<p>我知道我可能会迟到一点,但我想分享一下我使用NumPy阵列的解决方案:</p>
<pre class="lang-py prettyprint-override"><code>def get_level_peaks(v):
peaks = []
i = 1
while i < v.size-1:
pos_left = i
pos_right = i
while v[pos_left] == v[i] and pos_left > 0:
pos_left -= 1
while v[pos_right] == v[i] and pos_right < v.size-1:
pos_right += 1
is_lower_peak = v[pos_left] > v[i] and v[i] < v[pos_right]
is_upper_peak = v[pos_left] < v[i] and v[i] > v[pos_right]
if is_upper_peak or is_lower_peak:
peaks.append(i)
i = pos_right
peaks = np.array(peaks)
"""
# uncomment this part of the code
# to include first and last positions
first_pos, last_pos = 0, v.size-1
peaks = np.append([first_pos], peaks)
peaks = np.append(peaks, [last_pos])
"""
return peaks
</code></pre>
<ul>
<li>例1(<a href="https://i.stack.imgur.com/vDxIc.png" rel="nofollow noreferrer">see graph</a>):</li>
</ul>
^{pr2}$
<ul>
<li>例2(<a href="https://i.stack.imgur.com/H4gOK.png" rel="nofollow noreferrer">see graph</a>):</li>
</ul>
<pre><code>v = np.array([8, 2, 1, 0, 1, 2, 2, 5, 9, 3])
p = get_peaks(v)
print(v) # [8 2 1 0 1 2 2 5 9 3]
print(p) # [0 3 8 9] (peak indexes)
print(v[p]) # [8 0 9 3] (peak elements)
</code></pre>
<ul>
<li>例3(<a href="https://i.stack.imgur.com/uIRmt.png" rel="nofollow noreferrer">see graph</a>):</li>
</ul>
<pre><code>v = np.array([9, 8, 8, 8, 0, 8, 9, 9, 9, 6])
p = get_peaks(v)
print(v) # [9 8 8 8 0 8 9 9 9 6]
print(p) # [0 4 6 9] (peak indexes)
print(v[p]) # [9 0 9 6] (peak elements)
</code></pre>
<hr/>
<p>在示例3中,我们有一个从索引6到索引8的平坦的上限。在这种情况下,索引将始终指示平台的最左侧位置。如果要指明中间位置或最右侧位置,只需更改代码的以下部分:</p>
<pre class="lang-py prettyprint-override"><code> ...
if is_upper_peak or is_lower_peak:
peaks.append(i)
...
</code></pre>
<p>为此:</p>
<pre class="lang-py prettyprint-override"><code> ...
# middle position
if is_upper_peak or is_lower_peak:
peaks.append((pos_left + pos_right) // 2)
...
</code></pre>
<pre class="lang-py prettyprint-override"><code> ...
# rightmost position
if is_upper_peak or is_lower_peak:
peaks.append(pos_right)
...
</code></pre>