index error for find function

2014 3 6
3 个回答
回答已采纳
更新时间：2014 3 7
1 次查看（30 天）

0 个投票

I am trying to find the value of Signal_fft_hilbert at the values of newf. But i get the error:
??? Attempted to access Signal_fft_hilbert(14.6484); index must be a positive integer or logical.
I want to find the value at exactly 14.6484. How can I find the value?
newf=aa1(:)'
for s=1:length(newf)
ppks=find(Signal_fft_hilbert(newf(s)))
end

1 个评论
显示 -1更早的评论 隐藏 -1更早的评论

Dishant Arora
Dishant Arora 2014-3-6
You might have to go for interpolation to get the value at non- integral coordinates.

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Iain
Iain 2014-3-6

0 个投票

Ok, it looks like you've got a vector "Signal_fft_hilbert", which I'm going to call "v".
If want v(14.6484). The only way you can get v(14. ... ) is if you interpolate.
doc interp1 % will tell you all about one function for interpolation.
Alternatively, if you want the v that corresponds to something having a value you could do something like:
v( time == 14.684324)
or
v(find(time > 14.00))

4 个评论
显示 2更早的评论 隐藏 2更早的评论

ankita
ankita 2014-3-6
but 14.6484 isn't the only value. I have an array of such values in aa1. i want the v for all these values.
Iain
Iain 2014-3-6
You haven't told us everything we need. Let me use an analogy that tells you the method:
time = [0 0.1 0.2 0.3 0.4 .... ]; % or frequency, or number of bananas eaten, or "x"
voltage = [1 2 3 4.5 .... ]; % or current, or power, or apes fed, or "y".
Now, if you want to know y, when x = 0.05, then you need to "interpolate" or fill in the blanks.
interp1 is a simple method for automatically calculating the in-between parts, which needs to know, "x", "y" and the "x" you want to know y for.
If, instead, you have a list of times, which have known voltage, you can do something like:
times_i_want = [0 0.2 0.4 ... ];
for i = 1:numel(times_i_want)
voltage_at_time_i_want(i) = voltage(times_i_want(i) == time);
end
or...
voltage_at_time_i_want(i) = voltage(find(times_i_want(i) == time));
ankita
ankita 2014-3-7
Thanks alot Lain. I had 1 more doubt. This is my code
for k=1:length(peaks)
top1{k}=freq_s(locs(k))
for s1=1:length(peaks)
xx{s1}=top1{k}*s1;
end
[xx1]=[xx(:)]';
xx2=cell2mat(xx1);
aa1(k,:)=xx2;
In the above answers you can see my output aa1. Is there a method where I can get output like for each top1 with their subscripts included like top11 for i=1,top12 for i=2..... see my basic idea is to create different arrays of top1,which is a frequency, and show the values of corresponding Signal_fft_hilbert. The user can then pick which top1 values suit them and they will get the final Signal_fft_hilbert values. Any suggestions?
Chandrasekhar
Chandrasekhar 2014-3-7
编辑：Chandrasekhar 2014-3-7
hope this works replace
top1{k}=freq_s(locs(k)) with
evalin('base',['top1' num2str(k) '=freq_s(locs(k))';])
this will append the 'k' value to top1 every time and creates a new variable top1k in the workspace.

请先登录，再进行评论。

更多回答（2 个）

Chandrasekhar
Chandrasekhar 2014-3-6

0 个投票

k = find(newf == 14.6484);
ppks = Signal_fft_hilbert(k);

