I'm working on pointcloud lidar data. I want to calculate skewness and kurtosis to distinguish road from ground. My current intensity column looks like this in plot Is there an implementation on another language? I read
First, the skewness of the point cloud is calculated. If it is
greater than zero, peaks dominate the point cloud distribution as
shown in Table 1. Thus, the highest value of the point cloud is re-
moved by classifying it as an object point. To separate all ground
and object points, these steps are iteratively executed while the
skewness of the point cloud is greater than zero.
UPDATE
I tried
the following code but its an infinite loop
I'm getting StatsBase.skewness(poi.intensity) around 0.125
data = rand(Gamma(7.5, 1.0), 75)
th=maximum(data)
classification=rand(2:3,length(data))
poi=Table(intensity=data,classification=classification)
while sk > 0
poi.classification[findall(poi.intensity .== th)] .= 7 #object
th = th .- 1
sk = skewness(poi.intensity[findall(poi.classification .== 2)])
print(sk)
end
I have 3 questions:
1)
The confusion matrix for sklearn is as follows:
TN | FP
FN | TP
While when I'm looking at online resources, I find it like this:
TP | FP
FN | TN
Which one should I consider?
2)
Since the above confusion matrix for scikit learn is different than the one I find in other rescources, in a multiclass confusion matrix, what's the structure will be? I'm looking at this post here:
Scikit-learn: How to obtain True Positive, True Negative, False Positive and False Negative
In that post, #lucidv01d had posted a graph to understand the categories for multiclass. is that category the same in scikit learn?
3)
How do you calculate the accuracy of a multiclass? for example, I have this confusion matrix:
[[27 6 0 16]
[ 5 18 0 21]
[ 1 3 6 9]
[ 0 0 0 48]]
In that same post I referred to in question 2, he has written this equation:
Overall accuracy
ACC = (TP+TN)/(TP+FP+FN+TN)
but isn't that just for binary? I mean, for what class do I replace TP with?
The reason why sklearn has show their confusion matrix like
TN | FP
FN | TP
like this is because in their code, they have considered 0 to be the negative class and one to be positive class. sklearn always considers the smaller number to be negative and large number to positive. By number, I mean the class value (0 or 1). The order depends on your dataset and class.
The accuracy will be the sum of diagonal elements divided by the sum of all the elements.p The diagonal elements are the number of correct predictions.
As the sklearn guide says: "(Wikipedia and other references may use a different convention for axes)"
What does it mean? When building the confusion matrix, the first step is to decide where to put predictions and real values (true labels). There are two possibilities:
put predictions to the columns, and true labes to rows
put predictions to the rows, and true labes to columns
It is totally subjective to decide which way you want to go. From this picture, explained in here, it is clear that scikit-learn's convention is to put predictions to columns, and true labels to rows.
Thus, according to scikit-learns convention, it means:
the first column contains, negative predictions (TN and FN)
the second column contains, positive predictions (TP and FP)
the first row contains negative labels (TN and FP)
the second row contains positive labels (TP and FN)
the diagonal contains the number of correctly predicted labels.
Based on this information I think you will be able to solve part 1 and part 2 of your questions.
For part 3, you just sum the values in the diagonal and divide by the sum of all elements, which will be
(27 + 18 + 6 + 48) / (27 + 18 + 6 + 48 + 6 + 16 + 5 + 21 + 1 + 3 + 9)
or you can just use score() function.
The scikit-learn convention is to place predictions in columns and real values in rows
The scikit-learn convention is to put 0 by default for a negative class (top) and 1 for a positive class (bottom). the order can be changed using labels = [1,0].
You can calculate the overall accuracy in this way
M = np.array([[27, 6, 0, 16], [5, 18,0,21],[1,3,6,9],[0,0,0,48]])
M
sum of diagonal
w = M.diagonal()
w.sum()
99
sum of matrices
M.sum()
160
ACC = w.sum()/M.sum()
ACC
0.61875
I have been following Khan Academy videos to gain understanding of hypothesis testing, and I must confess that all my understanding thus far is based on that source.
