从 TimeDelta 到 Pandas 中的浮动天数
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/35502927/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
From TimeDelta to float days in Pandas
提问by alpagarou
I have a TimeDelta column with values that look like this:
我有一个 TimeDelta 列，其值如下所示：
2 days 21:54:00.000000000
2 天 21:54:00.000000000
I would like to have a float representing the number of days, let's say here 2+21/24 = 2.875, neglecting the minutes. Is there a simple way to do this ? I saw an answer suggesting
我想要一个代表天数的浮点数，假设这里是 2+21/24 = 2.875，忽略分钟。有没有一种简单的方法可以做到这一点？我看到一个答案建议
res['Ecart_lacher_collecte'].apply(lambda x: float(x.item().days+x.item().hours/24.))
But I get "AttributeError: 'str' object has no attribute 'item' "
但我得到“AttributeError: 'str' object has no attribute 'item'”
Numpy version is '1.10.4' Pandas version is u'0.17.1'
Numpy 版本是 '1.10.4' Pandas 版本是 u'0.17.1'
The columns has originally been obtained with:
这些列最初是通过以下方式获得的：
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')
in a first script. Then in a second one:
在第一个脚本中。然后在第二个：
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']
Maybe saving it to csv change their types back to string? The transformation I'm trying to do is in a third script.
也许将它保存到 csv 将它们的类型改回字符串？我正在尝试进行的转换是在第三个脚本中。
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
res.info():
资源信息（）：
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
回答by jpp
You can use pd.to_timedeltaor np.timedelta64to define a duration and divide by this:
您可以使用pd.to_timedelta或np.timedelta64定义持续时间并除以：
# 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')
回答by EdChum
You can use dt.total_secondsand divide this by the total number of seconds in a day, example:
您可以使用dt.total_seconds并将其除以一天中的总秒数，例如：
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
回答by sharinganSawant
Have you tried using this instead?
你试过用这个代替吗？
res['Ecart_lacher_collecte'].apply(lambda x: (x.total_seconds()//(3600*24)) + (x.total_seconds()%(3600*24)//3600)/24))
The first term is the Day ( 2 in your case ) The second term is the hour ratio neglecting the minutes ( 21/24 in your case)
第一项是天（在您的情况下为 2）第二项是忽略分钟的小时比率（在您的情况下为 21/24）
If you don't want the minutes and seconds data to be neglected, and rather need a ratio which considers all the seconds in the day, the code is as mentioned below:
如果您不想忽略分秒数据，而是需要一个考虑当天所有秒数的比率，则代码如下：
res['Ecart_lacher_collecte'].apply(lambda x: (x.total_seconds()/(3600*24))
