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By : user2952922
Date : November 19 2020, 03:59 PM
seems to work fine I hope I get you right:
assume you have a vector a, and you would like to get a normal distribution that represents it. code :
``````>>a = randn(2); //your vector
>>x=[-10:0.1:10];
>> y = gaussmf(x,[std(a) mean(a)]); //std(a)=sigma,mean(a)=mu
>> plot(x,y);
`````` ## Function to transform empirical distribution to a uniform distribution in Matlab?

By : angel
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I know the procedure of transforming one distribution to another by the use of CDF. However, I would like to know if there is existing function in Matlab which can perform this task?
code :
``````for t=1:nT
[f_CDFTemp,x_CDFTemp]=ecdf(uncon_noise_columndata_all_nModels_diff_t(:,1,t)); % compute CDF of empirical distribution
f_CDF(1:length(f_CDFTemp),t) = f_CDFTemp; % store the CDF of different distributions with unequal size in a new variable
x_CDF(1:length(x_CDFTemp),t) = x_CDFTemp;
b_unifdist=4*t;
[Noise.N, Noise.X]=hist((a_unifdist+(b_unifdist-a_unifdist).*f_CDF(:,t)),100); % generate the uniform distribution by using the CDF of empirical distribution as the CDF of the uniform distribution
generatedNoise(:,:,t)=emprand(Noise.X,nRows,nCol); % sample some random numbers from the uniform distribution generated above by using 'emrand' function
end
`````` ## Transform data to fit normal distribution

By : Tesla Liu
Date : March 29 2020, 07:55 AM
To fix the issue you can do May be what you are interested in is rank-based inverse normal transformation. Basically you rank the data first an them convert it to normal distribution:
code :
``````rank = tiedrank( data );
p = rank / ( length(rank) + 1 ); %# +1 to avoid Inf for the max point
newdata = norminv( p, 0, 1 );
`````` ## Transform a normal distribution to linear

By : nvalenton
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Since 0 < Your data < 1, what about using Math.log10 to calculate the base 10 logarithm of your data?
For example, Math.log10(0.999943) = -0.000024755491006670377 while Math.log10(0.99902 ) = -0.00042581727682710913, then if you like, you can multiple the result with 100000 and get the result -2.4 and -42, the variance is exaggerated and their relationship is not changed, because log10 is an monotone increasing function. ## Find a python transformation function or numpy matrix to transform skewed normal distribution to normal distribution

By : Evelyn Wang
Date : March 29 2020, 07:55 AM
this one helps. I've done one of 2 things:
Use box-cox transformations. This requires you find the appropriate power or lambda that transforms you data to having zero skew. Force a normal distribution.
code :
``````from scipy.stats import norm

df = pd.DataFrame(np.random.rand(1000), columns=['Uniform'])
df['Normal'] = norm.ppf((df.Uniform.rank() - .5) / len(df))
df.plot(kind='kde')
``````
``````df.skew()

Uniform    2.392991e-02
Normal     2.114051e-15
dtype: float64
`````` ## Transform some kind of exponential distribution into normal distribution

By : user2071787
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I have the following exponential distribution, generated with the following code: , You can try to see if this helps:
code :
``````from scipy.stats import boxcox
tags_ratio = boxcox(tags_ratio, 0.3)
plt.hist(tags_ratio)
plt.show()
`````` 