如果想进行测试时间,pandas默认使用今天的日期,因此可能的解决方案是使用Series.dt.date
,Timestamp.date
和Series.all
对它们进行测试,看看列的所有值是否匹配。
此外,还可以通过Series.dt.floor
删除时间后测试相同的日期数值:
df = pd.DataFrame({'a':['2019-01-01 12:23:10',
'2019-01-02 12:23:10'],
'b':['2019-01-01',
'2019-01-02'],
'c':['12:23:10',
'15:23:10'],
'd':['a','b']})
print (df)
a b c d
0 2019-01-01 12:23:10 2019-01-01 12:23:10 a
1 2019-01-02 12:23:10 2019-01-02 15:23:10 b
def check(col):
try:
dt = pd.to_datetime(df[col])
if (dt.dt.floor('d') == dt).all():
return ('Its a pure date field')
elif (dt.dt.date == pd.Timestamp('now').date()).all():
return ('Its a pure time field')
else:
return ('Its a Datetime field')
except:
return ('its not a datefield')
print (check('a'))
print (check('b'))
print (check('c'))
print (check('d'))
Its a Datetime field
Its a pure date field
Its a pure time field
its not a datefield
另一个想法是测试数字列,并默认返回非数字,以防止将数字转换为日期时间,但如果可能,所有日期时间都只包含今天的日期(
f
列),则测试时间是否与匹配模式
HH:MM:SS
或
H:MM:SS
的
Series.str.contains
不同:
df = pd.DataFrame({'a':['2019-01-01 12:23:10',
'2019-01-02'],
'b':['2019-01-01',
'2019-01-02'],
'c':['12:23:10',
'15:23:10'],
'd':['a','b'],
'e':[1,2],
'f':['2019-11-13 12:23:10',
'2019-11-13'],})
print (df)
a b c d e f
0 2019-01-01 12:23:10 2019-01-01 12:23:10 a 1 2019-11-13 12:23:10
1 2019-01-02 2019-01-02 15:23:10 b 2 2019-11-13
def check(col):
if np.issubdtype(df[col].dtype, np.number):
return ('its not a datefield')
try:
dt = pd.to_datetime(df[col])
if (dt.dt.floor('d') == dt).all():
return ('Its a pure date field')
elif df[col].str.contains(r"^\d{1,2}:\d{2}:\d{2}$").all():
return ('Its a pure time field')
else:
return ('Its a Datetime field')
except:
return ('its not a datefield')
print (check('a'))
print (check('b'))
print (check('c'))
print (check('d'))
print (check('e'))
print (check('f'))
Its a Datetime field
Its a pure date field
Its a pure time field
its not a datefield
its not a datefield
Its a Datetime field