Question

基本上，unicode格式将从一个列的datepicker和25-06-15 08:42:43.830000000 PM这种格式中获取我的数据框是：

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，例如以下格式(u'2015-06-23',u'2015-06-29')。如何比较这个unicode日期和recieved_date列。

我必须在这两个日期之间显示数据（将从datepicker获得）

Answer 1

将它们转换为日期时间。

{{1}}

Answer 2

我认为您首先需要转换dates to_datetime，然后转换received_date列并提取date。上次使用boolean indexing与mask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

但更快的修改PhilChang解决方案：

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]]

测试（len(df) == 40k）：

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

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)

如何使用python pandas比较unicode date u'2006-07-23'format和25-06-15 08：42：43.830000000 PM？

2 个答案: