鉴于这个pandas数据框有两列，'Values'和'Intervals'。如何获得第三列“MinMax”，指示该值是该区间内的最大值还是最小值？对我来说，挑战是间隔长度和间隔之间的距离不固定，因此我发布了这个问题。
+-------+----------+-----------+---------+
| index | Value | Intervals | Min/Max |
+-------+----------+-----------+---------+
| 0 | 1879.289 | np.nan | |
| 1 | 1879.281 | np.nan | |
| 2 | 1879.292 | 1 | |
| 3 | 1879.295 | 1 | |
| 4 | 1879.481 | 1 | |
| 5 | 1879.294 | 1 | |
| 6 | 1879.268 | 1 | min |
| 7 | 1879.293 | 1 | |
| 8 | 1879.277 | 1 | |
| 9 | 1879.285 | 1 | |
| 10 | 1879.464 | 1 | |
| 11 | 1879.475 | 1 | |
| 12 | 1879.971 | 1 | |
| 13 | 1879.779 | 1 | |
| 17 | 1879.986 | 1 | |
| 18 | 1880.791 | 1 | max |
| 19 | 1880.29 | 1 | |
| 55 | 1879.253 | np.nan | |
| 56 | 1878.268 | np.nan | |
| 57 | 1875.73 | 1 | |
| 58 | 1876.792 | 1 | |
| 59 | 1875.977 | 1 | min |
| 60 | 1876.408 | 1 | |
| 61 | 1877.159 | 1 | |
| 62 | 1877.187 | 1 | |
| 63 | 1883.164 | 1 | |
| 64 | 1883.171 | 1 | |
| 65 | 1883.495 | 1 | |
| 66 | 1883.962 | 1 | |
| 67 | 1885.158 | 1 | |
| 68 | 1885.974 | 1 | max |
| 69 | 1886.479 | np.nan | |
| 70 | 1885.969 | np.nan | |
| 71 | 1884.693 | 1 | |
| 72 | 1884.977 | 1 | |
| 73 | 1884.967 | 1 | |
| 74 | 1884.691 | 1 | min |
| 75 | 1886.171 | 1 | max |
| 76 | 1886.166 | np.nan | |
| 77 | 1884.476 | np.nan | |
| 78 | 1884.66 | 1 | max |
| 79 | 1882.962 | 1 | |
| 80 | 1881.496 | 1 | |
| 81 | 1871.163 | 1 | min |
| 82 | 1874.985 | 1 | |
| 83 | 1874.979 | 1 | |
| 84 | 1871.173 | np.nan | |
| 85 | 1871.973 | np.nan | |
| 86 | 1871.682 | np.nan | |
| 87 | 1872.476 | np.nan | |
| 88 | 1882.361 | 1 | max |
| 89 | 1880.869 | 1 | |
| 90 | 1882.165 | 1 | |
| 91 | 1881.857 | 1 | |
| 92 | 1880.375 | 1 | |
| 93 | 1880.66 | 1 | |
| 94 | 1880.891 | 1 | |
| 95 | 1880.377 | 1 | |
| 96 | 1881.663 | 1 | |
| 97 | 1881.66 | 1 | |
| 98 | 1877.888 | 1 | |
| 99 | 1875.69 | 1 | |
| 100 | 1875.161 | 1 | min |
| 101 | 1876.697 | np.nan | |
| 102 | 1876.671 | np.nan | |
| 103 | 1879.666 | np.nan | |
| 111 | 1877.182 | np.nan | |
| 112 | 1878.898 | 1 | |
| 113 | 1878.668 | 1 | |
| 114 | 1878.871 | 1 | |
| 115 | 1878.882 | 1 | |
| 116 | 1879.173 | 1 | max |
| 117 | 1878.887 | 1 | |
| 118 | 1878.68 | 1 | |
| 119 | 1878.872 | 1 | |
| 120 | 1878.677 | 1 | |
| 121 | 1877.877 | 1 | |
| 122 | 1877.669 | 1 | |
| 123 | 1877.69 | 1 | |
| 124 | 1877.684 | 1 | |
| 125 | 1877.68 | 1 | |
| 126 | 1877.885 | 1 | |
| 127 | 1877.863 | 1 | |
| 128 | 1877.674 | 1 | |
| 129 | 1877.676 | 1 | |
| 130 | 1877.687 | 1 | |
| 131 | 1878.367 | 1 | |
| 132 | 1878.179 | 1 | |
