1 
Msgtype Date ConvID   message
enquire 12/1 689  I want your car
reply   12/3 689  it is available
reply   12/4 689  rent please?
reply   12/6 689  $200
accept  12/8 689  please pay through CC
reply   12/8 689  thank you, what about fuel?
reply   12/8 689  you have to take care
enquire 12/3 690  Looking for car
reply   12/4 690  available
accept  12/5 690  paid
reply   12/6 690  thank you

このデータを ConvID でグループ化し、日付で並べ替えたいと考えています。「Msgtype」=その特定の ConvID を受け入れるまでの行が必要です。特定の ConvID の予約リクエストが受け入れられるまで、メッセージ データを分析することを目指します。したがって、ConvID = 689 の場合、"Msgtype" = 受け入れるまで行が必要です。「同意」の後の残りの行は必須ではありません。

例: ConvID = 689 の場合、これら 2 つは不要です。

    Msgtype Date ConvID   message
    reply   12/8 689  thank you, what about fuel?
    reply   12/8 689  you have to take care

同様に、ConvID = 690 の場合、この行は不要です。

Msgtype Date ConvID   message
 reply   12/6 690  thank you
4

2 に答える 2

1

私はあなたが使用できると思います：

mask1 = (df.Msgtype == 'accept')
mask = mask1.groupby([df.ConvID]).apply(lambda x: x.shift().fillna(False).cumsum()) == 0

print (df[mask].sort_values(['ConvID','Date']))
   Msgtype  Date  ConvID                message
0  enquire  12/1     689        I want your car
1    reply  12/3     689        it is available
2    reply  12/4     689           rent please?
3    reply  12/6     689                   $200
4   accept  12/8     689  please pay through CC
7  enquire  12/3     690        Looking for car
8    reply  12/4     690              available
9   accept  12/5     690                   paid

説明:

#mask where is 'accept'
mask1 = (df.Msgtype == 'accept')
print (mask1)
0     False
1     False
2     False
3     False
4      True
5     False
6     False
7     False
8     False
9      True
10    False
Name: Msgtype, dtype: bool

#per group shift, replace NaN by False and cumulative sum
print (mask1.groupby([df.ConvID]).apply(lambda x: x.shift().fillna(False).cumsum()))
0     0
1     0
2     0
3     0
4     0
5     1
6     1
7     0
8     0
9     0
10    1
Name: Msgtype, dtype: int32

#where output of groupby is 0 
mask = mask1.groupby([df.ConvID]).apply(lambda x: x.shift().fillna(False).cumsum()) == 0
print (mask)
0      True
1      True
2      True
3      True
4      True
5     False
6     False
7      True
8      True
9      True
10    False
Name: Msgtype, dtype: bool

#boolean indexing and sorting
print (df[mask].sort_values(['ConvID','Date']))
   Msgtype  Date  ConvID                message
0  enquire  12/1     689        I want your car
1    reply  12/3     689        it is available
2    reply  12/4     689           rent please?
3    reply  12/6     689                   $200
4   accept  12/8     689  please pay through CC
7  enquire  12/3     690        Looking for car
8    reply  12/4     690              available
9   accept  12/5     690                   paid
于 2016-10-31T18:07:51.677 に答える
0

簡単：

for name, grp in df.groupby('ConvID'):
    grp.sort_values('Date', inplace=True)
    accept_date = grp.loc[grp['Msgtype'] == 'accept', 'Date']
    req = grp[grp['Date'] < accept_date]
    # Or, you can use index, like so:
    # grp = grp.sort_values('Date').reset_index(drop=True)
    # req = grp.iloc[:grp[grp['Msgtype'] == 'accept'].index.values[0], :]

req分析に使用できる必要な行だけが含まれます。

于 2016-10-31T18:12:14.677 に答える