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pythonpandasdatetimedate-difference
Dario Federici
Dario Federici

Reputation: 1258

Python Pandas time difference from the start of every day

I've got the following data frame on pandas:

d = {'col_Date_Time': ['2020-08-01 00:00:00', 
                       '2020-08-01 00:10:00', 
                       '2020-08-01 00:15:00', 
                       '2020-08-01 00:19:00',
                       '2020-08-01 01:19:00',
                       '2020-08-02 00:00:00', 
                       '2020-08-02 00:15:00', 
                       '2020-08-02 00:35:00',
                       '2020-08-02 01:35:00']}

df = pd.DataFrame(data=d)
df = pd.to_datetime(df.col_Date_Time)

I want to add another column that contains the number of minutes from the start of each day.

So, the result in this case woud be:

       NAN 
       10 
       15 
       19 
       79
       NAN 
       15 
       35
       95

Upvotes: 0

Views: 518

Answers (3)

vrana95
vrana95

Reputation: 521

import pandas as pd
import numpy as np
df = pd.DataFrame({'col_Date_Time': ['2020-08-01 00:00:00', 
                       '2020-08-01 00:10:00', 
                       '2020-08-01 00:15:00', 
                       '2020-08-01 00:19:00', 
                        '2020-08-01 01:23:00',
                       '2020-08-02 00:00:00', 
                       '2020-08-02 00:15:00', 
                       '2020-08-02 00:35:00',
                        '2020-08-02 06:31:00']})
df['col_Date_Time'] = pd.to_datetime(df.col_Date_Time)
df['start_day_time_stamp']=list(map(lambda x: x.date(),df['col_Date_Time'])) 
df['mins_from_day_start']=((pd.to_datetime(df['col_Date_Time'])-pd.to_datetime(df['start_day_time_stamp'])).dt.total_seconds())/60
df

enter image description here

Upvotes: 1

Ricardo Rendich
Ricardo Rendich

Reputation: 651

You can truncate the column to days (.dt.floor('d')), subtract this to col_Date_Time, and save in another column:

df["DELTA"] = df.col_Date_Time - df.col_Date_Time.dt.floor('d')

If you want this like integer:

    df["DELTA2"] = df.DELTA.dt.seconds.div(60).astype(int) 
    col_Date_Time   DELTA   DELTA2
0   2020-08-01 00:00:00 00:00:00    0
1   2020-08-01 00:10:00 00:10:00    10
2   2020-08-01 00:15:00 00:15:00    15
3   2020-08-01 00:19:00 00:19:00    19
4   2020-08-01 01:19:00 01:19:00    79
5   2020-08-02 00:00:00 00:00:00    0
6   2020-08-02 00:15:00 00:15:00    15
7   2020-08-02 00:35:00 00:35:00    35
8   2020-08-02 01:35:00 01:35:00    95

Upvotes: 1

BENY
BENY

Reputation: 323326

Let us try

s = df.dt.minute.where(df.dt.date.duplicated())
Out[66]: 
0     NaN
1    10.0
2    15.0
3    19.0
4     NaN
5    15.0
6    35.0
Name: col_Date_Time, dtype: float64

Upvotes: 1

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