Ajouter une nouvelle colonne de valeurs timedelta dans un format particulier dans Pandas

Question

J'ai un dataframe et je veux ajouter une colonne contenant la différence de temps entre deux une autre colonne:

df[Diff] = df['End Time'] - df['Open Time'] 
df[Diff] 
0  0 days 01:25:40 
1  0 days 00:41:57 
2  0 days 00:21:47 
3  0 days 16:41:57 
4  0 days 04:32:00 
5  0 days 03:01:57 
6  0 days 01:37:56 
7  0 days 01:13:57 
8  0 days 01:07:56 
9  0 days 02:33:59 
10 29 days 18:33:53 
11 0 days 03:50:56 
12 0 days 01:57:56 

Je voudrais avoir cette colonne format '1h 25m', donc je l'ai essayé de calculer jours en heures:

diff = df['End Time'] - df['Open Time'] 
hours = diff.dt.days * 24 + diff.dt.components.hours 
minutes = diff.dt.components.minutes 

et reçu ces résultats:

0  1 
1  0 
2  0 
3  16 
4  4 
5  3 
6  1 
7  1 
8  1 
9  2 
10 714 
11  3 
12  1 
dtype: int64h 0  25 
1  41 
2  21 
3  41 
4  32 
5  1 
6  37 
7  13 
8  7 
9  33 
10 33 
11 50 
12 57 
Name: minutes, dtype: int64m 

Mais je ne peux pas exprimer ces résultats dans ce format dans la nouvelle colonne:

'{}h {}m'.format(hours,minutes)) 

Answer 1

Vous pouvez extraire les colonnes pertinentes, se convertir à l'aide strastype, et juste concat les au besoin Col..

c = (df['End Time'] - df['Open Time'])\ 
       .dt.components[['days', 'hours', 'minutes']] 
df['diff'] = (c.days * 24 + c.hours).astype(str) + 'h ' + c.minutes.astype(str) + 'm' 
df['diff'] 
0  1h 25m 
1  0h 41m 
2  0h 21m 
3  16h 41m 
4  4h 32m 
5  3h 1m 
6  1h 37m 
7  1h 13m 
8  1h 7m 
9  2h 33m 
10 714h 33m 
11  3h 50m 
12  1h 57m 
Name: diff, dtype: object 

Answer 2

Vous pouvez utiliser total_seconds pour convertir timedelta en secondes puis compter hours, minutes et aussi quelques secondes, ce qui est 10 fois plus rapide que l'utilisation dt.components:

s = diff.dt.total_seconds().astype(int) 

hours = s // 3600 
# remaining seconds 
s = s - (hours * 3600) 
# minutes 
minutes = s // 60 
# remaining seconds 
seconds = s - (minutes * 60) 

a = hours.astype(str) + 'h ' + minutes.astype(str) 
print (a) 
0  1h 25 
1  0h 41 
2  0h 21 
3  16h 41 
4  4h 32 
5  3h 1 
6  1h 37 
7  1h 13 
8  1h 7 
9  2h 33 
10 714h 33 
11  3h 50 
12  1h 57 
Name: Diff, dtype: object 

Zero comment solutions:

hours = diff.dt.days * 24 + diff.dt.components.hours 
minutes = diff.dt.components.minutes 

a = hours.astype(str) + 'h ' + minutes.astype(str) 
print (a) 
0  1h 25m 
1  0h 41m 
2  0h 21m 
3  16h 41m 
4  4h 32m 
5  3h 1m 
6  1h 37m 
7  1h 13m 
8  1h 7m 
9  2h 33m 
10 18h 33m 
11  3h 50m 
12  1h 57m 
dtype: object 

Autre:

a = pd.Series(['{0[0]}h {0[1]}m'.format(x) for x in zip(hours, minutes)]) 
print (a) 
0  1h 25m 
1  0h 41m 
2  0h 21m 
3  16h 41m 
4  4h 32m 
5  3h 1m 
6  1h 37m 
7  1h 13m 
8  1h 7m 
9  2h 33m 
10 714h 33m 
11  3h 50m 
12  1h 57m 
dtype: object 

minutage:

#13000 rows 
df = pd.concat([df]*1000).reset_index(drop=True) 

In [191]: %%timeit 
    ...: hours = diff.dt.days * 24 + diff.dt.components.hours 
    ...: minutes = diff.dt.components.minutes 
    ...: 
    ...: a = hours.astype(str) + 'h ' + minutes.astype(str) 
    ...: 
1 loop, best of 3: 483 ms per loop 

In [192]: %%timeit 
    ...: s = diff.dt.total_seconds().astype(int) 
    ...: 
    ...: hours = s // 3600 
    ...: # remaining seconds 
    ...: s = s - (hours * 3600) 
    ...: # minutes 
    ...: minutes = s // 60 
    ...: # remaining seconds 
    ...: seconds = s - (minutes * 60) 
    ...: 
    ...: a = hours.astype(str) + 'h ' + minutes.astype(str) 
    ...: 
10 loops, best of 3: 43.9 ms per loop 

In [193]: %%timeit 
    ...: hours = diff.dt.days * 24 + diff.dt.components.hours 
    ...: minutes = diff.dt.components.minutes 
    ...: s = pd.Series(['{0[0]}h {0[1]}m'.format(x) for x in zip(hours, minutes)]) 
    ...: 
1 loop, best of 3: 465 ms per loop 

---

#cᴏʟᴅsᴘᴇᴇᴅ solution 
In [194]: %%timeit 
    ...: c = diff.dt.components[['days', 'hours', 'minutes']] 
    ...: a = (c.days * 24 + c.hours).astype(str) + 'h ' + c.minutes.astype(str) + 'm' 
    ...: 
1 loop, best of 3: 208 ms per loop