Timedelta to String Type in Pandas Dataframe

timedelta to string type in pandas dataframe

It is possible by:

df['duration1'] = df['duration'].astype(str).str[-18:-10]

But solution is not general, if input is 3 days 05:01:11 it remove 3 days too.

So solution working only for timedeltas less as one day correctly.

More general solution is create custom format:

N = 10
np.random.seed(11230)
rng = pd.date_range('2017-04-03 15:30:00', periods=N, freq='13.5H')
df = pd.DataFrame({'duration': np.abs(np.random.choice(rng, size=N) - 
                                 np.random.choice(rng, size=N)) })  

df['duration1'] = df['duration'].astype(str).str[-18:-10]

def f(x):
    ts = x.total_seconds()
    hours, remainder = divmod(ts, 3600)
    minutes, seconds = divmod(remainder, 60)
    return ('{}:{:02d}:{:02d}').format(int(hours), int(minutes), int(seconds)) 

df['duration2'] = df['duration'].apply(f)
print (df)

         duration duration1  duration2
0 2 days 06:00:00  06:00:00   54:00:00
1 2 days 19:30:00  19:30:00   67:30:00
2 1 days 03:00:00  03:00:00   27:00:00
3 0 days 00:00:00  00:00:00    0:00:00
4 4 days 12:00:00  12:00:00  108:00:00
5 1 days 03:00:00  03:00:00   27:00:00
6 0 days 13:30:00  13:30:00   13:30:00
7 1 days 16:30:00  16:30:00   40:30:00
8 0 days 00:00:00  00:00:00    0:00:00
9 1 days 16:30:00  16:30:00   40:30:00

How to convert timedelta column in pandas to string

As long as your columns are datetimes, they have a .dt accessor with lots of useful datetime methods attached, including strftime, so you can do:

import pandas as pd

df = pd.DataFrame({'dates': pd.date_range('2015-01-01', '2015-01-10', freq='12H')})
df.dates.dt.strftime('%H:%S')

Apply change to timedelta to columns containing a given string

You can apply pandas.to_timedelta on all columns selected by filter and update the original dataframe:

df.update(df.filter(like='_SEL').apply(pd.to_timedelta, unit='s'))

NB. there is no output, the modification is inplace

updated dataframe:

         date         time_SEL     time_02_SEL      time_03_SEL  other
0  2022-01-01  0 days 09:39:16 2 days 16:47:12 39 days 17:27:03    756
1  2022-01-03  0 days 06:28:42 0 days 01:12:23  0 days 00:55:34    343
2  2022-02-01  1 days 10:13:52 0 days 00:54:02  0 days 06:30:23    434
3  2022-03-01 84 days 18:13:52 0 days 09:00:23  3 days 17:50:23  34324

update "TypeError: invalid type promotion"

ensure you have numbers:

(df.update(df.filter(like='_SEL')
             .apply(lambda c: pd.to_timedelta(pd.to_numeric(c, errors='coerce'),
                                              unit='s'))
)

Turn a string back into a datetime timedelta

Use pd.to_timedelta

pd.to_timedelta(df.iloc[:, 0])

0     0 days 00:00:57.416000
1     0 days 00:00:12.036000
2     0 days 16:46:23.127000
3    49 days 00:09:30.813000
4    50 days 00:39:31.306000
5    55 days 12:39:32.269000
6   -1 days +22:03:05.256000
Name: 0, dtype: timedelta64[ns]

Dataframe - mean of string type column with time values

You can use regex to extract hours and minutes. To calcualte the mean time in minutus:

h = df['time'].str.extract('(\d{1,2})h').fillna(0).astype(int)
m = df['time'].str.extract('(\d{1,2})m').fillna(0).astype(int)

(h * 60 + m).mean()

Result:

0    83.75
dtype: float64

---

Related Topics