I need to efficiently process very large 1D arrays extracting some statistics per bin and I have found very useful the function binned_statistic from scipy.stats as it includes a 'statistic' argument that works quite efficiently.
I would like to perform a 'count' function but without considering zero values.
I am working in parallel with sliding windows (pandas rolling function) over the same arrays and it work nicely to substitute zeroes to NaN, but this behavior is not shared to my case.
This is a toy example of what I am doing:
import numpy as np
import pandas as pd
from scipy.stats import binned_statistic
# As example with sliding windows, this returns just the length of each window:
a = np.array([1., 0., 0., 1.])
pd.Series(a).rolling(2).count() # Returns [1.,2.,2.,2.]
# You can make the count to do it only if not zero:
nonzero_a = a.copy()
nonzero_a[nonzero_a==0.0]='nan'
pd.Series(nonzero_a).rolling(2).count() # Returns [1.,1.,0.,1.]
# However, with binned_statistic I am not able to do anything similar:
binned_statistic(range(4), a, bins=2, statistic='count')[0]
binned_statistic(range(4), nonzero_a, bins=2, statistic='count')[0]
binned_statistic(range(4), np.array([1., False, None, 1.], bins=2, statistic='count')[0]
All the previous runs provide the same output: [2., 2.] but I am expecting [1., 1.].
The only option found is to pass a custom function but it performs considerably worst than the implemented functions with real cases.
binned_statistic(range(4), a, bins=2, statistic=np.count_nonzero)
I have found and easy way to replicate the nonzero count transforming the array to 0-1 and applying sum:
# Transform all non-zero to 1s
a = np.array([1., 0., 0., 2.])
nonzero_a = a.copy()
nonzero_a[nonzero_a>0.0]=1.0 # nonzero_a = [1., 0., 0., 1.]
binned_statistic(np.arange(len(nonzero_a)), nonzero_a, bins=bins, statistic='sum')[0] # Returns [1.0, 1.0]
Related
I am trying to multiply two Gaussian distributions to obtain posterior for GMM data. In order to do that, I am trying to use .prob() function from tf.contrib.distributions.MultivariateNormalDiag, but every time I am getting the same error, even if I am providing the argument with float64.
I am using TensorFlow 1.8 version.
x = tf.placeholder(tf.float64, [None,2], name="input")
likelihood = tf.contrib.distributions.MultivariateNormalDiag(loc = [0., 0., 0.], scale_diag= [1., 1., 1.])
y_LL = likelihood.prob(x).eval()
TypeError: Input had dtype <dtype: 'float32'> but expected <dtype: 'float64'>.
I am confused whether I am doing it the wrong way, or what? Can someone please help me with this?
For this example, you are using x as a tf.float64. Unless you explicitly specify, tensorflow will auto-convert list inputs to tf.float32. You want to do something like (not executable code, but demonstrating you need to signal float64):
import numpy as np
likelihood = tf.contrib.distributions.MultivariateNormalDiag(loc=np.float64([0., 0., 0.]), scale_diag=np.float64([1., 1., 1.]))
y_LL = likelihood.prob(x).eval()
I have a 150x4 matrix X which I created from a pandas dataframe using the following code:
X = df_new.as_matrix()
I have to normalize it using this function:
I know that Uj is the mean val of j, and that σ j is the standard deviation of j, but I don't understand what j is. I'm having a little trouble understanding what the bar on X is, and I'm confused by the commas in the equation (I don't know if they have any significance or not).
Can anyone help me understand what this equation means so I can then write the normalization using sklearn?
You don't actually need to write code for the normalization yourself - it comes ready with sklearn.preprocessing.scale.
Here is an example from the docs:
>>> from sklearn import preprocessing
>>> import numpy as np
>>> X_train = np.array([[ 1., -1., 2.],
... [ 2., 0., 0.],
... [ 0., 1., -1.]])
>>> X_scaled = preprocessing.scale(X_train)
>>> X_scaled
array([[ 0. ..., -1.22..., 1.33...],
[ 1.22..., 0. ..., -0.26...],
[-1.22..., 1.22..., -1.06...]])
When used with the default setting axis=0, the mormalization happens column-wise (i.e. for each column j, as in your equestion). As a result, it is easy to confirm that scaled data has zero mean and unit variance:
>>> X_scaled.mean(axis=0)
array([ 0., 0., 0.])
>>> X_scaled.std(axis=0)
array([ 1., 1., 1.])
The indexes for matrix X are row (i) and column (j). Hence, X,j means column j of matrix X. I.e. normalize each column of matrix X to z-scores.
You can do that using pandas:
df_new_zscores = (df_new - df_new.mean()) / df_new.std()
I do not know pandas but I think that the equation means that the normalized matrix is given by
You subtract the empirical mean and devide by the empirical standard deviation per column.
You sometimes use this for Principal Component Analysis.
I have a 2D Numpy array, in which I want to normalise each column to zero mean and unit variance. Since I'm primarily used to C++, the method in which I'm doing is to use loops to iterate over elements in a column and do the necessary operations, followed by repeating this for all columns. I wanted to know about a pythonic way to do so.
