I am trying to find the cosine similarity of a list of strings. I used sklearn tfidf vector to convert the text into a numerical vector first and then used the pairwise cosine_similarity api to find the score for each string pair.
The strings seem similar, but I am getting a weird answer. The first and third value in the string array are similar except the word TRENTON, but the cosine similarity is 0. Similarly, the 1st,3rd and 4th string are the same, except for a space between GREEN and CHILLI and the cosine similarity is zero. Isn't that strange?
My code:
from sklearn.metrics import pairwise_kernels
from sklearn.metrics.pairwise import cosine_similarity
from sklearn.feature_extraction.text import TfidfVectorizer
tfidf_vectorizer=TfidfVectorizer()
values =['GREENCHILLI TRENTON'
,'GREENCHILLI'
,'GREEN CHILLI'
,'GREEN CHILLI']
X_train_counts = tfidf_vectorizer.fit_transform(values)
similarities = cosine_similarity(X_train_counts)
print(similarities)
Output
[[1. 0.6191303 0. 0. ]
[0.6191303 1. 0. 0. ]
[0. 0. 1. 1. ]
[0. 0. 1. 1. ]]
coma (,) missing between last two GREEN CHILLI so tfidf is treating them as only 3 records not 4.
If you correct it you should see below cosine similarity
[[1. 0.6191303 0. 0. ]
[0.6191303 1. 0. 0. ]
[0. 0. 1. 1. ]
[0. 0. 1. 1. ]]
How to interpret the above matrix: The value in the nth row are cosine similarities of that tfidf vector with all other vectors (in sequential order). So all the diagonal will be 1 because every vector is similar to itself.
The first and third value in the string array values is similar except the word Trenton but cosine similarity is 0.
Similarly, 1st,3rd and 4th strings are same only space between GREEN and CHILLI and the cosine similarity is zero. isn't it strange?
It is not as strange as you might think. You will only get a non-zero cosine similarity if you have exact word matches between the strings that you compare. I will try to explain what happens:
When the TF-IDF vectorizer creates vectors from your list of strings, it starts by making a list of all words that occur.
So in your case, the list would look like this:
GREENCHILLI
TRENTON
GREEN
CHILLI
Now, every word becomes an axis in a coordinate system that the algorithm uses. All axes are perpendicular to each other.
So when you compare 'GREENCHILLI TRENTON' with 'GREEN CHILLI', the algorithm makes two vectors. One from 'GREENCHILLI TRENTON' that has a component parallel to 'GREENCHILLI' and a component parallel to 'TRENTON'. The vector from the string 'GREEN CHILI' has components in 'GREEN' and 'CHILLI' direction of your coordinate system. When you calculate the dot product between the two you will get a zero. So the cosine similarity is zero as well.
So the gap in 'GREEN CHILLI' makes all the difference, when you compare it to 'GREENCHILLI'. The letters don't matter anymore, once the vectorizer made its coordinate system based on all the words it found in your list, because it identifies 'GREENCHILLI', 'GREEN' and 'CHILLI' as different words and makes them into perpendicular axes in its reference coordinate system.
Hope that makes it more clear. I suggest reading the following article series for a more in-depth understanding of whats going on:
http://blog.christianperone.com/2011/09/machine-learning-text-feature-extraction-tf-idf-part-i/
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'm am doing PCA and I am interested in which original features were most important. Let me illustrate this with an example:
import numpy as np
from sklearn.decomposition import PCA
X = np.array([[1,-1, -1,-1], [1,-2, -1,-1], [1,-3, -2,-1], [1,1, 1,-1], [1,2,1,-1], [1,3, 2,-0.5]])
print(X)
Which outputs:
[[ 1. -1. -1. -1. ]
[ 1. -2. -1. -1. ]
[ 1. -3. -2. -1. ]
[ 1. 1. 1. -1. ]
[ 1. 2. 1. -1. ]
[ 1. 3. 2. -0.5]]
Intuitively, one could already say that feature 1 and feature 4 are not very important due to their low variance. Let's apply pca on this set:
pca = PCA(n_components=2)
pca.fit_transform(X)
comps = pca.components_
Output:
array([[ 0. , 0.8376103 , 0.54436943, 0.04550712],
[-0. , 0.54564656, -0.8297757 , -0.11722679]])
This output represents the importance of each original feature for each of the two principal components (see this for reference). In other words, for the first principal component, feature 2 is most important, then feature 3. For the second principal component, feature 3 looks most important.
The question is, which feature is most important, which one second most etc? Can I use the component_ attribute for this? Or am I wrong and is PCA not the correct method for doing such analyses (and should I use a feature selection method instead)?
The component_ attribute is not the right spot to look for feature importance. The loadings in the two arrays (i.e. the two componments PC1 and PC2) tell you how your original matrix is transformed by each feature (taken together, they form a rotational matrix). But they don't tell you how much each component contributes to describing the transformed feature space, so you don't know yet how to compare the loadings across the two components.
However, the answer that you linked actually tells you what to use instead: the explained_variance_ratio_ attribute. This attribute tells you how much of the variance in your feature space is explained by each principal component:
In [5]: pca.explained_variance_ratio_
Out[5]: array([ 0.98934303, 0.00757996])
This means that the first prinicpal component explaines almost 99 percent of the variance. You know from components_ that PC1 has the highest loading for the second feature. It follows, therefore, that feature 2 is the most important feature in your data space. Feature 3 is the next most important feature, as it has the second highest loading in PC1.
In PC2, the absolute loadings are nearly swapped between feature 2 and feature 3. But as PC2 explains next to nothing of the overall variance, this can be neglected.
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()
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.