I have a problem like this :
syms al_1 be_1
simplify(log([ exp((4*pi*(al_1*1i - be_1*2i))/5), exp((4*pi*(al_1*1i - be_1*1i))/5)]))
but the matlab output instead of removing log and exp, keeps both
ans =
[ log(exp((4*pi*(al_1*1i - be_1*2i))/5)), log(exp((4*pi*(al_1*1i - be_1*1i))/5))]
is there a work around for this ?
Thanks horchler for your comment.
This is the solution,
simplify(log([ exp((4*pi*(al_1*1i - be_1*2i))/5), exp((4*pi*(al_1*1i - be_1*1i))/5)]),'IgnoreAnalyticConstraints', true )
ans =
[ (4*pi*(al_1*1i - be_1*2i))/5, (4*pi*(al_1*1i - be_1*1i))/5]
Related
data = {'desc': ['ADRIAN PETER - ANN 80020355787C - 11 Baillon Pass.pdf', 'AILEEN MARCUS - ANC 800E15432922 - 5 Mandarin Way.pdf',
'AJITH SINGH - ANN 80020837750 - 11 Berkeley Loop.pdf', 'ALEX MARTIN-CURTIS - ANC 80021710355 - 26 Dovedale St.pdf',
'Alice.Smith\Jodee - Karen - ANE 80020428377 - 58 Harrisdale Dr.pdf']}
df = pd.DataFrame(data, columns = ['desc'])
df
From the data frame, I want to create a new column called ID, and in that ID, I want to have only those values starting after ANN, ANC or ANE. So I am expecting a result as below.
ID
80020355787C
800E15432922
80020837750
80021710355
80020428377
I tried running the code below, but it did not get the desired result. Appreciate your help on this.
df['id'] = df['desc'].str.extract(r'\-([^|]+)\-')
You can use - AN[NCE] (800[0-9A-Z]+) -, where:
AN[NCE] matches literally AN followed by N or C or E;
800[0-9A-Z]+ matches literally 800 followed by one or more characters between 0 and 9 or between A and Z.
>>> df['desc'].str.extract(r'- AN[NCE] (800[0-9A-Z]+) -')
0
0 80020355787C
1 800E15432922
2 80020837750
3 80021710355
4 80020428377
If not all your ids start with "800", you can just remove it from the pattern.
In the first scenario link the solution provided was excellent and it did work. However I tried to make it work with another function and I ended up with nothing close to what is expected. My code so far:
yn = [-1.20449 -1.14398 -1.02273 -0.962285 -0.90203 -0.841474 -0.780881...
-0.720346 -0.659896 -0.579599 -0.539505 -0.478662 -0.418963 -0.35859...
-0.299039 -0.238886 -0.179108 -0.118999 -0.058841 -0.006249 -0.06189...
0.006332 0.04056 0.11813 0.1776723 0.238403 0.29827 0.358396...
0.418149 0.4786 0.478154 0.538114 0.53862 0.598954 0.659804...
0.720267 0.781026 0.8412 0.901548 0.962022 1.022567 1.083291...
1.143653];
xn = linspace(-22,22,43)';
yn = yn';
fn = fit(xn,yn,'poly1')
figure()
hold all
plot(fn,xn,yn)
yn_fit = f(xn);
error = yn - yn_fit;
figure()
plot(xn, error,'k')
I would appreciate any help.
