Reputation: 9081
A similar question has been asked here before
But when I tried all the available solutions it was giving me error.
Code:
print sum(data['Activity_Duration'],datetime.timedelta())
#import operator
#print reduce(operator.add, data['Activity_Duration'])
Error:
OverflowError
1 #print sum(data['Activity_Duration'],datetime.timedelta())
2 import operator
----> 3 print reduce(operator.add, data['Activity_Duration'])OverflowError: long too big to convert
Am i missing something or can we come up with a more scalable solution?
Info: My data has 436746 rows.
I am working on 8 gig machine and data size is 650MB
Upvotes: 1
Views: 6823
Reputation: 862661
I think you need sum
:
print (df['Activity_Duration'].sum())
Sample:
import pandas as pd
start = pd.to_datetime('2015-02-24')
end = pd.to_datetime('2016-04-25')
rng = pd.date_range(start, end, freq='6D')
start = pd.to_datetime('2015-02-26')
end = pd.to_datetime('2016-04-27')
rng1 = pd.date_range(start, end, freq='6D')
df = pd.DataFrame({'Date1': rng, 'Date2': rng1})
df['Activity_Duration'] = df.Date2 - df.Date1
print (df)
Date1 Date2 Activity_Duration
0 2015-02-24 2015-02-26 2 days
1 2015-03-02 2015-03-04 2 days
2 2015-03-08 2015-03-10 2 days
3 2015-03-14 2015-03-16 2 days
4 2015-03-20 2015-03-22 2 days
5 2015-03-26 2015-03-28 2 days
6 2015-04-01 2015-04-03 2 days
7 2015-04-07 2015-04-09 2 days
8 2015-04-13 2015-04-15 2 days
9 2015-04-19 2015-04-21 2 days
...
...
print (df['Activity_Duration'].sum())
144 days 00:00:00
If need output in float
:
import numpy as np
df['Activity_Duration'] = (df.Date2 - df.Date1) / np.timedelta64(1, 'D')
print (df)
Date1 Date2 Activity_Duration
0 2015-02-24 2015-02-26 2.0
1 2015-03-02 2015-03-04 2.0
2 2015-03-08 2015-03-10 2.0
3 2015-03-14 2015-03-16 2.0
4 2015-03-20 2015-03-22 2.0
...
...
...
print (df['Activity_Duration'].sum())
144.0
Another solution is dt.days
- output is int
:
print (df['Activity_Duration'].dt.days.sum())
144
Upvotes: 4