如何找到两个不同列表中每个元素的平均值答案

【问题标题】：How to find the average of each element in two different lists如何找到两个不同列表中每个元素的平均值
【发布时间】：2021-06-18 20:32:58
【问题描述】：

这些是我的数组

image =  [[0,0,0,0,0,0,0,0,0,0],
          [0,0,0,0,0,0,0,0,0,0],
          [0,0,0,0,0,0,0,0,0,0],
          [0,0,0,0,0,0,0,0,0,0],
          [0,0,0,0,0,0,0,0,0,0],
          [0,0,1,1,1,1,1,1,0,0],
          [0,0,1,1,1,1,1,1,0,0],
          [0,0,1,1,1,1,1,1,0,0],
          [0,0,0,0,0,0,0,0,0,0],
          [0,0,0,0,0,0,0,0,0,0] 
           ]
image1 = [[0,0,0,0,0,0,0,0,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0] 
           ]

我想从不同的数组中获取每个元素的平均值并列出新的平均值，请帮助

这是我到目前为止所做的：

【问题讨论】：

请发布您目前尝试过的代码 - 请参阅 How to Ask 和 minimal reproducible example。
ok 添加代码
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标签： python arrays list average

【解决方案1】：

import numpy as np
ans = (np.array(image) + np.array(image1))/2.0
ans = ans.tolist()

但这似乎是一个重复的问题 Average values in two Numpy arrays

【讨论】：

【解决方案2】：

Numpy 忍者代码。

import numpy as np
image =  [[0,0,0,0,0,0,0,0,0,0],
          [0,0,0,0,0,0,0,0,0,0],
          [0,0,0,0,0,0,0,0,0,0],
          [0,0,0,0,0,0,0,0,0,0],
          [0,0,0,0,0,0,0,0,0,0],
          [0,0,1,1,1,1,1,1,0,0],
          [0,0,1,1,1,1,1,1,0,0],
          [0,0,1,1,1,1,1,1,0,0],
          [0,0,0,0,0,0,0,0,0,0],
          [0,0,0,0,0,0,0,0,0,0] 
           ]
image1 = [[0,0,0,0,0,0,0,0,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0],
          [0,0,1,0,0,0,1,1,0,0] 
           ]
images = [image,image1]
np.array(images).mean(axis=0)

【讨论】：

【解决方案3】：

除了其他答案之外，您的代码将按以下方式运行：

imagi = []
for x, y in zip(img, img1):
    z = (x + y) / 2
    imagi.append(z)

【讨论】：