在随时间变化的概率数据框中返回第一列名称，其中每行的值 <.5

Question

给定一个 pandas 数据框，如下所示，其中列名是时间，行是每个主题，值是概率返回列名（或时间）第一次概率小于每个主题的 50在数据框中。 概率总是从 1-0 递减。我尝试循环遍历数据帧，但计算效率不高。

科目编号	0	1	2	3	4	5	6	7	…	669	670	671
1	1	0.997913	0.993116	0.989017	0.976157	0.973078	0.968056	0.963685	…	0.156092	0.156092	0.156092
2	1	0.990335	0.988685	0.983145	0.964912	0.958	0.952	0.946995	…	0.148434	0.148434	0.148434
3	1	0.996231	0.990571	0.985775	0.976809	0.972736	0.969633	0.966116	…	0.17037	0.17037	0.17037
4	1	0.997129	0.994417	0.991054	0.978795	0.974216	0.96806	0.963039	…	0.15192	0.15192	0.15192
5	1	0.997728	0.993598	0.986641	0.98246	0.977371	0.972874	0.96816	…	0.154545	0.154545	0.154545
6	1	0.998134	0.995564	0.989901	0.986941	0.982313	0.972951	0.969645	…	0.17473	0.17473	0.17473
7	1	0.995681	0.994131	0.990401	0.974494	0.967941	0.961859	0.956636	…	0.144753	0.144753	0.144753
8	1	0.997541	0.994904	0.991941	0.983389	0.979375	0.973158	0.966358	…	0.158763	0.158763	0.158763
9	1	0.992253	0.989064	0.979258	0.955747	0.948842	0.942899	0.935784	…	0.150291	0.150291	0.150291

目标 Output

科目编号	时间概率 <.05
1	100
2	99
3	34
4	19
5	600
6	500
7	222
8	111
9	332

Answer 1

由于概率总是下降的，你可以这样做：

>>> df.set_index("subject id").gt(.98).sum(1)
subject id
1    4
2    4
3    4
4    4
5    5
6    6
7    4
8    5
9    3
dtype: int64

注意：我使用的是.98而不是.5 ，因为我只使用了一部分数据。

---

使用的数据

{'subject id': {0: 1, 1: 2, 2: 3, 3: 4, 4: 5, 5: 6, 6: 7, 7: 8, 8: 9},
 '0': {0: 1, 1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 1, 8: 1},
 '1': {0: 0.997913,
  1: 0.990335,
  2: 0.996231,
  3: 0.997129,
  4: 0.997728,
  5: 0.998134,
  6: 0.995681,
  7: 0.997541,
  8: 0.992253},
 '2': {0: 0.993116,
  1: 0.988685,
  2: 0.990571,
  3: 0.994417,
  4: 0.993598,
  5: 0.995564,
  6: 0.994131,
  7: 0.994904,
  8: 0.989064},
 '3': {0: 0.989017,
  1: 0.983145,
  2: 0.985775,
  3: 0.991054,
  4: 0.986641,
  5: 0.989901,
  6: 0.990401,
  7: 0.991941,
  8: 0.979258},
 '4': {0: 0.976157,
  1: 0.964912,
  2: 0.976809,
  3: 0.978795,
  4: 0.98246,
  5: 0.986941,
  6: 0.974494,
  7: 0.983389,
  8: 0.955747},
 '5': {0: 0.973078,
  1: 0.958,
  2: 0.972736,
  3: 0.974216,
  4: 0.977371,
  5: 0.982313,
  6: 0.967941,
  7: 0.979375,
  8: 0.948842},
 '6': {0: 0.968056,
  1: 0.952,
  2: 0.969633,
  3: 0.96806,
  4: 0.972874,
  5: 0.972951,
  6: 0.961859,
  7: 0.973158,
  8: 0.942899},
 '7': {0: 0.963685,
  1: 0.946995,
  2: 0.966116,
  3: 0.963039,
  4: 0.96816,
  5: 0.969645,
  6: 0.956636,
  7: 0.966358,
  8: 0.935784}}

Answer 2

如果我理解正确，我认为这就是您要寻找的：

df.where(df.lt(.5)).idxmax(axis=1)