擅长:python、mysql、java
<p>您可以尝试以下方法:</p>
<pre><code>import math
def get_next_k(arr,k,idx):
return arr[k*idx:k*idx+k]
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
k = 3
no_of_iters = int(math.ceil(float(len(arr))/k))
for i in range(no_of_iters):
print(get_next_k(arr,k,i)
</code></pre>
<p>输出</p>
<pre><code>[1, 2, 3]
[4, 5, 6]
[7, 8, 9]
[10]
</code></pre>