擅长:python、mysql、java
<p>我认为当你考虑迭代索引而不是列表上的项目时,它更容易:</p>
<pre><code>from itertools import product
d = [[1, 2, 3], [7, 9, 8], [4, 5, 6]]
# generate all indices
x_len = range(len(d))
y_len = range(len(d[0]))
indices = product(x_len, y_len)
# select maximal item by index
key = lambda x: d[x[0]][x[1]]
max_index = max(indices, key=key)
print(max_index)
</code></pre>