熊猫groupby得到一个平均的日子

Question

我有一个数据框，这是从阅读csv的结果。它包含一个日期时间列和与事件相关的数据。我需要用每20分钟的统计数据来计算平均一天，在下面的代码中，我使用'mean'作为示例。

编辑： 我的数据是观察结果。这意味着并非所有箱都有数据。但是，在计算平均值时，必须考虑这个零计数：mean = count /＃days

此代码有效，但这是要走的路吗？它看起来对我来说很复杂，我不知道我是否真的需要在一天中的某个时间给我们一个BinID和不可能的组。

import pandas as pd 

# Create dataframe 
data = {'date': pd.date_range('2017-01-01 00:30:00', freq='10min', periods=282), 
     'i/o': ['in', 'out'] * 141} 
df = pd.DataFrame(data) 

# Add ones 
df['move'] = 1 

# I did try: 
# 1) 
# df['time'] = df['date'].dt.time 
# df.groupby(['i/o', pd.Grouper(key='time', freq='20min')]) 
# This failed with groupby, so should I use my own bins then??? 

# 2) 
# Create 20 minutes bins 
# df['binID'] = df['date'].dt.hour*3 + df['date'].dt.minute//20 
# averageDay = df.groupby(['i/o', 'binID']).agg(['count', 'sum', 'mean']) 
# 
# Well, bins with zero moves aren't their. 
# So 'mean' can't be used as well as other functions that 
# need the number of observations. Resample and reindex then??? 

# Resample 
df2 = df.groupby(['i/o', pd.Grouper(key='date', freq='20min')]).agg('sum') 

# Reindex and reset (for binID and groupby) 
levels = [['in', 'out'], 
      pd.date_range('2017-01-01 00:00:00', freq='20min', periods=144)] 
newIndex = pd.MultiIndex.from_product(levels, names=['i/o', 'date']) 
df2 = df2.reindex(newIndex, fill_value=0).reset_index() 

# Create 20 minutes bins 
df2['binID'] = df2['date'].dt.hour*3 + df2['date'].dt.minute//20 

# Average day 
averageDay2 = df2.groupby(['i/o', 'binID']).agg(['count', 'sum', 'mean']) 
print(averageDay2) 

Answer 1

IIUC：

In [124]: df.groupby(['i/o',df.date.dt.hour*3 + df.date.dt.minute//20]) \ 
      .agg(['count','sum','mean']) 
Out[124]: 
      move 
     count sum mean 
i/o date 
in 0  1 1 1 
    1  2 2 1 
    2  2 2 1 
    3  2 2 1 
    4  2 2 1 
    5  2 2 1 
    6  2 2 1 
    7  2 2 1 
    8  2 2 1 
    9  2 2 1 
...  ... .. ... 
out 62  2 2 1 
    63  2 2 1 
    64  2 2 1 
    65  2 2 1 
    66  2 2 1 
    67  2 2 1 
    68  2 2 1 
    69  2 2 1 
    70  2 2 1 
    71  1 1 1 

[144 rows x 3 columns]