<p>如果可以对数据进行预处理以删除重复的数字,并且只保留1个唯一的数字,那么也可以对平台使用相同的算法。因此,您可以将示例<code>[1, 2, 2, 2, 1]</code>转换为<code>[1, 2, 1]</code>,并应用相同的算法。在</p>
<p>编辑:
代码:</p>
<pre><code>from itertools import groupby
def process_data(data):
return [list(val for num in group) for val, group in groupby(data)]
def peaks(arr):
#print(arr)
posPeaks = {
"pos": [],
"peaks": [],
}
startFound = False
n = 0
while startFound == False:
if arr[n][0] == arr[n+1][0]:
n += 1
else:
startFound = True
endFound = False
m = len(arr) - 1
while endFound == False:
if arr[m][0] == arr[m-1][0]:
m -= 1
else:
endFound = True
for i in range(n+1, m):
if arr[i][0] == arr[i-1][0]:
None
elif arr[i][0] >= arr[i-1][0] and arr[i][0] >= arr[i+1][0]:
pos = sum([len(arr[idx]) for idx in range(i)])
posPeaks["pos"].append(pos) #.append(i)
posPeaks["peaks"].append(arr[i][0])
return posPeaks
print(peaks(process_data([0, 1, 2, 5, 1, 0])))
print(peaks(process_data([1, 2, 2, 2, 1])))
print(peaks(process_data([1, 2, 2, 2, 3])))
</code></pre>
<p>输出:</p>
^{pr2}$