來自分組鄰居的熊貓時間三角洲

我有一組日期。我想從它們的前向鄰居中減去它們以獲得它們之間的增量。我的代碼是這樣的：來自分組鄰居的熊貓時間三角洲

import pandas, numpy, StringIO 


txt = '''ID,DATE 
002691c9cec109e64558848f1358ac16,2003-08-13 00:00:00 
002691c9cec109e64558848f1358ac16,2003-08-13 00:00:00 
0088f218a1f00e0fe1b94919dc68ec33,2006-05-07 00:00:00 
0088f218a1f00e0fe1b94919dc68ec33,2006-06-03 00:00:00 
00d34668025906d55ae2e529615f530a,2006-03-09 00:00:00 
00d34668025906d55ae2e529615f530a,2006-03-09 00:00:00 
0101d3286dfbd58642a7527ecbddb92e,2007-10-13 00:00:00 
0101d3286dfbd58642a7527ecbddb92e,2007-10-27 00:00:00 
0103bd73af66e5a44f7867c0bb2203cc,2001-02-01 00:00:00 
0103bd73af66e5a44f7867c0bb2203cc,2008-01-20 00:00:00 
''' 
df = pandas.read_csv(StringIO.StringIO(txt)) 
df = df.sort('DATE') 
df.DATE = pandas.to_datetime(df.DATE) 
grouped = df.groupby('ID') 
df['X_SEQUENCE_GAP'] = pandas.concat([g['DATE'].sub(g['DATE'].shift(), fill_value=0) for title,g in grouped])

我越來越難以理解的結果。所以，我要走了，我有一個邏輯錯誤。是

我得到的結果如下：

       ID    DATE  X_SEQUENCE_GAP 
0 002691c9cec109e64558848f1358ac16 2003-08-13 00:00:00 12277 days, 00:00:00 
1 002691c9cec109e64558848f1358ac16 2003-08-13 00:00:00    00:00:00 
3 0088f218a1f00e0fe1b94919dc68ec33 2006-06-03 00:00:00 27 days, 00:00:00 
2 0088f218a1f00e0fe1b94919dc68ec33 2006-05-07 00:00:00 13275 days, 00:00:00 
5 00d34668025906d55ae2e529615f530a 2006-03-09 00:00:00 13216 days, 00:00:00 
4 00d34668025906d55ae2e529615f530a 2006-03-09 00:00:00    00:00:00 
6 0101d3286dfbd58642a7527ecbddb92e 2007-10-13 00:00:00 13799 days, 00:00:00 
7 0101d3286dfbd58642a7527ecbddb92e 2007-10-27 00:00:00 14 days, 00:00:00 
9 0103bd73af66e5a44f7867c0bb2203cc 2008-01-20 00:00:00 2544 days, 00:00:00 
8 0103bd73af66e5a44f7867c0bb2203cc 2001-02-01 00:00:00 11354 days, 00:00:00

我期待爲exapme 0和1人同時具有0的結果。任何幫助最受讚賞。

來源

2013-04-15 BigHandsome

也許這個錯誤對某人來說並不難理解。發佈錯誤可以幫助我們更多。 – gustavodidomenico

升級到0.11.0rc1，並看看新的文檔和這些食譜：http://pandas.pydata.org/pandas-docs/dev/cookbook.html#miscellaneous，猜測你正在使用0.10.1， timedeltas有很多很好的變化 – Jeff

這是0.11rc1（我不認爲在以前的版本將工作）當你轉移日期，第一個是在NAT（如楠，但對於日期時間/ timedeltas）

In [27]: df['X_SEQUENCE_GAP'] = grouped.apply(lambda g: g['DATE']-g['DATE'].shift()) 

In [30]: df.sort() 
Out[30]: 
           ID    DATE  X_SEQUENCE_GAP 
0 002691c9cec109e64558848f1358ac16 2003-08-13 00:00:00     NaT 
1 002691c9cec109e64558848f1358ac16 2003-08-13 00:00:00   00:00:00 
2 0088f218a1f00e0fe1b94919dc68ec33 2006-05-07 00:00:00     NaT 
3 0088f218a1f00e0fe1b94919dc68ec33 2006-06-03 00:00:00 27 days, 00:00:00 
4 00d34668025906d55ae2e529615f530a 2006-03-09 00:00:00     NaT 
5 00d34668025906d55ae2e529615f530a 2006-03-09 00:00:00   00:00:00 
6 0101d3286dfbd58642a7527ecbddb92e 2007-10-13 00:00:00     NaT 
7 0101d3286dfbd58642a7527ecbddb92e 2007-10-27 00:00:00 14 days, 00:00:00 
8 0103bd73af66e5a44f7867c0bb2203cc 2001-02-01 00:00:00     NaT 
9 0103bd73af66e5a44f7867c0bb2203cc 2008-01-20 00:00:00 2544 days, 00:00:00

然後你可以填充（但你必須做一個numpy錯誤的這種類型轉換，將固定在0.12）。

In [57]: df['X_SEQUENCE_GAP'].sort_index().astype('timedelta64[ns]').fillna(0) 
Out[57]: 
0    00:00:00 
1    00:00:00 
2    00:00:00 
3  27 days, 00:00:00 
4    00:00:00 
5    00:00:00 
6    00:00:00 
7  14 days, 00:00:00 
8    00:00:00 
9 2544 days, 00:00:00 
Name: X_SEQUENCE_GAP, dtype: timedelta64[ns]

來源

2013-04-15 19:59:33 Jeff

來自分組鄰居的熊貓時間三角洲

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