Suppose that we have following array:
0.196238259763928
0.0886250228175519
0.417543614272817
0.182403230538167
0.136500793051860
0.389922187581014
0.0344012946153299
0.381603315802419
0.0997542838649466
0.274807632628596
0.601652859233616
0.209431489000677
0.396925294300794
0.0351587496999554
0.177321874549738
0.369200511917405
0.287108838007101
0.477076452316346
0.127558716868438
0.792431584110476
0.0459982776925879
0.612598437936600
0.228340227044324
0.190267907472804
0.564751537228850
0.00269368929400299
0.940538666131177
0.101588565140294
0.426175626669060
0.600215481734847
0.127859067121782
0.985881201195063
0.0945679498528667
0.950077461673118
0.415212985598547
0.467423473845033
1.24336273213410
0.0848695928658021
1.84522775800633
0.289288949281834
1.38792131632743
1.73186592736729
0.554254947026916
3.46075557122590
0.0872957577705428
4.93259798197976
2.03544238985229
3.71059303259615
8.47095716918618
0.422940369071662
25.2287636895831
4.14535369056670
63.7312173032838
152.080907190007
1422.19492782494
832.134744027851
0.0220089962114756
60.8238733887811
7.71053463387430
10.4151913932115
11.3141744831953
0.988978595613829
8.65598040591953
0.219820300144944
3.92785491164888
2.28370963778411
1.60232807621444
2.51086405960291
0.0181622519984990
2.27469230188760
0.487809730727909
0.961063613990814
1.90435488292485
0.515640996120482
1.25933693517960
0.0953200831348589
1.52851575480462
0.582109930768162
0.933543409438383
0.717947488528521
0.0445235241119612
1.21157308704582
0.0942421028083462
0.536069075206508
0.821400666720535
0.308956823975938
1.28706199713640
0.0339217632187507
1.19575886464231
0.0853733920496230
0.736744959694641
0.635218502184121
0.262305581223588
0.986899895695809
0.0398800891449550
0.758792061180657
0.134279188964854
0.442531129290843
0.542782326712391
0.377221037448628
0.704787750202814
0.224180325609783
0.998785634315287
0.408055416702400
0.329684702125840
0.522384453408780
0.154542718256493
0.602294251721841
0.240357912028348
0.359040779285709
0.525224294805813
0.427539247203335
0.624034405807298
0.298184846094056
0.498659616687732
0.0962076792277457
0.430092706132805
0.656212420735658
0.278310520474744
0.866037361133916
0.184971060800812
0.481149730712771
0.624405636807668
0.382388147099945
0.435350646037440
0.216499523971397
1.22960953802959
0.330841706900755
0.891793067878849
0.628241046456751
0.278687691121678
1.06358076764171
0.365652714373067
1.34921178081181
0.652888708375276
0.861138633227739
1.02878577330537
0.591174450919664
1.93594290806582
0.497631035062465
1.14486512201656
0.978067581547298
0.948931658572253
2.01004088022982
0.917415940349743
2.24124811810385
1.42691656876436
2.15636037453584
1.92812357585099
1.12786835077183
4.81721425534142
1.70055431306602
4.87939454466131
3.90293284926105
5.16542230018432
10.5783535493504
1.74023535081791
27.0572221453758
7.78813114379733
69.2528169436690
167.769806437531
1490.03057130613
869.247150795648
3.27543244752518
62.3527480644562
9.74192115073051
13.6074209231800
10.5686495478844
7.70239986387120
9.62850426896699
9.85304975304259
7.09026325332085
12.8782040428502
16.3163128995995
7.00070066635845
74.1532966917877
4.80506505312457
1042.52337489620
1510.37374385290
118.514435606795
80.7915675273571
2.96352221859211
27.7825124315786
1.55102367292252
8.66382951478539
5.02910503820560
1.25219344189599
7.72195587189507
0.356973215117373
6.06702456628919
1.01953617014621
2.76489896186652
3.35353608882459
0.793376336025486
4.90341095941571
0.00742857354167949
5.07665716731356
1.16863474789604
4.47635486149688
4.33050121578669
2.42974020115261
9.79494608790444
0.0568839453395247
22.9153086380666
4.48791386399205
59.6962194708933
97.8636220152072
1119.97978883924
806.144299041605
7.33252581243942
57.0699524267842
0.900104994068117
15.2791339483160
3.31266162202546
3.20809490583211
5.36617545130941
0.648122925703121
3.90480316969632
0.0338850542128927
2.58828964019220
0.543604662856673
1.16385064506181
1.01835324272839
0.172915006573539
1.55998411282069
0.00221570175453666
1.14803074836796
0.0769335878967426
0.421762398811163
0.468260146832541
0.203765185125597
0.467641715366303
0.00142988680149041
0.698088976126660
0.0413316717103625
0.190548157914037
0.504713663418641
0.325697764871308
0.375910057283262
0.123307135682793
0.331115262928959
0.00263961045860704
0.204555648718379
0.139008751575803
0.182936666944843
0.154943314848474
0.0840483576044629
0.293075999812128
0.00306911699543199
0.272993318570981
0.0864711337990886
0.280495615619829
0.0910123210559269
0.148399626645134
0.141945002415500
0.0512001531781583
0.0295283557338525
In MATLAB it is very easy to find peaks using findpeaks, like so:
[pxx_peaks,location] = findpeaks(Pxx);
If we plot pxx_peaks, we get
plot(pxx_peaks)
Of course, besides these peaks, there are smaller peaks which are not shown on the picture, but my goal is to find all peaks which are 95-96% above all other peaks.
