081['Time'], format = "%d/%m/%Y%H:%M:%S.%f" ))
delta=datetime.timedelta(minutes=5)
i=0
j=0
start=
end=
while trades14081['Date'][i]==trades14081['Date'][j] and j<len(trades14081):
if trades14081['DateTime'][j]-trades14081['DateTime'][i]<delta:
j=j+1
else:
start.append(i)
end.append(j)
j=j+1
while trades14081['Date'][i]==trades14081['Date'][j] and j<len(trades14081):
j=j+1
i=j
for i in range(len(start)):
trades14081=trades14081.drop(trades14081.index[start[i]:end[i]])
But I keep on getting this error:
KeyError: 19996
> 12 l.append(j)
> 13 j=j+1
> ---> 14 while trades14081['Date'][i]==trades14081['Date'][j]:
> 15 j=j+1
> 16 i=j
19996 is the length of my dataframe trades14081.
Any ideas?
python python-3.x pandas pandas-groupby timedelta
edited Nov 11 at 12:05
jpp
90.4k2052101
asked Nov 11 at 10:34
Nada Baili
193
I have a dataframe with columns: Date of a transaction , Time of the transaction and Price. I want to drop the first and last 5 minutes in each day.
Here is an example:
----------------------------------------
Date | Time | Price
----------------------------------------
03/03/2014 | 09:36:36.814 | 43.90
---------------------------------------
03/03/2014 | 09:37:02.381 | 43.40
---------------------------------------
03/03/2014 | 09:41:02.381 | 43.40
---------------------------------------
03/03/2014 | 09:50:02.381 | 43.40
---------------------------------------
I want to get this output:
----------------------------------------
Date | Time | Price
---------------------------------------
03/03/2014 | 09:50:02.381 | 43.40
---------------------------------------
I need to do this for each day of time serie.
I tried this code:
trades14081.insert(2,'DateTime',pd.to_datetime(trades14081['Date']+trades14081['Time'], format = "%d/%m/%Y%H:%M:%S.%f" ))
delta=datetime.timedelta(minutes=5)
i=0
j=0
start=
end=
while trades14081['Date'][i]==trades14081['Date'][j] and j<len(trades14081):
if trades14081['DateTime'][j]-trades14081['DateTime'][i]<delta:
j=j+1
else:
start.append(i)
end.append(j)
j=j+1
while trades14081['Date'][i]==trades14081['Date'][j] and j<len(trades14081):
j=j+1
i=j
for i in range(len(start)):
trades14081=trades14081.drop(trades14081.index[start[i]:end[i]])
But I keep on getting this error:
KeyError: 19996
> 12 l.append(j)
> 13 j=j+1
> ---> 14 while trades14081['Date'][i]==trades14081['Date'][j]:
> 15 j=j+1
> 16 i=j
19996 is the length of my dataframe trades14081.
Any ideas?
python python-3.x pandas pandas-groupby timedelta
python python-3.x pandas pandas-groupby timedelta
edited Nov 11 at 12:05
jpp
90.4k2052101
asked Nov 11 at 10:34
Nada Baili
193
edited Nov 11 at 12:05
jpp
90.4k2052101
asked Nov 11 at 10:34
Nada Baili
193
edited Nov 11 at 12:05
jpp
90.4k2052101
edited Nov 11 at 12:05
jpp
90.4k2052101
edited Nov 11 at 12:05
jpp
90.4k2052101
90.4k2052101
asked Nov 11 at 10:34
Nada Baili
193
asked Nov 11 at 10:34
Nada Baili
193
asked Nov 11 at 10:34
Nada Baili
193
193
Can you share sample input and expected output? That'll make it clearer for us.
– Mayank Porwal
Nov 11 at 11:00
This problem domain looks like it's more suited to something like Observables. RxJS/Marble diagrams rxmarbles.com - egghead.io and angular university have good course around this if you were coming at it from a JS perspective. Perhaps github.com/ReactiveX/RxPY would work - have never used it..
– JGFMK
Nov 11 at 11:02
add a comment |
Can you share sample input and expected output? That'll make it clearer for us.
– Mayank Porwal
Nov 11 at 11:00
This problem domain looks like it's more suited to something like Observables. RxJS/Marble diagrams rxmarbles.com - egghead.io and angular university have good course around this if you were coming at it from a JS perspective. Perhaps github.com/ReactiveX/RxPY would work - have never used it..
– JGFMK
Nov 11 at 11:02
Can you share sample input and expected output? That'll make it clearer for us.
– Mayank Porwal
Nov 11 at 11:00
Can you share sample input and expected output? That'll make it clearer for us.
– Mayank Porwal
Nov 11 at 11:00
This problem domain looks like it's more suited to something like Observables. RxJS/Marble diagrams rxmarbles.com - egghead.io and angular university have good course around this if you were coming at it from a JS perspective. Perhaps github.com/ReactiveX/RxPY would work - have never used it..
– JGFMK
Nov 11 at 11:02
This problem domain looks like it's more suited to something like Observables. RxJS/Marble diagrams rxmarbles.com - egghead.io and angular university have good course around this if you were coming at it from a JS perspective. Perhaps github.com/ReactiveX/RxPY would work - have never used it..
– JGFMK
Nov 11 at 11:02
add a comment |
1 Answer
1
active
oldest
votes
2
groupby + Boolean indexing
You can and should avoid Python-level loops. Here you can use groupby:
# convert strings to timedelta
df['Time'] = pd.to_timedelta(df['Time'])
# define offset from start to omit
offset = pd.Timedelta(minutes=5)
# apply Boolean filter to dataframe
res = df.loc[df['Time'] > df.groupby('Date')['Time'].transform('min') + offset]
print(res)
Date Time Price
4 03/03/2014 09:40:00 41
5 03/03/2014 09:46:00 42
answered Nov 11 at 12:02
jpp
90.4k2052101
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
groupby + Boolean indexing
You can and should avoid Python-level loops. Here you can use groupby:
# convert strings to timedelta
df['Time'] = pd.to_timedelta(df['Time'])
# define offset from start to omit
offset = pd.Timedelta(minutes=5)
# apply Boolean filter to dataframe
res = df.loc[df['Time'] > df.groupby('Date')['Time'].transform('min') + offset]
print(res)
Date Time Price
4 03/03/2014 09:40:00 41
5 03/03/2014 09:46:00 42
answered Nov 11 at 12:02
jpp
90.4k2052101
add a comment |
2
groupby + Boolean indexing
You can and should avoid Python-level loops. Here you can use groupby:
# convert strings to timedelta
df['Time'] = pd.to_timedelta(df['Time'])
# define offset from start to omit
offset = pd.Timedelta(minutes=5)
# apply Boolean filter to dataframe
res = df.loc[df['Time'] > df.groupby('Date')['Time'].transform('min') + offset]
print(res)
Date Time Price
4 03/03/2014 09:40:00 41
5 03/03/2014 09:46:00 42
answered Nov 11 at 12:02
jpp
90.4k2052101
add a comment |
2
2
2
groupby + Boolean indexing
You can and should avoid Python-level loops. Here you can use groupby:
# convert strings to timedelta
df['Time'] = pd.to_timedelta(df['Time'])
# define offset from start to omit
offset = pd.Timedelta(minutes=5)
# apply Boolean filter to dataframe
res = df.loc[df['Time'] > df.groupby('Date')['Time'].transform('min') + offset]
print(res)
Date Time Price
4 03/03/2014 09:40:00 41
5 03/03/2014 09:46:00 42
answered Nov 11 at 12:02
jpp
90.4k2052101
groupby + Boolean indexing
You can and should avoid Python-level loops. Here you can use groupby:
# convert strings to timedelta
df['Time'] = pd.to_timedelta(df['Time'])
# define offset from start to omit
offset = pd.Timedelta(minutes=5)
# apply Boolean filter to dataframe
res = df.loc[df['Time'] > df.groupby('Date')['Time'].transform('min') + offset]
print(res)
Date Time Price
4 03/03/2014 09:40:00 41
5 03/03/2014 09:46:00 42
answered Nov 11 at 12:02
jpp
90.4k2052101
answered Nov 11 at 12:02
jpp
90.4k2052101
answered Nov 11 at 12:02
jpp
90.4k2052101
answered Nov 11 at 12:02
jpp
90.4k2052101
90.4k2052101
add a comment |
add a comment |
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Can you share sample input and expected output? That'll make it clearer for us.
– Mayank Porwal
Nov 11 at 11:00
This problem domain looks like it's more suited to something like Observables. RxJS/Marble diagrams rxmarbles.com - egghead.io and angular university have good course around this if you were coming at it from a JS perspective. Perhaps github.com/ReactiveX/RxPY would work - have never used it..
– JGFMK
Nov 11 at 11:02