我有一个排序数组（如下所示），我可以使用哪些方法来计算等效目标响应，从而计算每个值的平均值？例如将与target = 15.26相对应的所有值相加（在其他试验中不是相同的值）。
我考虑过循环查找第3列中的类似数字，但肯定有一个更简单的解决方案。
非常感谢， 新手
response target
103 4.556049 15.260
55 10.549498 15.260
31 18.497221 15.260
130 13.275155 15.260
93 6.621331 15.260
35 7.653972 15.260
149 15.808224 15.260
84 4.556049 15.260
113 8.996922 15.260
152 18.282948 15.260
162 14.606229 15.260
60 5.588690 15.260
57 7.653972 15.260
159 20.902759 15.260
23 11.645647 15.260
40 21.385003 25.367
76 19.298551 25.367
71 17.724806 25.367
70 11.639967 25.367
178 36.359849 25.367
65 16.947743 25.367
32 25.811419 25.367
52 27.309781 25.367
26 27.160049 25.367
179 34.706590 25.367
54 22.157935 25.367
119 13.888544 25.367
123 39.826426 25.367
147 36.674288 25.367
151 33.022869 25.367
175 46.078035 39.536
4 34.567184 39.536
45 34.130893 39.536
8 51.218523 39.536
42 35.367997 39.536
78 26.210535 39.536
157 43.627062 39.536
116 22.879751 39.536
102 25.996262 39.536
135 45.632451 39.536
126 34.580608 39.536
125 40.390764 39.536
30 35.767629 39.536
86 33.775664 39.536
94 30.904922 39.536
88 68.897857 59.655
177 64.219992 59.655
171 63.937565 59.655
74 54.867877 59.655
163 39.521796 59.655
75 68.286534 59.655
148 70.700332 59.655
115 47.631128 59.655
167 55.704317 59.655
80 51.786992 59.655
0 54.931901 59.655
12 46.967656 59.655
10 62.339037 59.655
3 64.174040 59.655
43 66.948747 59.655
44 75.237758 89.387
101 82.690846 89.387
27 74.046606 89.387
105 94.395834 89.387
108 63.940663 89.387
112 111.211880 89.387
161 70.394671 89.387
29 116.047222 89.387
164 86.483221 89.387
96 83.612994 89.387
51 90.062871 89.387
49 99.448547 89.387
120 64.238064 89.387
121 99.623064 89.387
136 87.784865 89.387
53 99.587954 119.710
90 99.497598 119.710
155 117.134593 119.710
2 118.382540 119.710
87 123.984619 119.710
173 126.473800 119.710
124 128.213801 119.710
62 104.233807 119.710
142 151.525160 119.710
77 84.349268 119.710
46 128.496744 119.710
137 135.726266 119.710
6 106.980116 119.710
109 135.305464 119.710
56 146.565384 119.710
21 149.950898 155.140
144 138.947073 155.140
132 157.788645 155.140
25 138.816444 155.140
98 159.238989 155.140
97 136.179079 155.140
18 160.264919 155.140
92 106.939843 155.140
50 133.825173 155.140
156 160.650610 155.140
169 164.086207 155.140
7 120.081751 155.140
82 144.995253 155.140
73 148.619307 155.140
160 155.345932 155.140
154 286.343698 241.970
20 238.666653 241.970
17 243.265521 241.970
61 233.941803 241.970
67 225.647113 241.970
134 238.871632 241.970
141 257.964136 241.970
39 237.710944 241.970
106 267.179426 241.970
158 288.864375 241.970
104 219.470369 241.970
38 221.280073 241.970
36 216.673977 241.970
128 255.494058 241.970
91 222.512530 241.970
9 248.174697 281.250
143 339.346073 281.250
165 319.828122 281.250
166 339.152453 281.250
172 311.936161 281.250
14 229.413155 281.250
153 362.308915 281.250
117 334.014030 281.250
99 266.162791 281.250
85 307.998184 281.250
118 322.768051 281.250
22 247.992436 281.250
100 282.320528 281.250
24 308.043620 281.250
48 277.614265 281.250
89 324.674307 312.340
34 319.110436 312.340
83 294.367320 312.340
107 256.297453 312.340
5 330.217008 312.340
127 394.634200 312.340
66 335.137544 312.340
63 303.852646 312.340
16 336.398915 312.340
133 401.600397 312.340
176 336.454678 312.340
122 367.271789 312.340
1 330.173121 312.340
140 389.322293 312.340
33 306.170925 312.340
170 463.588300 365.130
68 354.929661 365.130
174 511.082051 365.130
41 407.971277 365.130
81 352.324308 365.130
19 455.697372 365.130
95 347.397060 365.130
13 374.191002 365.130
15 471.887121 365.130
146 420.940734 365.130
114 365.869462 365.130
138 466.096069 365.130
11 421.345013 365.130
139 451.122771 365.130
111 358.154084 365.130
129 513.113772 415.210
131 465.486811 415.210
58 498.471436 415.210
59 480.860257 415.210
64 435.301676 415.210
37 401.883341 415.210
28 520.876652 415.210
69 402.135305 415.210
145 514.131956 415.210
47 506.972655 415.210
72 402.655756 415.210
79 402.615483 415.210
150 519.844011 415.210
168 504.783972 415.210
110 435.998192 415.210
mean_plain = []
for i in range(0,11):
mean_plain.append([a[i][1].response.mean(),a[i][0]])
坚持到这里，如此接近但不是正确的格式（理想情况下我希望原始列目标和响应
答案 0 :(得分：1)
继续使用groupby的UlfR想法，这很好用
a=list(data.groupby('target'))
a[0].response.mean()
只需循环完成并立即存储值
答案 1 :(得分：0)
最简单的事情就是如果您拥有数据库中的数据。如果数据已经来自数据库，或者您可以将其放在数据库中，则可以方便地执行以下操作：
select target, count(*), avg(col1), avg(response)
from your_table
group by target
如果您无法访问数据库，那么您可以尝试https://www.sqlite.org/，如果您渴望了解数据库使用的基础知识。