I have tried like this:
>> average = mean(pxx_peaks);
>> stand = std(pxx_peaks);
>> final_peaks = pxx_peaks( pxx_peaks > average + 3*stand );
The result of this is
>> final_peaks
final_peaks =
1.0e+03 *
1.4222
1.4900
1.5104
1.1200
but how to return their corresponding locations? I want to write it as one m-file, so please help me
EDIT
also please help me in this question: can I parameterize the confidence interval? For instance instead of 95%, I want to find peaks that are 60% above then other peaks, is it possible?
Note that 3σ ≈ 99.73%
As for your first question, it's easy, you just have to keep track of the locations in the same way as you do for the peaks:
inds = pxx_peaks > mean(pxx_peaks) + 3*std(pxx_peaks);
final_peaks = pxx_peaks(inds);
final_locations = location(inds);
plot(Pxx), hold on
plot(final_locations, final_peaks, 'r.')
As for your second question, that's a little more complicated. If you want to formulate it like you say, you'll have to convert a desired percentage to the correct number of σ. That involves an integration of the standard normal, and a root finding:
%// Convert confidence interval percentage to number-of-sigmas
F = #(P) fzero(#(sig) quadgk(#(x) exp(-x.^2/2),-sig,+sig)/sqrt(2*pi) - P/100, 1);
% // Repeat with the desired percentage
inds = pxx_peaks > mean(pxx_peaks) + F(63)*std(pxx_peaks); %// 63%
final_peaks = pxx_peaks(inds);
final_locations = location(inds);
plot(final_locations, final_peaks, 'r.')
I have a question about level-set implementation.
In the article "Distance Regularized Level Set Evolution and Its Application to Image Segmentation" Chunming Li, Chenyang Xu you can find a diffusion equation:
(sorry, but i'm not allowed to post images :( )
(14) page 4 from pdf
For dp(s) = 1-1/s the implementation is [MATLAB]:
distRegTerm = 4*del2(phi)-curvature;
where:
[phi_x,phi_y]=gradient(phi);
s=sqrt(phi_x.^2 + phi_y.^2);
smallNumber=1e-10;
Nx=phi_x./(s+smallNumber);
Ny=phi_y./(s+smallNumber);
curvature=div(Nx,Ny);
and it is OK, because for that dp the equation is (15)
I don't understand why for (10)
where:
p(s) is (16)
the code is:
distRegTerm=distReg_p2(phi);
where:
function f = distReg_p2(phi)
[phi_x,phi_y]=gradient(phi);
s=sqrt(phi_x.^2 + phi_y.^2);
a=(s>=0) & (s<=1);
b=(s>1);
ps=a.*sin(2*pi*s)/(2*pi)+b.*(s-1);
dps=((ps~=0).*ps+(ps==0))./((s~=0).*s+(s==0));
f = div(dps.*phi_x - phi_x, dps.*phi_y - phi_y) + 4*del2(phi);
I don't understand the last line of this function.
Thanks
In Matlab, I have a IDPOLY object which I wish to "present" and store the resulting text in a text file.
eg if a is an IDPOLY object and I do
>> present(a)
I get
Discrete-time IDPOLY model: A(q)y(t) = B(q)u(t) + e(t)
A(q) = 1 - 1.31 q^-1 + 0.2425 q^-2 - 0.431 q^-3 + 0.4987 q^-4
B1(q) = 0.01357 + 0.04006 q^-1 - 0.04489 q^-2 + 0.007757 q^-3
- 0.1761 q^-4 + 0.06396 q^-5 - 0.2874 q^-6 + 0.3835 q^-7
B2(q) = -0.006397
Estimated using ARX with focus on data set iddata_est_shift
Loss function 0.0617185 and FPE 0.061879
Sampling interval: 0.025
Created: 24-Nov-2010 13:05:10
Last modified: 24-Nov-2010 13:06:56
Does anyone know how to capture this text as I'd like to write it to a text file in a log file. There are no returned arguments from present.
You can try using the diary command.
I've found the evalc() command which does exactly what I need.
evalc
Evaluate MATLAB expression with capture
Syntax
T = evalc(S)
[T, X, Y, Z, ...] = evalc(S)
and pass the expression as a string.