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By : Christopher Lemoine
Date : October 17 2020, 03:08 PM
I think the issue was by ths following , I would use: scipy.signal.find_peaks_cwt().
From its documentation: code :
``````import numpy as np
import scipy as sp

import matplotlib.pyplot as plt

import scipy.signal

print('array size: ', y_arr.shape)

arr_size = len(y_arr)
expected_num = 30

expected_width = arr_size // expected_num // 2
print('expected width of peaks: ', expected_width)

peaks = sp.signal.find_peaks_cwt(y_arr, np.linspace(2, expected_width, 10))

print('num peaks: ', len(peaks))
print('peaks: ', peaks)

plt.plot(y_arr)
for peak in peaks:
plt.axvline(peak)
plt.show()
`````` ## Finding local maxima in a 2D array

By : Thibault Laurent
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I am pretty sure this cannot be solved in less than O(n^2) comparisons. Assume a chess board 2d matrix where all the white squares are 1 and blacks are 0. It wiil will have O(n^2) solutions and each solution requires at least one comparison. ## Finding the local maxima in a 1D array

By : Kannan TN
Date : March 29 2020, 07:55 AM
This might help you Is there an easy way to find the local maxima in a 1D array?
code :
``````maxes = []
for (var i = 1; i < a.length - 1; ++i) {
if (a[i-1] < a[i] && a[i] > a[i+1])
maxes.push(a[i])
}
`````` ## C++ Finding local maxima in 2d array

By : vitthal
Date : March 29 2020, 07:55 AM
it fixes the issue You are overwriting your (single) float localmax.
There are two solutions to your problem:
code :
``````int main()
{
float array = { { 1, 22 }, { 2, 15 }, { 3, 16 }, { 4, 14 }, { 5, 13 }, {6,19}, {7,12} };
int i;
float before = 0, after = 0, localmax = 0;
int Index = 0;

for (i = 0; i<7; i++)
{
if ((array[i] >= before) && (array[i] >= after))
{
before = array[i-1];
after = array[i + 1];
localmax = array[i];
cout << "A local maxima is: " << localmax << endl;
Index = i;
}
}

_getch();
return 0;
}
``````
``````    int main()
{
float array = { { 1, 22 }, { 2, 15 }, { 3, 16 }, { 4, 14 }, { 5, 13 }, {6,19}, {7,12} };
int i;
float before = 0, after = 0, localmax = 0;
int Index = 0;
std::vector<float> localMaxVector;

for (i = 0; i<7; i++)
{
if ((array[i] >= before) && (array[i] >= after))
{
before = array[i-1];
after = array[i + 1];
localMaxVector.push_back(array[i]);
Index = i;
}
}

cout << "The local maxima in the array are " << endl;
for( std::vector<float>::const_iterator i = localMaxVector.begin(); i != localMaxVector.end(); ++i)
std::cout << *i << ' ';
_getch();
return 0;
}
`````` ## finding global maxima of a function from comparing each processor's local maxima using MPI ring topology

By : iamdevops
Date : March 29 2020, 07:55 AM
it should still fix some issue For this sort of thing, better using either MPI_Reduce() or MPI_Allreduce() with MPI_MAX as operator. The former will compute the max over the values exposed by all processes and give the result to the "root" process only, while the later will do the same, but give the results to all processes.
code :
``````// Only process of rank 0 get the global max
MPI_Reduce( &local_max, &global_max, 1, MPI_DOUBLE, MPI_MAX, 0, MPI_COMM_WORLD );
// All processes get the global max
MPI_Allreduce( &local_max, &global_max, 1, MPI_DOUBLE, MPI_MAX, MPI_COMM_WORLD );
// All processes get the global max, stored in place of the local max
// after the call ends - this might be the most interesting one for you
MPI_Allreduce( MPI_IN_PLACE, &max, 1, MPI_DOUBLE, MPI_MAX, MPI_COMM_WORLD );
``````
``````if (rank==0)
{
max_store = max;
}
else if (rank==1)
{
max_store = max;
}
else if (rank ==2)
{
max_store = max;
}
else if (rank ==3)
{
max_store = max;
}
``````
``````if ( rank < 4 && rank >= 0 ) {
max_store[rank] = max;
}
`````` ## Finding local maxima within an array with conditions [JS]

By : Victor
Date : March 29 2020, 07:55 AM
Hope this helps I'm struggling to find a solution the following code challenge that meets all of the requirements and could use some help: , You could add a while loop for getting the end of a plateau.
code :
``````function getLocalMaxima(array) {
return array.reduce(function (r, v, i, a) {
var j = i;
while (v === a[++j]);
if (a[i - 1] < v && (a[i + 1] < v || a[i + 1] === v && a[j] < v)) {
r.pos.push(i);
r.peaks.push(v);
}
return r;
}, { pos: [], peaks: []});
}

var ex1 = [5, 5, 2, 1, 4, 2, 6, 2, 1, 2, 7, 7],  // { pos: [4, 6], peaks:[4, 6] }
ex2 = [3, 2, 3, 6, 4, 1, 2, 3, 2, 1, 2, 3],  // { pos: [3, 7], peaks: [6, 3]}
plateau = [1, 2, 2, 2, 1];                   // { pos: , peaks }

console.log(getLocalMaxima(ex1));
console.log(getLocalMaxima(ex2));
console.log(getLocalMaxima(plateau));``````
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