pandas中DataFrame中两列日期相减，并去除单位days,及timedelta64理解

1.DataFrame两列直接相减

data['L'] = data['LOAD_TIME']-data['FFP_DATE']

在pandas中两列直接相减会出现如下错误：

TypeError: unsupported operand type(s) for -: 'str' and 'str'

2.使用to_datetime函数

不能直接相减，针对日期pandas有专门函数进行处理：

data['L'] = pd.to_datetime(data['LOAD_TIME'])-pd.to_datetime(data['FFP_DATE'])
'''
输出为：
    L    R    F       M         C
0     2706 days    1  210  580717  0.961639
1     2597 days    7  140  293678  1.252314
2     2615 days   11  135  283712  1.254676
3     2047 days   97   23  281336  1.090870
4     1816 days    5  152  309928  0.970658
...         ...  ...  ...     ...       ...
62974 3249 days   89    2     368  0.710000
62975 1961 days  121    2     368  0.670000
62976 1362 days   39    2    1062  0.225000
62977  466 days  464    2     904  0.250000
62978 1082 days  282    2     760  0.280000

[62044 rows x 5 columns]
--finish--
'''

3.去除结果中出现的days，精度到月，或者天数，小时等

如果向上面写，的确可以处理两列相减，但是结果中会有days，去除方法使用timedelta64函数：

'''精度到月'''
data['L'] = data['L'].map(lambda x: x/np.timedelta64(1,'M'))
'''
#或者是这样写，结果会细微差别
data['L'] = data['L'].map(lambda x: x/np.timedelta64(30,'D'))
'''
'''结果输出为
                L    R    F       M         C
0       88.905316    1  210  580717  0.961639
1       85.324134    7  140  293678  1.252314
2       85.915522   11  135  283712  1.254676
3       67.253948   97   23  281336  1.090870
4       59.664469    5  152  309928  0.970658
...           ...  ...  ...     ...       ...
62974  106.745518   89    2     368  0.710000
62975   64.428428  121    2     368  0.670000
62976   44.748352   39    2    1062  0.225000
62977   15.310376  464    2     904  0.250000
62978   35.548985  282    2     760  0.280000

[62044 rows x 5 columns]
'''

data['L'] = data['L'].map(lambda x: x/np.timedelta64(1,'D'))
'''
输出小时
data['L'] = data['L'].map(lambda x: x/np.timedelta64(1,'h'))
输出天数
            L    R    F       M         C
0      2706.0    1  210  580717  0.961639
1      2597.0    7  140  293678  1.252314
2      2615.0   11  135  283712  1.254676
3      2047.0   97   23  281336  1.090870
4      1816.0    5  152  309928  0.970658
...       ...  ...  ...     ...       ...
62974  3249.0   89    2     368  0.710000
62975  1961.0  121    2     368  0.670000
62976  1362.0   39    2    1062  0.225000
62977   466.0  464    2     904  0.250000
62978  1082.0  282    2     760  0.280000

[62044 rows x 5 columns]
'''

4.timedelta64理解

官方解释：Datetimes and Timedeltas work together to provide ways for simple datetime calculations.

timedelta64其实就是能将时间转换为是以年，月或者日，小时等为基础单位。在已有的日期时间上可以与之加减运算，较为方便，不需要自己将两个运算的时间转换一致。看几个例子：

np.datetime64('2009-01-01') - np.datetime64('2008-01-01')
np.datetime64('2009') + np.timedelta64(20, 'D')
np.datetime64('2011-06-15T00:00') + np.timedelta64(12, 'h')
np.timedelta64(1,'W') / np.timedelta64(1,'D')
'''
输出：
366 days
2009-01-21
2011-06-15T12:00
7.0
'''

timedelta64单位及含义：

'''
Here are the date units:

Code	Meaning	Time span (relative)	Time span (absolute)
Y	year	+/- 9.2e18 years	[9.2e18 BC, 9.2e18 AD]
M	month	+/- 7.6e17 years	[7.6e17 BC, 7.6e17 AD]
W	week	+/- 1.7e17 years	[1.7e17 BC, 1.7e17 AD]
D	day	+/- 2.5e16 years	[2.5e16 BC, 2.5e16 AD]
And here are the time units:

Code	Meaning	Time span (relative)	Time span (absolute)
h	hour	    +/- 1.0e15 years	[1.0e15 BC, 1.0e15 AD]
m	minute	    +/- 1.7e13 years	[1.7e13 BC, 1.7e13 AD]
s	second	    +/- 2.9e11 years	[2.9e11 BC, 2.9e11 AD]
ms	millisecond	+/- 2.9e8 years	[ 2.9e8 BC, 2.9e8 AD]
us	microsecond	+/- 2.9e5 years	[290301 BC, 294241 AD]
ns	nanosecond	+/- 292 years	[ 1678 AD, 2262 AD]
ps	picosecond	+/- 106 days	[ 1969 AD, 1970 AD]
fs	femtosecond	+/- 2.6 hours	[ 1969 AD, 1970 AD]
as	attosecond	+/- 9.2 seconds	[ 1969 AD, 1970 AD]
'''

详解请参阅官方文档：https://docs.scipy.org/doc/numpy/reference/arrays.datetime.html?highlight=timedelta64

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