計算列表中存儲的2D numpy數組的平均值

Question

A：

[array([[0.10865657, 0.10638294, 0.10471012, 0.09508586, 0.09283491],
        [0.10892282, 0.10664408, 0.10496752, 0.09531553, 0.09305617],
        [0.11664   , 0.1143077 , 0.11259081, 0.1026154 , 0.10025029],
        [0.11626453, 0.11392252, 0.11219875, 0.10217754, 0.09980005]]),
 array([[0.04213751, 0.04178241, 0.04158858, 0.04331489, 0.04447674],
        [0.04213751, 0.04178241, 0.04158858, 0.04331489, 0.04447674],
        [0.04267657, 0.04255925, 0.04253528, 0.04520177, 0.04655534],
                         ...

我可以做a[0].mean ，我會得到想要的結果。 我想用for循環來完成'a'的整個長度。

我努力了：

mean_all = []

for i in len(dist):
    mean = dist[i].mean
    mean_all.append(mean)

TypeError：“ int”對象不可迭代

Answer 1

首先， dist[0].mean返回一個函數，而不是均值。 通常，您需要dist[0].mean() 。

您可以使用列表推導輕松避免for循環：

from numpy import array

dist = [array([[0.10865657, 0.10638294, 0.10471012, 0.09508586, 0.09283491],
               [0.10892282, 0.10664408, 0.10496752, 0.09531553, 0.09305617],
               [0.11664   , 0.1143077 , 0.11259081, 0.1026154 , 0.10025029],
               [0.11626453, 0.11392252, 0.11219875, 0.10217754, 0.09980005]]),
        array([[0.04213751, 0.04178241, 0.04158858, 0.04331489, 0.04447674],
               [0.04213751, 0.04178241, 0.04158858, 0.04331489, 0.04447674],
               [0.04267657, 0.04255925, 0.04253528, 0.04520177, 0.04655534]])]

mean_all = [dist[i].mean() for i in range(len(dist))]

print(mean_all)
[0.10536720549999998, 0.04307523133333334]

---

如果您確實要使用for循環，請使用以下命令：

mean_all = []
for i in range(len(dist)):
    mean = dist[i].mean()
    mean_all.append(mean)

print(mean_all)
[0.10536720549999998, 0.04307523133333334]

Answer 2

使用使用range（）的正確格式

 for i in range(len(dist))