Let class_input_data be my 2D array. I can get the column mean as:
column_mean = numpy.sum(class_input_data, axis = 0)/class_input_data.shape[0]
I then subtract the mean from all columns by:
class_input_data = class_input_data - column_mean
By now, the data should be zero mean. However, the value of:
numpy.sum(class_input_data, axis = 0)
isn't equal to 0, implying that I have done something wrong in my normalisation. By isn't equal to 0, I don't mean very small numbers which can be attributed to floating point inaccuracies.
Something like:
import numpy as np
eg_array = 5 + (np.random.randn(10, 10) * 2)
normed = (eg_array - eg_array.mean(axis=0)) / eg_array.std(axis=0)
normed.mean(axis=0)
Out[14]:
array([ 1.16573418e-16, -7.77156117e-17, -1.77635684e-16,
9.43689571e-17, -2.22044605e-17, -6.09234885e-16,
-2.22044605e-16, -4.44089210e-17, -7.10542736e-16,
4.21884749e-16])
normed.std(axis=0)
Out[15]: array([ 1., 1., 1., 1., 1., 1., 1., 1., 1., 1.])
Consider singular value decomposition M=USV*. Then the eigenvalue decomposition of M* M gives M* M= V (S* S) V*=VS* U* USV*. I wish to verify this equality with numpy by showing that the eigenvectors returned by eigh function are the same as those returned by svd function:
import numpy as np
np.random.seed(42)
# create mean centered data
A=np.random.randn(50,20)
M= A-np.array(A.mean(0),ndmin=2)
# svd
U1,S1,V1=np.linalg.svd(M)
S1=np.square(S1)
V1=V1.T
# eig
S2,V2=np.linalg.eigh(np.dot(M.T,M))
indx=np.argsort(S2)[::-1]
S2=S2[indx]
V2=V2[:,indx]
# both Vs are in orthonormal form
assert np.all(np.isclose(np.linalg.norm(V1,axis=1), np.ones(V1.shape[0])))
assert np.all(np.isclose(np.linalg.norm(V1,axis=0), np.ones(V1.shape[1])))
assert np.all(np.isclose(np.linalg.norm(V2,axis=1), np.ones(V2.shape[0])))
assert np.all(np.isclose(np.linalg.norm(V2,axis=0), np.ones(V2.shape[1])))
assert np.all(np.isclose(S1,S2))
assert np.all(np.isclose(V1,V2))
The last assertion fails. Why?
Just play with small numbers to debug your problem.
Start with A=np.random.randn(3,2) instead of your much larger matrix with size (50,20)
In my random case, I find that
v1 = array([[-0.33872745, 0.94088454],
[-0.94088454, -0.33872745]])
and for v2:
v2 = array([[ 0.33872745, -0.94088454],
[ 0.94088454, 0.33872745]])
they only differ for a sign, and obviously, even if normalized to have unit module, the vector can differ for a sign.
Now if you try the trick
assert np.all(np.isclose(V1,-1*V2))
for your original big matrix, it fails... again, this is OK. What happens is that some vectors have been multiplied by -1, some others haven't.
A correct way to check for equality between the vectors is:
assert allclose(abs((V1*V2).sum(0)),1.)
and indeed, to get a feeling of how this works you can print this quantity:
(V1*V2).sum(0)
that indeed is either +1 or -1 depending on the vector:
array([ 1., -1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1.,
1., -1., 1., 1., 1., -1., -1.])
EDIT: This will happen in most cases, especially if starting from a random matrix. Notice however that this test will likely fail if one or more eigenvalues has an eigenspace of dimension larger than 1, as pointed out by #Sven Marnach in his comment below:
There might be other differences than just vectors multiplied by -1.
If any of the eigenvalues has a multi-dimensional eigenspace, you
might get an arbitrary orthonormal basis of that eigenspace, and to
such bases might be rotated against each other by an arbitraty
unitarian matrix
I have a data structure that looks like a list values and I am trying to compute the (x,y) 2d hermite functions from them using numpy. I'm trying to use as many numpy arrays as possible due to the performance boost you get from getting to Fortran as quickly as possible (I'm expecting x to be in practice many thousands of 3-arrays). Specifically, my code looks like this:
x = np.array([[1., 2., 3.], [4., 5., 6.]])
coefs = np.array([[[1., 0.],[0., 1.]], [[0., 1.], [1., 0.]]])
z = np.array([0., 0.])
z[:] = hermval2d(x[:,0], x[:,1], coefs[:])
This returns an error about the shape of hermval2d, which according to just running the hermval2d function instead of assigning it:
In [XX]: hermval2d(x[:,0], x[:,1], coefs[:])
Out[XX]:
array([[ 9., 81.],
[ 6., 18.]])
I would expect the hermval2d to be a scalar for every x, y, and coefficient matrix, which is what you would expect from the documentation. So what am I missing here? What's the score?
It's right there in the docs :)
hermval2d(x, y, c)
[...]
The shape of the result will be c.shape[2:] + x.shape
In your case this seems to return the Hermite values for x and y evaluated for each ith 2d array in c[:,:,i].