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如何计算每个迭代的平均值?

我想计算每个迭代中的平均适应度候选人,但我不知道怎么做。

import pandas as pd
import numpy as np

  while iteration < n_iterations:
        print('iteration     fitness_candidate')
        for i in range(n_particles):

            temp = []
            fitness_cadidate = fitness_function(particle_position_vector[i])
            print(iteration,' ', -(fitness_cadidate))

            temp.append(iteration)
            temp.append(particle_position_vector[i])
            temp.append(-(fitness_cadidate))
            ls.append(temp)

        iteration = iteration + 1

ls = pd.DataFrame(ls)

正如您可以看到的,每个迭代生成几个合适的候选者。所以我只需要计算迭代中适应度候选的平均值。如果它有4次迭代,那么它需要产生4个平均值。 输出:

iteration     fitness_candidate
0            20.24475
0            15.720000000000002
0            16.242250000000002
0            11.0975
0            20.923250000000007
0            15.720000000000002
0            22.924500000000002
0            17.472250000000003
0            24.247250000000005
0            24.305750000000003
iteration     fitness_candidate
1            21.72342
1            16.798420000000004
1            19.321920000000002
1            10.945920000000001
1            21.601420000000008
1            17.598920000000003
1            23.202420000000007
1            20.55192
1            24.124920000000003
1            24.305750000000003
iteration     fitness_candidate
2            22.801840000000002
2            19.47784
2            21.601090000000003
2            15.597339999999999
2            22.279590000000002
2            19.878089999999997
2            23.080090000000002
2            22.152920000000005
2            24.402840000000005
2            24.305750000000003
iteration     fitness_candidate
3            23.050510000000006
3            20.52701
3            21.44951
3            17.447010000000002
3            22.12801
3            19.72651
3            22.528260000000003
3            22.001340000000003
3            24.402840000000005
3            24.00259

问题来源StackOverflow 地址:/questions/59466834/how-to-calculate-average-for-each-iteration

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kun坤 2019-12-25 09:40:09 1020 分享 版权
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