17 个评论
显示 15更早的评论 隐藏 15更早的评论

ankita
ankita 2014-3-6
aa1 has a set of values. I want to find the Signal_fft_hilbert of all those values.
Chandrasekhar
Chandrasekhar 2014-3-6
ok, but the index should be a scalar value. is there any relation between newf and Signal_fft_hilber?
ankita
ankita 2014-3-6
newf is the frequency and Signal_fft_hilbert gives the amplitude. I want to find the amplitude of these particular frequencies.
Chandrasekhar
Chandrasekhar 2014-3-6
does frequency at index1 of newf correspond to amplitude at index 1 of signal_fft_hilbert?
ankita
ankita 2014-3-6
I tried this.
k= find(newf==14.6484)
abs(Signal_fft_hilbert(k))
Output was
k =
Empty matrix: 1-by-0
ans =
Empty matrix: 1-by-0
Chandrasekhar
Chandrasekhar 2014-3-6
k= find(newf==14.6484) this statement should give k = ,because the first element of newf is 14.6484
ankita
ankita 2014-3-6
ya but it is giving me empty matrix 1-by-0
Chandrasekhar
Chandrasekhar 2014-3-6
Please find the data attached that i am using
ankita
ankita 2014-3-6
Ah. I am not able to open this file
Chandrasekhar
Chandrasekhar 2014-3-6
copy the file and load it into your matlab workspace. load('Data.mat'); have you done the same?
ankita
ankita 2014-3-6
ya i loaded your file and got index as 1. but it shows empty matrix in my program.
Chandrasekhar
Chandrasekhar 2014-3-6
Ok, you can use this data now. It is the same data that you had posted. what is the next step you have to do once you get the K value?
ankita
ankita 2014-3-6
I have to find the corresponding Signal_fft_hilbert values for the corresponding frequencies. But why is the output coming for your data and not mine?
Chandrasekhar
Chandrasekhar 2014-3-6
can you save your data as .mat file and attach it. I will check it. Also let me know do you have data for Signal_fft_hilbert or it is only a graph that you have.
ankita
ankita 2014-3-6
anki.m is my code and hilbertData.m is the Signal_fft_hilbert values
Chandrasekhar
Chandrasekhar 2014-3-6
the code(anki.m) doesnt execute. it is giving errors.
ankita
ankita 2014-3-6
maybe because of the file in the first 2-3 lines. you can check the code from line 136. there the Signal_fft_hilbert is calculated.

请先登录，再进行评论。

Niklas Nylén
Niklas Nylén 2014-3-6
编辑：Niklas Nylén 2014-3-6

0 个投票

I misunderstood the question, removed my response but kept the answer to not lose the comments below.

3 个评论
显示 1更早的评论 隐藏 1更早的评论

ankita
ankita 2014-3-6
I tried this already.Gives an error: ??? Subscript indices must either be real positive integers or logicals.
Chandrasekhar
Chandrasekhar 2014-3-6
can you share the data of aa1 and Signal_fft_hilbert
ankita
ankita 2014-3-6
aa1 =
14.6484 29.2969 43.9453 58.5938 73.2422 87.8906 102.5391
51.2695 102.5391 153.8086 205.0781 256.3477 307.6172 358.8867
65.9180 131.8359 197.7539 263.6719 329.5898 395.5078 461.4258
80.5664 161.1328 241.6992 322.2656 402.8320 483.3984 563.9648
95.2148 190.4297 285.6445 380.8594 476.0742 571.2891 666.5039
109.8633 219.7266 329.5898 439.4531 549.3164 659.1797 769.0430
124.5117 249.0234 373.5352 498.0469 622.5586 747.0703 871.5820
newf =
Columns 1 through 7
14.6484 51.2695 65.9180 80.5664 95.2148 109.8633 124.5117
Columns 8 through 14
29.2969 102.5391 131.8359 161.1328 190.4297 219.7266 249.0234
Columns 15 through 21
43.9453 153.8086 197.7539 241.6992 285.6445 329.5898 373.5352
Columns 22 through 28
58.5938 205.0781 263.6719 322.2656 380.8594 439.4531 498.0469
Columns 29 through 35
73.2422 256.3477 329.5898 402.8320 476.0742 549.3164 622.5586
Columns 36 through 42
87.8906 307.6172 395.5078 483.3984 571.2891 659.1797 747.0703
Columns 43 through 49
102.5391 358.8867 461.4258 563.9648 666.5039 769.0430 871.5820
this is the plot of Signal_fft_hilbert

请先登录，再进行评论。

类别

在 帮助中心 和 File Exchange 中查找有关 Creating, Deleting, and Querying Graphics Objects 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by