Pandas如何按列按间隔分割数据框

我有一个巨大的数据框，其中有一个名为的日期时间类型列dt，数据框排序基于dt已经。我想根据以下内容将数据帧拆分为多个数据帧dt，每个数据帧包含行1 hr range.

Split

   dt                    text
0  20160811 11:05        a
1  20160811 11:35        b
2  20160811 12:03        c
3  20160811 12:36        d
4  20160811 12:52        e
5  20160811 14:32        f

into

   dt                    text
0  20160811 11:05        a
1  20160811 11:35        b
2  20160811 12:03        c

   dt                    text
0  20160811 12:36        d
1  20160811 12:52        e

   dt                    text 
0  20160811 14:32        f

你需要groupby http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.groupby.html通过列的第一个值的差异dt转换成hour by astype http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.astype.html:

S = pd.to_datetime(df.dt)
for i, g in df.groupby([(S - S[0]).astype('timedelta64[h]')]):
        print (g.reset_index(drop=True))

               dt text
0  20160811 11:05    a
1  20160811 11:35    b
2  20160811 12:03    c
               dt text
0  20160811 12:36    d
1  20160811 12:52    e
               dt text
0  20160811 14:32    f

List comprehension解决方案：

S = pd.to_datetime(df.dt)

print ((S - S[0]).astype('timedelta64[h]'))
0    0.0
1    0.0
2    0.0
3    1.0
4    1.0
5    3.0
Name: dt, dtype: float64

L = [g.reset_index(drop=True) for i, g in df.groupby([(S - S[0]).astype('timedelta64[h]')])]

print (L[0])
               dt text
0  20160811 11:05    a
1  20160811 11:35    b
2  20160811 12:03    c

print (L[1])
               dt text
0  20160811 12:36    d
1  20160811 12:52    e

print (L[2])
               dt text
0  20160811 14:32    f

旧的解决方案，由hour:

您可以使用groupby http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.groupby.html by dt.hour http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.dt.hour.html，但首先需要转换dt to_datetime http://pandas.pydata.org/pandas-docs/stable/generated/pandas.to_datetime.html:

for i, g in df.groupby([pd.to_datetime(df.dt).dt.hour]):
    print (g.reset_index(drop=True))

               dt text
0  20160811 11:05    a
1  20160811 11:35    b
               dt text
0  20160811 12:03    c
1  20160811 12:36    d
2  20160811 12:52    e
               dt text
0  20160811 14:32    f

List comprehension解决方案：

L = [g.reset_index(drop=True) for i, g in df.groupby([pd.to_datetime(df.dt).dt.hour])]

print (L[0])
               dt text
0  20160811 11:05    a
1  20160811 11:35    b

print (L[1])
               dt text
0  20160811 12:03    c
1  20160811 12:36    d
2  20160811 12:52    e

print (L[2])
               dt text
0  20160811 14:32    f

Or use list comprehension带转换柱dt to datetime:

df.dt = pd.to_datetime(df.dt)
L =[g.reset_index(drop=True) for i, g in df.groupby([df['dt'].dt.hour])]

print (L[1])
                   dt text
0 2016-08-11 12:03:00    c
1 2016-08-11 12:36:00    d
2 2016-08-11 12:52:00    e

print (L[2])
                   dt text
0 2016-08-11 14:32:00    f

如果需要分割dates and hours:

#changed dataframe for testing
print (df)
               dt text
0  20160811 11:05    a
1  20160812 11:35    b
2  20160813 12:03    c
3  20160811 12:36    d
4  20160811 12:52    e
5  20160811 14:32    f

serie = pd.to_datetime(df.dt)
for i, g in df.groupby([serie.dt.date, serie.dt.hour]):
    print (g.reset_index(drop=True))
               dt text
0  20160811 11:05    a
               dt text
0  20160811 12:36    d
1  20160811 12:52    e
               dt text
0  20160811 14:32    f
               dt text
0  20160812 11:35    b
               dt text
0  20160813 12:03    c