Findpeaks the reason why my index is floating

Question

Life is Wonderful 2023-5-3

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1957244-findpeaks-the-reason-why-my-index-is-floating

评论： Star Strider 2023-5-3

Hi there,

I have a question about the findpeaks function. To my surprise, my location index is receiving a float, which is causing all of my calculations to be incorrect.

Please assist me in fixing the coding error. Expectation is that the peak value's position should be an integer number.

dctCoef = [8429.999786
2771.087211
607.721985
3118.100821
1112.991245
1013.033319
350.156147
368.204981
822.470058
971.052424
1472.593343
406.350435
893.834342
612.080075
-13.394000
138.237194
61.093121
127.999481
222.903100
313.873293];
minHeigth = mean(round(abs(dctCoef / 2.^8))) + std(round(abs(dctCoef / 2.^8)))*8e-1;  
[pks,locs] = findpeaks(round(abs(dctCoef / 2.^8)),20,'MinPeakHeight',minHeigth);
locs
locs = 0.1500
pks
pks = 12
% Manual analysis
fprintf('%20s|%20s|%20s|\n--------------------+--------------------+--------------------+\n', ...
        'indx','raw_data', 'scaled_data');
                indx|            raw_data|         scaled_data|
--------------------+--------------------+--------------------+
for i = 1:size(dctCoef,1)
        fprintf('%20d|%20.6f|%20.6f|\n',i,dctCoef(i), round(abs(dctCoef(i)/ 2.^8)))
end
                   1|         8429.999786|           33.000000|
                   2|         2771.087211|           11.000000|
                   3|          607.721985|            2.000000|
                   4|         3118.100821|           12.000000|
                   5|         1112.991245|            4.000000|
                   6|         1013.033319|            4.000000|
                   7|          350.156147|            1.000000|
                   8|          368.204981|            1.000000|
                   9|          822.470058|            3.000000|
                  10|          971.052424|            4.000000|
                  11|         1472.593343|            6.000000|
                  12|          406.350435|            2.000000|
                  13|          893.834342|            3.000000|
                  14|          612.080075|            2.000000|
                  15|          -13.394000|            0.000000|
                  16|          138.237194|            1.000000|
                  17|           61.093121|            0.000000|
                  18|          127.999481|            0.000000|
                  19|          222.903100|            1.000000|
                  20|          313.873293|            1.000000|

2 个评论
显示无隐藏无

Stephen23 2023-5-3

You are supplying the sample rate as the 2nd input: "If you specify a sample rate, Fs, then locs is a numeric vector of time instants with a time difference of 1/Fs between consecutive samples."

Source: https://www.mathworks.com/help/signal/ref/findpeaks.html#bufbbs1-locs

Your example seems to be consistent with the documentation.

Life is Wonderful 2023-5-3

Thank you for helping me out

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Star Strider 2023-5-3

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1957244-findpeaks-the-reason-why-my-index-is-floating#answer_1228479

在 MATLAB Online 中打开

The findpeaks function is interpreting the ‘20’ argument as a sampling frequency, and so is returning the position of the peak as the coordinate of the signal sampled at that frequency (20 cycles per x-coordinate unit).

Remove the ‘20’ and you get the result you expect —

dctCoef = [8429.999786
2771.087211
607.721985
3118.100821
1112.991245
1013.033319
350.156147
368.204981
822.470058
971.052424
1472.593343
406.350435
893.834342
612.080075
-13.394000
138.237194
61.093121
127.999481
222.903100
313.873293];
minHeigth = mean(round(abs(dctCoef / 2.^8))) + std(round(abs(dctCoef / 2.^8)))*8e-1;  
[pks,locs] = findpeaks(round(abs(dctCoef / 2.^8)),'MinPeakHeight',minHeigth);
locs
locs = 4
pks
pks = 12
% Manual analysis
fprintf('%20s|%20s|%20s|\n--------------------+--------------------+--------------------+\n', ...
        'indx','raw_data', 'scaled_data');
                indx|            raw_data|         scaled_data|
--------------------+--------------------+--------------------+
for i = 1:size(dctCoef,1)
        fprintf('%20d|%20.6f|%20.6f|\n',i,dctCoef(i), round(abs(dctCoef(i)/ 2.^8)))
end
                   1|         8429.999786|           33.000000|
                   2|         2771.087211|           11.000000|
                   3|          607.721985|            2.000000|
                   4|         3118.100821|           12.000000|
                   5|         1112.991245|            4.000000|
                   6|         1013.033319|            4.000000|
                   7|          350.156147|            1.000000|
                   8|          368.204981|            1.000000|
                   9|          822.470058|            3.000000|
                  10|          971.052424|            4.000000|
                  11|         1472.593343|            6.000000|
                  12|          406.350435|            2.000000|
                  13|          893.834342|            3.000000|
                  14|          612.080075|            2.000000|
                  15|          -13.394000|            0.000000|
                  16|          138.237194|            1.000000|
                  17|           61.093121|            0.000000|
                  18|          127.999481|            0.000000|
                  19|          222.903100|            1.000000|
                  20|          313.873293|            1.000000|