Question

在下面的代码中，我声明三个列表只是为了查找三个变量的均值。代码可以更简洁吗？例如，使用一个列表。

import numpy as np
rmse_train_arr=[]
wt_error_train_arr=[]
rmse_test_arr=[]
for i in range(5):
    rmse_train = i
    wt_error_train=i+5
    rmse_test=i+6

    rmse_train_arr.append(rmse_train)
    wt_error_train_arr.append(wt_error_train)
    rmse_test_arr.append(rmse_test)

print(np.array(rmse_train_arr).mean(),np.array(wt_error_train_arr).mean(),np.array(rmse_test_arr).mean())

for i in range(5)只是为了简化此处的目的，实际上，您可以假设您在for循环中不知道rmse_train，wt_error_train等的确切值。目的仅仅是在末尾找到自己的意思。

Answer 1

认为这是保留列表的更好方法：

rmse_train_mean = 0
wt_error_train_mean = 0
rmse_test_mean = 0
counter = 0  # if you don't know how many elements there are

for i in range(5):
    rmse_train = i
    wt_error_train = i + 5
    rmse_test = i + 6

    rmse_train_mean += rmse_train
    wt_error_train_mean += wt_error_train
    rmse_test_mean += rmse_test

    counter += 1

print(rmse_train_mean / counter, wt_error_train_mean / counter, rmse_test_mean / counter)

输出

2.0 7.0 8.0

至少有一种方法可以保留一个列表，但前提是必须这样：

import numpy as np

one = []

for i in range(5):
    rmse_train = i
    wt_error_train = i + 5
    rmse_test = i + 6

    one.extend([rmse_train, wt_error_train, rmse_test])

result = np.mean([one[::3], one[1::3], one[2::3]], axis=1)

输出

[2. 7. 8.]

第一种方法可以一次性完成所有操作，并且不会在内存中保留多个（或一个）列表。

使用一个列表查找三个变量的均值

1 个答案: