我查看了其他几个相关问题here，here和here，但没有一个问题与我发生过完全相同的问题。
我正在使用Pandas版本0.16.2。我在Pandas数据框中有几列，dtype为datetime64 [ns]：
In [6]: date_list = ["SubmittedDate","PolicyStartDate", "PaidUpDate", "MaturityDate", "DraftDate", "CurrentValuationDate", "DOB", "InForceDate"]
In [11]: data[date_list].head()
Out[11]:
SubmittedDate PolicyStartDate PaidUpDate MaturityDate DraftDate \
0 NaT 2002-11-18 NaT 2041-03-04 NaT
1 NaT 2015-01-13 NaT NaT NaT
2 NaT 2014-10-15 NaT NaT NaT
3 NaT 2009-08-27 NaT NaT NaT
4 NaT 2007-04-19 NaT 2013-10-01 NaT
CurrentValuationDate DOB InForceDate
0 2015-04-30 1976-03-04 2002-11-18
1 NaT 1949-09-27 2015-01-13
2 NaT 1947-06-15 2014-10-15
3 2015-07-30 1960-06-07 2009-08-27
4 2010-04-21 1950-10-01 2007-04-19
这些最初是字符串格式（例如'1976-03-04'），我使用以下方式转换为datetime对象：
In [7]: for datecol in date_list:
...: data[datecol] = pd.to_datetime(data[datecol], coerce=True, errors = 'raise')
以下是每个列的dtypes：
In [8]: for datecol in date_list:
print data[datecol].dtypes
返回：
datetime64[ns]
datetime64[ns]
datetime64[ns]
datetime64[ns]
datetime64[ns]
datetime64[ns]
datetime64[ns]
datetime64[ns]
到目前为止，这么好。但我想要做的是为每个列创建一个新列，从特定日期开始以天为单位（作为整数）。
In [13]: current_date = pd.to_datetime("2015-07-31")
我第一次跑这个：
In [14]: for i in date_list:
....: data[i+"InDays"] = data[i].apply(lambda x: current_date - x)
但是，当我检查返回列的dtype时：
In [15]: for datecol in date_list:
....: print data[datecol + "InDays"].dtypes
我得到了这些：
object
timedelta64[ns]
object
timedelta64[ns]
object
timedelta64[ns]
timedelta64[ns]
timedelta64[ns]
我不知道为什么他们三个是对象，什么时候应该是timedeltas。我接下来要做的是：
In [16]: for i in date_list:
....: data[i+"InDays"] = data[i+"InDays"].dt.days
这种方法适用于timedelta列。但是，由于其中三列不是timedeltas，我收到此错误：
AttributeError: Can only use .dt accessor with datetimelike values
我怀疑这三列中有一些值阻止Pandas将它们转换为timedeltas。我无法弄清楚如何弄清楚这些价值观可能是什么。
答案 0 :(得分：1)
出现此问题是因为您有三列仅包含NaT值，这会导致在您对其应用条件时将这些列视为对象。
你应该在apply部分放置某种条件，默认为NaT时的某个时间点。示例 -
for i in date_list:
data[i+"InDays"] = data[i].apply(lambda x: current_date - x if x is not pd.NaT else pd.Timedelta(0))
或者如果你不能做到这一点，你应该把你想做的条件 - data[i+"InDays"] = data[i+"InDays"].dt.days，只有在系列的dtype允许的情况下才能接受。
或者更简单的方法来更改apply部分以直接获得您想要的内容 -
for i in date_list:
data[i+"InDays"] = data[i].apply(lambda x: (current_date - x).days if x is not pd.NaT else x)
这会输出 -
In [110]: data
Out[110]:
SubmittedDate PolicyStartDate PaidUpDate MaturityDate DraftDate \
0 NaT 2002-11-18 NaT 2041-03-04 NaT
1 NaT 2015-01-13 NaT NaT NaT
2 NaT 2014-10-15 NaT NaT NaT
3 NaT 2009-08-27 NaT NaT NaT
4 NaT 2007-04-19 NaT 2013-10-01 NaT
CurrentValuationDate DOB InForceDate SubmittedDateInDays \
0 2015-04-30 1976-03-04 2002-11-18 NaT
1 NaT 1949-09-27 2015-01-13 NaT
2 NaT 1947-06-15 2014-10-15 NaT
3 2015-07-30 1960-06-07 2009-08-27 NaT
4 2010-04-21 1950-10-01 2007-04-19 NaT
PolicyStartDateInDays PaidUpDateInDays MaturityDateInDays DraftDateInDays \
0 4638 NaT -9348 NaT
1 199 NaT NaN NaT
2 289 NaT NaN NaT
3 2164 NaT NaN NaT
4 3025 NaT 668 NaT
CurrentValuationDateInDays DOBInDays InForceDateInDays
0 92 14393 4638
1 NaN 24048 199
2 NaN 24883 289
3 1 20142 2164
4 1927 23679 3025
如果您希望将NaT更改为NaN，则可以使用 -
for i in date_list:
data[i+"InDays"] = data[i].apply(lambda x: (current_date - x).days if x is not pd.NaT else np.NaN)
示例/演示 -
In [114]: for i in date_list:
.....: data[i+"InDays"] = data[i].apply(lambda x: (current_date - x).days if x is not pd.NaT else np.NaN)
.....:
In [115]: data
Out[115]:
SubmittedDate PolicyStartDate PaidUpDate MaturityDate DraftDate \
0 NaT 2002-11-18 NaT 2041-03-04 NaT
1 NaT 2015-01-13 NaT NaT NaT
2 NaT 2014-10-15 NaT NaT NaT
3 NaT 2009-08-27 NaT NaT NaT
4 NaT 2007-04-19 NaT 2013-10-01 NaT
CurrentValuationDate DOB InForceDate SubmittedDateInDays \
0 2015-04-30 1976-03-04 2002-11-18 NaN
1 NaT 1949-09-27 2015-01-13 NaN
2 NaT 1947-06-15 2014-10-15 NaN
3 2015-07-30 1960-06-07 2009-08-27 NaN
4 2010-04-21 1950-10-01 2007-04-19 NaN
PolicyStartDateInDays PaidUpDateInDays MaturityDateInDays \
0 4638 NaN -9348
1 199 NaN NaN
2 289 NaN NaN
3 2164 NaN NaN
4 3025 NaN 668
DraftDateInDays CurrentValuationDateInDays DOBInDays InForceDateInDays
0 NaN 92 14393 4638
1 NaN NaN 24048 199
2 NaN NaN 24883 289
3 NaN 1 20142 2164
4 NaN 1927 23679 3025