Now, the following videos talk about z-score/hypothesis testing:
Hypothesis Testing
Z-statistic vs T-statistic
Now, coming to my doubts, which is all about the denominator in the z-score:
For the z-score formula which is: z = (x – μ) / σ,
we use this directly when the standard deviation of the population(σ), is known.
But when its unknown, and we use a sampling distribution,
then we have z = (x – μ) / (σ / √n); and we estimate σ with σs ; where σs is the standard deviation of the sample, and n is the sample size.
Then z score = (x – μ) / (σs / √n). Why are dividing by √n, when σs is already known?
Even in the video, Hypothesis Testing - Sal divides the sample's standard deviation by √n. Why are we doing this, when σs is directly given?
Please help me understand.
I tried applying this on the following question, and faced the problems below:
Question : Yardley designed new perfumes. Yardley company claimed that an average new
perfume bottle lasts 300 days. Another company randomly selects 35 new perfume bottles from
Yardley for testing. The sampled bottles last an average of 190 days, with a
standard deviation of 50 days. If the Yardley's claim were true,
what is the probability that 35 randomly selected bottles would have an average
life of no more than 190 days ?
So, the above question, when I do the following:
z = (190-300)/(50/√35), we get z = -13.05, which is not a possible score, since
z score should be between +-3.
And when I do, z = (190-110)/50, or rather z = (x – μ) / σ, I seem to be getting an acceptable answer over here.
Please help me figure out what I am missing.
I think the origin of the 1/\sqrt{n} is simply whether you're calculating the standard deviation of the lifetime of a single bottle, or the standard deviation of the (sample) mean of a set of bottles.
The question indicates that 50 days is the standard deviation of the lifetimes of the set of 35 bottles. That implies that the estimated mean age (190 days) will have a margin of error of about 50/\sqrt{35} days. Assuming that this similar margin of error applied to the claimed 300-day lifetime, one can calculate the probability that a set of 35 bottles would be measured to be 190 days or less, using the complementary error function.
Your z=-13.05 looks about right, implying that it is extremely unlikely that claimed 300-day lifetime is consistent with that seen in the 35-bottle experiment.
This is an extension to the question asked in the forums a few years ago:
Excel produces scatter diagrams for sets of pair values. It also gives the option of producing a best fit trendline and formula for the trendline. It also produces bubble diagrams which take into consideration a weight provided with each value. However, the weight has no influence on the trendline or formula. Here is an example set of values, with their mappings and weights.
Value Map Weight
0 1 10
1 2 10
2 5 10
3 5 20
4 6 20
5 1 1
I have used the formula that brettDJ offered:
=INDEX(LINEST(B2:B7*C2:C7^0.5,IF({1,0},1,A2:A7)*C2:C7^0.5,TRUE,TRUE),3,1)
However, I could not understand why we used the ^0.5 here to sqrt the weights.
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I am doing my graduation project in the field of computer vision, and i have only taken one course in statistics that discussed very basic concepts, and now i am facing more difficulty in rather advanced topics, so i need help (book, tutorial, course, ..etc) to grasp and review the basic ideas and concepts in statistics and then dive into the details (statistical details) used in computer vision.
