我正在處理一個時間序列數(shù)據(jù)幀，它看起來像這樣，除了超過數(shù)千行。我想創(chuàng)建一個新列，枚舉具有相同“符號”值的行塊。即第 0 行將是 0，第 1 行到第 23 行將是 1，第 24 行到第 30 行將是 2 等等......（時間順序很重要）實現(xiàn)這一點的最pythonic 方法是什么？先感謝您 Date sign0 2011-01-27 11 2011-01-28 -12 2011-01-31 -13 2011-02-01 -14 2011-02-02 -15 2011-02-07 -16 2011-02-08 -17 2011-02-09 -18 2011-02-10 -19 2011-02-11 -110 2011-02-14 -111 2011-02-15 -112 2011-02-16 -113 2011-02-17 -114 2011-02-18 -115 2011-02-21 -116 2011-02-22 -117 2011-02-23 -118 2011-02-24 -119 2011-02-25 -120 2011-02-28 -121 2011-03-01 -122 2011-03-02 -123 2011-03-03 -124 2011-03-04 125 2011-03-07 126 2011-03-08 127 2011-03-09 128 2011-03-10 129 2011-03-11 130 2011-03-14 131 2011-03-15 -132 2011-03-16 -133 2011-03-17 -134 2011-03-18 -135 2011-03-21 -136 2011-03-22 -137 2011-03-23 -138 2011-03-24 -139 2011-03-25 -140 2011-03-28 -141 2011-03-29 142 2011-03-30 1
2 回答
哆啦的時光機
TA貢獻1779條經(jīng)驗 獲得超6個贊
您可以使用以下方法獲取cumsum符號更改的位置diff：
df['new_column'] = (df.sign.diff()!=0).cumsum()-1
>>> df
Date sign new_column
0 2011-01-27 1 0
1 2011-01-28 -1 1
2 2011-01-31 -1 1
3 2011-02-01 -1 1
4 2011-02-02 -1 1
5 2011-02-07 -1 1
6 2011-02-08 -1 1
7 2011-02-09 -1 1
8 2011-02-10 -1 1
9 2011-02-11 -1 1
10 2011-02-14 -1 1
11 2011-02-15 -1 1
12 2011-02-16 -1 1
13 2011-02-17 -1 1
14 2011-02-18 -1 1
15 2011-02-21 -1 1
16 2011-02-22 -1 1
17 2011-02-23 -1 1
18 2011-02-24 -1 1
19 2011-02-25 -1 1
20 2011-02-28 -1 1
21 2011-03-01 -1 1
22 2011-03-02 -1 1
23 2011-03-03 -1 1
24 2011-03-04 1 2
25 2011-03-07 1 2
26 2011-03-08 1 2
27 2011-03-09 1 2
28 2011-03-10 1 2
29 2011-03-11 1 2
30 2011-03-14 1 2
31 2011-03-15 -1 3
32 2011-03-16 -1 3
33 2011-03-17 -1 3
34 2011-03-18 -1 3
35 2011-03-21 -1 3
36 2011-03-22 -1 3
37 2011-03-23 -1 3
38 2011-03-24 -1 3
39 2011-03-25 -1 3
40 2011-03-28 -1 3
41 2011-03-29 1 4
42 2011-03-30 1 4
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