如何比较unicode日期u'；2006-07-23'；格式和25-06-15 08:42:43.830000000 PM使用python熊猫？

基本上，unicode格式将从datepicker获得，并且

25-06-15 08:42:43.830000000 PM

此格式来自一列 我的数据帧是：

query,status,received_date
a,closed,25-06-15 08:42:43.830000000 PM
b,pending,27-06-15 08:42:43.830000000 PM
ab,closed,28-06-15 08:42:43.830000000 PM
bb,pending,29-06-15 08:42:43.830000000 PM

我将从datepicker获得两个日期，格式如下

（u'2015-06-23'，u'2015-06-29'）

。如何比较此unicode日期和Received_date列

---

我必须显示这两个日期之间的数据（将从datepicker获取）

将它们转换为datetime

dates = (u'2015-06-23',u'2015-06-29')
df = df.set_index('received_date')
df.index = pd.DatetimeIndex(df.index)
df[dates[0]:dates[1]]

---

我想您首先需要转换

日期

，然后再转换列

接收日期

，并提取。最后与

掩码一起使用

进行过滤：

#datetimes changed for better testing
print df
  query   status                   received_date
0     a   closed  20-06-15 08:42:43.830000000 PM
1     b  pending  27-06-15 08:42:43.830000000 PM
2    ab   closed  28-06-15 08:42:43.830000000 PM
3    bb  pending  30-06-15 08:42:43.830000000 PM

dates = (u'2015-06-23',u'2015-06-29')
dates = pd.to_datetime(dates).date
print dates
[datetime.date(2015, 6, 23) datetime.date(2015, 6, 29)]

df['received_date'] = pd.to_datetime(df['received_date']).dt.date
print df
  query   status received_date
0     a   closed    2015-06-20
1     b  pending    2015-06-27
2    ab   closed    2015-06-28
3    bb  pending    2015-06-30

print (df['received_date'] > dates[0]) & (df['received_date'] < dates[1])
0    False
1     True
2     True
3    False
Name: received_date, dtype: bool

df = df[(df['received_date'] > dates[0]) & (df['received_date'] < dates[1])]
print df
  query   status received_date
1     b  pending    2015-06-27
2    ab   closed    2015-06-28

测试（

len（df）=40k

）：

测试代码：

#length is 40k
df = pd.concat([df]*10000).reset_index(drop=True)
df1 = df.copy()
df2 = df.copy()

def a(df):
    dates = (u'2015-06-23',u'2015-06-29')
    df = df.set_index('received_date')
    df.index = pd.DatetimeIndex(df.index)
    return df[dates[0]:dates[1]]


def b(df):
    dates = (u'2015-06-23',u'2015-06-29')
    dates = pd.to_datetime(dates).date
    df['received_date'] = pd.to_datetime(df['received_date']).dt.date
    df = df[(df['received_date'] > dates[0]) & (df['received_date'] < dates[1])]
    return df

def c(df):
    dates = (u'2015-06-23',u'2015-06-29')
    df['received_date'] = pd.to_datetime(df['received_date'])
    df = df.set_index('received_date')
    return df[dates[0]:dates[1]]

print a(df)
print b(df1)
print c(df2)

#长度为40k
df=pd.concat（[df]*10000）。重置索引（drop=True）
df1=df.copy（）
df2=df.copy（）
def a（df）：
日期=（u'2015-06-23'，u'2015-06-29'）
df=df.set\u索引（'received\u date'）
df.index=pd.DatetimeIndex（df.index）
返回日期[0]：日期[1]]
def b（df）：
日期=（u'2015-06-23'，u'2015-06-29'）
日期=pd.to_日期时间（日期）。日期
df['received_date']=pd.to_datetime（df['received_date']）.dt.date
df=df[（df['received_date']>日期[0]）和（df['received_date']<日期[1]）]
返回df
def c（df）：
日期=（u'2015-06-23'，u'2015-06-29'）
df['received_date']=pd.to_datetime（df['received_date']）
df=df.set\u索引（'received\u date'）
返回日期[0]：日期[1]]
打印a（df）
打印b（df1）
打印c（df2）

我必须比较unicode日期和接收日期列，并显示这些日期之间的数据如果我选择u'2015-06-23'，u'2015-06-29'意味着它将显示这些日期之间的数据

In [569]: %timeit a(df)
1 loops, best of 3: 12.2 s per loop

In [570]: %timeit b(df1)
10 loops, best of 3: 92.3 ms per loop

In [571]: %timeit c(df2)
100 loops, best of 3: 6.57 ms per loop

#length is 40k
df = pd.concat([df]*10000).reset_index(drop=True)
df1 = df.copy()
df2 = df.copy()

def a(df):
    dates = (u'2015-06-23',u'2015-06-29')
    df = df.set_index('received_date')
    df.index = pd.DatetimeIndex(df.index)
    return df[dates[0]:dates[1]]


def b(df):
    dates = (u'2015-06-23',u'2015-06-29')
    dates = pd.to_datetime(dates).date
    df['received_date'] = pd.to_datetime(df['received_date']).dt.date
    df = df[(df['received_date'] > dates[0]) & (df['received_date'] < dates[1])]
    return df

def c(df):
    dates = (u'2015-06-23',u'2015-06-29')
    df['received_date'] = pd.to_datetime(df['received_date'])
    df = df.set_index('received_date')
    return df[dates[0]:dates[1]]

print a(df)
print b(df1)
print c(df2)