I have created a weighted graph of k-Neighbors using scikit-learn, I'm wondering if there is any way to plot it as a graph.
Here is the result of computation in form of array which I want to plot:
array([[0. , 2.08243189, 0. , 3.42661108],
[2.08243189, 0. , 3.27141008, 0. ],
[0. , 3.27141008, 0. , 1.57294787],
[0. , 3.29779083, 1.57294787, 0. ]])
I just need to get some visualization of data, that's all I need.
More details about the array:
Each row represents a node and each column represents the weight of connectivity of that node with the other nodes.
For example: second column of first row (2.08243189) is the weight of connectivity from first node to second node.
Another example: second row, second column (0): the weight of connectivity from node 2 to itself.
The numbers represents euclidean distance.
Are you talking about something simple like this where the size of the point gives a visual indication of the relative weight compared to the other values? Assume the array is named ar:
for i in range(len(ar)):
for j in range(len(ar)):
v = ar[i,j]
plt.scatter(i+1,j+1,lw=0,s=10**v)
plt.grid(True)
plt.xlabel('Row')
plt.ylabel('Column')
ticks = list(range(1,1+len(ar)))
plt.xticks(ticks)
plt.yticks(ticks)
Related
I am working with two-dimensional data
X = array([[5.40310335, 0. ],
[6.86136114, 6.56225717],
[0. , 0. ],
...,
[5.88838732, 0. ],
[6.0003473 , 0. ],
[6.25971331, 0. ]])
looking for clusters, using euclidean distance, i run affinity propagation from scikit learn with this raw data as follows
af = AffinityPropagation(damping=.9, max_iter=300, random_state=0).fit(X)
obtaining as a result 9 clusters.
I understand that when you want to use another distance you have to enter the negative distance matrix, and use affintity = 'precomputed' as it follows
af_c = AffinityPropagation(damping=.9, max_iter=300,
affinity='precomputed', random_state=0).fit(distM)
if as distM I use the Euclidean distance matrix calculated as follows
distM_E = -distance_matrix(X,X)
np.fill_diagonal(distM, np.median(distM))
completing the diagonal with the median since it is a predefined preference value also in the method.
Using this I am getting 34 clusters as a result and I would expect to have 9 as if working with the default distance. I don't know if I'm interpreting the way of entering the distance matrix correctly or if the library does something different when one uses everything predefined.
I would appreciate any help.
I would like to calculate the gradients of the output of a neural network with respect to the input. I have the following tensors:
Input: (num_timesteps, features)
Output: (num_timesteps, 1)
For the gradients from the inputs to the entire output vector I can use the following:
tf.gradients(Output, Input)
Since I would like to compute the gradients for every single timesample I would like to calculate
tf.gradients(Output[i], Input)
for every i.
What is the best way to do that?
First up, I suppose you mean the gradient of Output with respect to the Input.
Now, the result of both of these calls:
dO = tf.gradients(Output, Input)
dO_i = tf.gradients(Output[i], Input) (for any valid i)
will be a list with a single element - a tensor with the same shape as Input, namely a [num_timesteps, features] matrix. Also, if you sum all matrices dO_i (over all valid i) is exactly the matrix dO.
With this in mind, back to your question. In many cases, individual rows from the Input are independent, meaning that Output[i] is calculated only from Input[i] and doesn't know other inputs (typical case: batch processing without batchnorm). If that is your case, then dO is going to give you all individual components dO_i at once.
This is because each dO_i matrix is going to look like this:
[[ 0. 0. 0.]
[ 0. 0. 0.]
...
[ 0. 0. 0.]
[ xxx xxx xxx] <- i-th row
[ 0. 0. 0.]
...
[ 0. 0. 0.]]
All rows are going to be 0, except for the i-th one. So just by computing one matrix dO, you can easily get every dO_i. This is very efficient.
However, if that's not your case and all Output[i] depend on all inputs, there's no way to extract individual dO_i just from their sum. You have no other choice other than calculate each gradient separately: just iterate over i and execute tf.gradients.
I have computed a jaccard similarity matrix with Python. I want to cluster highest similarities to lowest, however, no matter what linkage function I use it produces the same dendrogram! I have a feeling that the function assumes that my matrix is of original data, but I have already computed the first similarity matrix. Is there any way to pass this similarity matrix through to the dendrogram so it plots correctly? Or am I going to have to output the matrix and simply do it with R. Passing through the original raw data is not possible, as I am computing similarities of words. Thanks for the help!
Here is some code:
SimMatrix = [[ 0.,0.09259259, 0.125 , 0. , 0.08571429],
[ 0.09259259, 0. , 0.05555556, 0. , 0.05128205],
[ 0.125 , 0.05555556, 0. , 0.03571429, 0.05882353],
[ 0. , 0. , 0.03571429, 0. , 0. ],
[ 0.08571429, 0.05128205, 0.05882353, 0. , 0. ]]
linkage = hcluster.complete(SimMatrix) #doesnt matter what linkage...
dendro = hcluster.dendrogram(linkage) #same plot for all types?
show()
If you run this code, you will see a dendrogram that is completely backwards. No matter what linkage type I use, it produces the same dendrogram. This intuitively can not be correct!
