擅长:python、mysql、java
<p>使用for循环:</p>
<pre><code>def nzeros(a, n):
#Create a numpy array of zeros of length equal to n
b = np.zeros(n)
#Create a numpy array of zeros of same length as array a
c = np.zeros(len(a), dtype=int)
for i in range(0,len(a) - n):
if (b == a[i : i+n]).all(): #Check if array b is equal to slice in a
c[i+n-1] = 1
return c
</code></pre>
<p>样本输出:</p>
^{pr2}$