CODEWITHSUNDEEP
pythonpandasdata-science
Fragan
Fragan

Reputation: 892

Fill timestamp gaps in large dataset

I have a dataset with like 100K+ rows, one column on this dataset is a Datetime column, let's name it A.

My Dataset is sorted by column A.

I want to "Fill gaps" of my Dataset, i.e : if i have these two rows following each others :

0  2019-03-13 08:12:20
1  2019-03-13 08:12:25

I want to make add missing seconds between them, as a result, i'll have this :

0  2019-03-13 08:12:20
1  2019-03-13 08:12:21
2  2019-03-13 08:12:22
3  2019-03-13 08:12:23
4  2019-03-13 08:12:24
5  2019-03-13 08:12:25

I don't want to generate rows between two rows if they have different day, month or year.

So if have these two consecutive rows :

0  2019-03-13 08:12:20
1  2019-03-15 08:12:21

I won't add anything.

I can't also generate rows if the time difference between my two rows is greater than 2 hours.

So if have these two consecutive rows :

0  2019-03-13 08:12:20
1  2019-03-15 11:12:21

I won't add anything.

Here's an example to illustrate what i want :

df=pd.DataFrame({'A': ["2019-03-13 08:12:20", "2019-03-13 08:12:25", "2019-03-20 08:17:23", "2019-03-22 08:17:25", "2019-03-22 11:12:20", "2019-03-22 11:12:23", "2019-03-24 12:33:23"]})
                     A
0  2019-03-13 08:12:20
1  2019-03-13 08:12:25
2  2019-03-20 08:17:23
3  2019-03-22 08:17:25
4  2019-03-22 11:12:20
5  2019-03-22 11:12:23
6  2019-03-24 12:33:23

At the end, i want to have this result :

                      A
0   2019-03-13 08:12:20
1   2019-03-13 08:12:21
2   2019-03-13 08:12:22
3   2019-03-13 08:12:23
4   2019-03-13 08:12:24
5   2019-03-13 08:12:25
6   2019-03-20 08:17:23
7   2019-03-22 08:17:25
8   2019-03-22 11:12:20
9   2019-03-22 11:12:21
10  2019-03-22 11:12:22
11  2019-03-22 11:12:23
12  2019-03-24 12:33:23

I tried with this :

import pandas as pd

df=pd.DataFrame({'A': ["2019-03-13 08:12:20", "2019-03-13 08:12:25", "2019-03-20 08:17:23", "2019-03-22 08:17:25", "2019-03-22 11:12:20", "2019-03-22 11:12:23", "2019-03-24 12:33:23"]})
df['A']=pd.to_datetime(df['A'])
fill = [pd.date_range(df.iloc[i]['A'], df.iloc[i+1]['A'], freq='S') for i in range(len(df)-1) if (df.iloc[i+1]['A']-df.iloc[i]['A']).total_seconds()<=7200]
dates = [item for sublist in fill for item in sublist]
df=df.set_index('A').join(pd.DataFrame(index=pd.Index(dates, name='A')), how='outer').reset_index()
print(df)

It's doing the job, but it's slow, is there any faster way to do this ?

Upvotes: 2

Views: 377

Answers (1)

Ben.T
Ben.T

Reputation: 29635

You can create a column with a group number where the difference between two consecutive rows are below 2 hours, using diff and cumsum. Then set_index the column A to be able to resample per group and reset_index to select the column you want.

df['gr'] = df.A.diff().gt(pd.Timedelta(hours=2)).cumsum()
df_output = df.set_index('A').groupby('gr', as_index=False).resample('s').sum().reset_index()[['A']]
print (df_output)
                     A
0  2019-03-13 08:12:20
1  2019-03-13 08:12:21
2  2019-03-13 08:12:22
3  2019-03-13 08:12:23
4  2019-03-13 08:12:24
5  2019-03-13 08:12:25
6  2019-03-20 08:17:23
7  2019-03-22 08:17:25
8  2019-03-22 11:12:20
9  2019-03-22 11:12:21
10 2019-03-22 11:12:22
11 2019-03-22 11:12:23
12 2019-03-24 12:33:23

Upvotes: 2

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