我有一个看起来像这样的值的TimeDelta列：
2天21：54：00.000000000
我想有代表天数的浮动，让我们在这里说的2 +二十四分之二十一= 2.875，而忽略了分钟。有没有一种简单的方法来做到这一点？我看到一个答案提示
res['Ecart_lacher_collecte'].apply(lambda x: float(x.item().days+x.item().hours/24.))
但我得到“AttributeError的：‘海峡’对象有没有属性‘项目’”
NumPy的版本是“1.10.4”熊猫版u'0.17.1'
列最初已经获得了：
lac['DateHeureLacher'] = pd.to_datetime(lac['Date lacher']+' '+lac['Heure lacher'],format='%d/%m/%Y %H:%M:%S')
cap['DateCollecte'] = pd.to_datetime(cap['Date de collecte']+' '+cap['Heure de collecte'],format='%d/%m/%Y %H:%M:%S')
在第一个脚本。然后在第二之一：
res = pd.merge(lac, cap, how='inner', on=['Loc'])
res['DateHeureLacher'] = pd.to_datetime(res['DateHeureLacher'],format='%Y-%m-%d %H:%M:%S')
res['DateCollecte'] = pd.to_datetime(res['DateCollecte'],format='%Y-%m-%d %H:%M:%S')
res['Ecart_lacher_collecte'] = res['DateCollecte'] - res['DateHeureLacher']
也许它保存到csv改变他们都会以字符串？我试图做的转型是在第三脚本。
Sexe_x PiegeLacher latL longL Loc Col_x DateHeureLacher Nb envolees PiegeCapture latC longC Col_y Sexe_y Effectif DateCollecte DatePose Ecart_lacher_collecte Dist_m
M Q0-002 1629238 237877 H Rouge 2011-02-04 17:15:00 928 Q0-002 1629238 237877 Rouge M 1 2011-02-07 15:09:00 2011-02-07 12:14:00 2 days 21:54:00.000000000 0
M Q0-002 1629238 237877 H Rouge 2011-02-04 17:15:00 928 Q0-002 1629238 237877 Rouge M 4 2011-02-07 12:14:00 2011-02-07 09:42:00 2 days 18:59:00.000000000 0
M Q0-002 1629238 237877 H Rouge 2011-02-04 17:15:00 928 Q0-003 1629244 237950 Rouge M 1 2011-02-07 15:10:00 2011-02-07 12:16:00 2 days 21:55:00.000000000 75
热水.info():
Sexe_x 922 non-null object
PiegeLacher 922 non-null object
latL 922 non-null int64
longL 922 non-null int64
Loc 922 non-null object
Col_x 922 non-null object
DateHeureLacher 922 non-null object
Nb envolees 922 non-null int64
PiegeCapture 922 non-null object
latC 922 non-null int64
longC 922 non-null int64
Col_y 922 non-null object
Sexe_y 922 non-null object
Effectif 922 non-null int64
DateCollecte 922 non-null object
DatePose 922 non-null object
Ecart_lacher_collecte 922 non-null object
Dist_m 922 non-null int64
您可以使用dt.total_seconds并以秒为单位，每天例如总数除以本：
In [25]:
df = pd.DataFrame({'dates':pd.date_range(dt.datetime(2016,1,1, 12,15,3), periods=10)})
df
Out[25]:
dates
0 2016-01-01 12:15:03
1 2016-01-02 12:15:03
2 2016-01-03 12:15:03
3 2016-01-04 12:15:03
4 2016-01-05 12:15:03
5 2016-01-06 12:15:03
6 2016-01-07 12:15:03
7 2016-01-08 12:15:03
8 2016-01-09 12:15:03
9 2016-01-10 12:15:03
In [26]:
df['time_delta'] = df['dates'] - pd.datetime(2015,11,6,8,10)
df
Out[26]:
dates time_delta
0 2016-01-01 12:15:03 56 days 04:05:03
1 2016-01-02 12:15:03 57 days 04:05:03
2 2016-01-03 12:15:03 58 days 04:05:03
3 2016-01-04 12:15:03 59 days 04:05:03
4 2016-01-05 12:15:03 60 days 04:05:03
5 2016-01-06 12:15:03 61 days 04:05:03
6 2016-01-07 12:15:03 62 days 04:05:03
7 2016-01-08 12:15:03 63 days 04:05:03
8 2016-01-09 12:15:03 64 days 04:05:03
9 2016-01-10 12:15:03 65 days 04:05:03
In [27]:
df['total_days_td'] = df['time_delta'].dt.total_seconds() / (24 * 60 * 60)
df
Out[27]:
dates time_delta total_days_td
0 2016-01-01 12:15:03 56 days 04:05:03 56.170174
1 2016-01-02 12:15:03 57 days 04:05:03 57.170174
2 2016-01-03 12:15:03 58 days 04:05:03 58.170174
3 2016-01-04 12:15:03 59 days 04:05:03 59.170174
4 2016-01-05 12:15:03 60 days 04:05:03 60.170174
5 2016-01-06 12:15:03 61 days 04:05:03 61.170174
6 2016-01-07 12:15:03 62 days 04:05:03 62.170174
7 2016-01-08 12:15:03 63 days 04:05:03 63.170174
8 2016-01-09 12:15:03 64 days 04:05:03 64.170174
9 2016-01-10 12:15:03 65 days 04:05:03 65.170174
您可以使用pd.to_timedelta或np.timedelta64通过这种方法来定义的持续时间和鸿沟：
# set up as per @EdChum
df['total_days_td'] = df['time_delta'] / pd.to_timedelta(1, unit='D')
df['total_days_td'] = df['time_delta'] / np.timedelta64(1, 'D')
你有没有使用这种替代试过吗？
res['Ecart_lacher_collecte'].apply(lambda x: (x.total_seconds()//(3600*24)) + (x.total_seconds()%(3600*24)//3600)/24))
第一项是日（2你的情况）中的第二项是小时比率忽略分钟（21/24你的情况）
如果您不想分秒数据被忽略，而需要考虑其在当天的所有秒的比例，如提及下面的代码：
res['Ecart_lacher_collecte'].apply(lambda x: (x.total_seconds()/(3600*24))