| 133 | 1877.696 | 1 | |
| 134 | 1877.665 | 1 | min |
| 135 | 1877.667 | np.nan | |
| 136 | 1878.678 | np.nan | |
| 137 | 1878.661 | 1 | max |
| 138 | 1878.171 | 1 | |
| 139 | 1877.371 | 1 | |
| 140 | 1877.359 | 1 | |
| 141 | 1878.381 | 1 | |
| 142 | 1875.185 | 1 | min |
| 143 | 1875.367 | np.nan | |
| 144 | 1865.492 | np.nan | |
| 145 | 1865.495 | 1 | max |
| 146 | 1866.995 | 1 | |
| 147 | 1866.672 | 1 | |
| 148 | 1867.465 | 1 | |
| 149 | 1867.663 | 1 | |
| 150 | 1867.186 | 1 | |
| 151 | 1867.687 | 1 | |
| 152 | 1867.459 | 1 | |
| 153 | 1867.168 | 1 | |
| 154 | 1869.689 | 1 | |
| 155 | 1869.693 | 1 | |
| 156 | 1871.676 | 1 | |
| 157 | 1873.174 | 1 | min |
| 158 | 1873.691 | np.nan | |
| 159 | 1873.685 | np.nan | |
+-------+----------+-----------+---------+
在下面的第三列中，您可以看到每个区间的最大值和最小值。
{{1}}
答案 0 :(得分：2)
isnull = data.iloc[:, 1].isnull()
minmax = data.groupby(isnull.cumsum()[~isnull])[0].agg(['idxmax', 'idxmin'])
data.loc[minmax['idxmax'], 'MinMax'] = 'max'
data.loc[minmax['idxmin'], 'MinMax'] = 'min'
data.MinMax = data.MinMax.fillna('')
print(data)
0 1 MinMax
0 1879.289 NaN
1 1879.281 NaN
2 1879.292 1.0
3 1879.295 1.0
4 1879.481 1.0
5 1879.294 1.0
6 1879.268 1.0 min
7 1879.293 1.0
8 1879.277 1.0
9 1879.285 1.0
10 1879.464 1.0
11 1879.475 1.0
12 1879.971 1.0
13 1879.779 1.0
14 1879.986 1.0
15 1880.791 1.0 max
16 1880.290 1.0
17 1879.253 NaN
18 1878.268 NaN
19 1875.730 1.0 min
20 1876.792 1.0
21 1875.977 1.0
22 1876.408 1.0
23 1877.159 1.0
24 1877.187 1.0
25 1883.164 1.0
26 1883.171 1.0
27 1883.495 1.0
28 1883.962 1.0
29 1885.158 1.0
.. ... ... ...
85 1877.687 1.0
86 1878.367 1.0
87 1878.179 1.0
88 1877.696 1.0
89 1877.665 1.0 min
90 1877.667 NaN
91 1878.678 NaN
92 1878.661 1.0 max
93 1878.171 1.0
94 1877.371 1.0
95 1877.359 1.0
96 1878.381 1.0
97 1875.185 1.0 min
98 1875.367 NaN
99 1865.492 NaN
100 1865.495 1.0 min
101 1866.995 1.0
102 1866.672 1.0
103 1867.465 1.0
104 1867.663 1.0
105 1867.186 1.0
106 1867.687 1.0
107 1867.459 1.0
108 1867.168 1.0
109 1869.689 1.0
110 1869.693 1.0
111 1871.676 1.0
112 1873.174 1.0 max
113 1873.691 NaN
114 1873.685 NaN
[115 rows x 3 columns]
答案 1 :(得分：1)
data.columns=['Value','Interval']
data['Ingroup'] = (data['Interval'].notnull() + 0)
Use data['Interval'].notnull() to separate the groups...
Use cumsum() to number them with `groupno`...
Use groupby(groupno)..
Finally you want something using apply/idxmax/idxmin to label the max/min
But of course a for-loop as you suggested is the non-Pythonic but possibly simpler hack.