Here's the solution. Turns out the SimMatrix needs to be first converted into a condensed matrix (the diagonal, upper right or bottom left, of this matrix).
You can see this in the code below:
import scipy.spatial.distance as ssd
distVec = ssd.squareform(SimMatrix)
linkage = hcluster.linkage(1 - distVec)
dendro = hcluster.dendrogram(linkage)
show()
I have used Statsmodels to generate a OLS linear regression model to predict a dependent variable based on about 10 independent variables. The independent variables are all categorical.
I am interested in looking closer at the significance of the coefficients for one of the independent variables. There are 4 categories, so 3 coefficients -- each of which are highly significant. I would also like to look at the significance of the trend across all 3 categories. From my (limited) understanding, this is often done using a Wald Test and comparing all of the coefficients to 0.
How exactly is this done using Statsmodels? I see there is a Wald Test method for the OLS function. It seems you have to pass in values for all of the coefficients when using this method.
My approach was the following...
First, here are all of the coefficients:
np.array(lm.params) = array([ 0.21538725, 0.05675108, 0.05020252, 0.08112228, 0.00074715,
0.03886747, 0.00981819, 0.19907263, 0.13962354, 0.0491201 ,
-0.00531318, 0.00242845, -0.0097336 , -0.00143791, -0.01939182,
-0.02676771, 0.01649944, 0.01240742, -0.00245309, 0.00757727,
0.00655152, -0.02895381, -0.02027537, 0.02621716, 0.00783884,
0.05065323, 0.04264466, -0.13068456, -0.15694931, -0.25518566,
-0.0308599 , -0.00558183, 0.02990139, 0.02433505, -0.01582824,
-0.00027538, 0.03170669, 0.01130944, 0.02631403])
I am only interested in params 2-4 (which are the 3 coefficients of interest).
coeffs = np.zeros_like(lm.params)
coeffs = coeffs[1:4] = [0.05675108, 0.05020252, 0.08112228]
Checking to make sure this worked:
array([ 0. , 0.05675108, 0.05020252, 0.08112228, 0. ,
0. , 0. , 0. , 0. , 0. ,
0. , 0. , 0. , 0. , 0. ,
0. , 0. , 0. , 0. , 0. ,
0. , 0. , 0. , 0. , 0. ,
0. , 0. , 0. , 0. , 0. ,
0. , 0. , 0. , 0. , 0. ,
0. , 0. , 0. , 0. ])
Looks good, now to run in the test!
lm.wald_test(coeffs) =
<class 'statsmodels.stats.contrast.ContrastResults'>
<F test: F=array([[ 13.11493673]]), p=0.000304699208434, df_denom=1248, df_num=1>
Is this the correct approach? I could really use some help!
A linear hypothesis has the form R params = q where R is the matrix that defines the linear combination of parameters and q is the hypothesized value.
In the simple case where we want to test whether some parameters are zero, the R matrix has a 1 in the column corresponding to the position of the parameter and zeros everywhere else, and q is zero, which is the default. Each row specifies a linear combination of parameters, which defines a hypothesis as part of the overall or joint hypothesis.
In this case, the simplest way to get the restriction matrix is by using the corresponding rows of an identity matrix
R = np.eye(len(results.params))[1:4]
Then, lm.wald_test(R) will provide the test for the joint hypothesis that the 3 parameters are zero.
A simpler way to specify the restriction is by using the names of the parameters and defining the restrictions by a list of strings.
The model result classes also have a new method wald_test_terms which automatically generates the wald tests for terms in the design matrix where the hypothesis includes several parameters or columns, as in the case of categorical explanatory variables or of polynomial explanatory variables. This is available in statsmodels master and will be in the upcoming 0.7 release.
There are a couple of ball-bounce related questions on stackoverflow that i've looked through, however none of them seem to get me past my predicament. I have a turtle cursor defined by a transformation matrix that intersects a line in 3d space. What I want is to rotate the cursor, that is, the transformation matrix, at the point of intersection so that it's new direction matches the reflection vector. I have functions that will get both the reflection vector R from the incident vector V and the normal of the reflecting line N. I normalize each before evaluating:
N,V=unit_vector(N),unit_vector(V)
R = -2*(np.dot(V,N))*N - V
R=unit_vector(R)
My transformation matrix, T is in a numpy array:
array([[ -0.84923515, -0.6 , 0. , 3.65341878],
[ 0.52801483, -0.84923515, 0. , 25.12882224],
[ 0. , 0. , 1. , 0. ],
[ 0. , 0. , 0. , 1. ]])
How can I transform T by R to get the correct direction vector? I've found and used the R2_vect function from here to get a rotation matrix from one vector to another but only a few of the resulting reflections appear correct when i send them to vtk to render. I'm asking about this here because I seem to be reaching the limit of what I can remember from my already shaky linear algebra. Thanks for any information.
A little extra research clarified things: the first 3 columns of the transformation matrix represent 3 orthonormal vectors ( x1, x2, x3 ) and the 4th column represents the coordinates in space of the cursor at given time interval. the final row contains no data, it's just there to keep the matrix square. rotating the vectors was just a matter of removing the last row of T, taking the 3x3 rotation matrix from my listed function R and rotating each vector: R.dot(x1), R.dot(x2), R.dot(x3) Then I just had to composite the values back into a 4x4 matrix.