我查看了其他几个相关的问题 here 、 here 和 here ，但没有一个和我遇到过完全相同的问题。
我正在使用 Pandas 版本 0.16.2。我在 Pandas 数据框中有几列，数据类型为 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')，我使用以下方法将其转换为日期时间对象:
In [7]: for datecol in date_list:
...: data[datecol] = pd.to_datetime(data[datecol], coerce=True, errors = 'raise')
以下是每一列的 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]
到现在为止还挺好。但是我想要做的是为这些列中的每一个创建一个新列，给出从某个日期开始的天数(作为整数)。
In [13]: current_date = pd.to_datetime("2015-07-31")
我首先运行了这个:
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
我得到这些:
object
timedelta64[ns]
object
timedelta64[ns]
object
timedelta64[ns]
timedelta64[ns]
timedelta64[ns]
我不知道为什么其中三个是对象，而它们应该是 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。我不知道如何计算出这些值可能是什么。
最佳答案
出现此问题是因为您有三列只有 NaT 值，这会导致在您对其应用条件时将这些列视为对象。
您应该在 apply 部分中放置某种条件，以在 NaT 的情况下默认为某个 timedelta 。例子 -
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，如果你不能做到以上几点，你应该在你想做的地方放一个条件 -
data[i+"InDays"] = data[i+"InDays"].dt.days ，只有当系列的 dtype 允许时才接受它。或者更改
apply 部分以直接获得您想要的内容的更简单方法是 -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)
这将输出 -
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，您可以使用 -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
关于python - 从日期时间操作创建 TimeDeltas 的 Pandas 错误，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/32137330/