You can calculate False Positives/False negatives, etc with this Confusion Matrix PyTorch example:
import torch
def confusion(prediction, truth):
""" Returns the confusion matrix for the values in the `prediction` and `truth`
tensors, i.e. the amount of positions where the values of `prediction`
and `truth` are
- 1 and 1 (True Positive)
- 1 and 0 (False Positive)
- 0 and 0 (True Negative)
- 0 and 1 (False Negative)
"""
confusion_vector = prediction / truth
# Element-wise division of the 2 tensors returns a new tensor which holds a
# unique value for each case:
# 1 where prediction and truth are 1 (True Positive)
# inf where prediction is 1 and truth is 0 (False Positive)
# nan where prediction and truth are 0 (True Negative)
# 0 where prediction is 0 and truth is 1 (False Negative)
true_positives = torch.sum(confusion_vector == 1).item()
false_positives = torch.sum(confusion_vector == float('inf')).item()
true_negatives = torch.sum(torch.isnan(confusion_vector)).item()
false_negatives = torch.sum(confusion_vector == 0).item()
return true_positives, false_positives, true_negatives, false_negatives
You could use nn.BCEWithLogitsLoss (remove the sigmoid therefore) and set the pos_weight > 1 to increase the recall. Or further optimize it with using Dice Coefficients to penalize the model for false positives, with something like:
def Dice(y_true, y_pred):
"""Returns Dice Similarity Coefficient for ground truth and predicted masks."""
#print(y_true.dtype)
#print(y_pred.dtype)
y_true = np.squeeze(y_true)/255
y_pred = np.squeeze(y_pred)/255
y_true.astype('bool')
y_pred.astype('bool')
intersection = np.logical_and(y_true, y_pred).sum()
return ((2. * intersection.sum()) + 1.) / (y_true.sum() + y_pred.sum() + 1.)
IOU Calculations Explained
Count true positives (TP)
Count false positives (FP)
Count false negatives (FN)
Intersection = TP
Union = TP + FP + FN
IOU = Intersection/Union
The left side is our ground truth, while the right side contains our predictions. The highlighted cells on the left side note which class we are looking at for statistics on the right side. The highlights on the right side note true positives in a cream color, false positives in orange, and false negatives in yellow (note that all others are true negatives — they are predicted as this individual class, and should not be based on the ground truth).
For Class 0, only the top row of the 4x4 matrix should be predicted as zeros. This is a rather simplified version of a real ground truth. In reality, the zeros could be anywhere in the matrix. On the right side, we see 1,0,0,0, meaning the first is a false negative, but the other three are true positives (aka 3 for Intersection as well). From there, we need to find anywhere else where zero was falsely predicted, and we note that happens once on the second row, and twice on the fourth row, for a total of three false positives.
To get the union, we add up TP (3), FP (3) and FN (1) to get seven. The IOU for this class, therefore, is 3/7.
If we do this for all the classes and average the IOUs, we get:
Mean IOU = [(3/7) + (2/6) + (3/4) + (1/6)] / 4 = 0.420
You will also want to learn how to pull the statistics for mAP (Mean Avg Precision):
https://www.youtube.com/watch?v=pM6DJ0ZZee0
https://towardsdatascience.com/breaking-down-mean-average-precision-map-ae462f623a52#1a59
https://medium.com/#hfdtsinghua/calculate-mean-average-precision-map-for-multi-label-classification-b082679d31be
Compute Covariance Matrixes
The variance of a variable describes how much the values are spread. The covariance is a measure that tells the amount of dependency between two variables.
A positive covariance means that the values of the first variable are large when values of the second variables are also large. A negative covariance means the opposite: large values from one variable are associated with small values of the other.
The covariance value depends on the scale of the variable so it is hard to analyze it. It is possible to use the correlation coefficient that is easier to interpret. The correlation coefficient is just the normalized covariance.
A positive covariance means that large values of one variable are associated with big values from the other (left). A negative covariance means that large values of one variable are associated with small values of the other one (right).
The covariance matrix is a matrix that summarises the variances and covariances of a set of vectors and it can tell a lot of things about your variables. The diagonal corresponds to the variance of each vector:
A matrix A and its matrix of covariance. The diagonal corresponds to the variance of each column vector. Let’s check with the formula of the variance:
With n the length of the vector, and x̄ the mean of the vector. For instance, the variance of the first column vector of A is:
This is the first cell of our covariance matrix. The second element on the diagonal corresponds of the variance of the second column vector from A and so on.
Note: the vectors extracted from the matrix A correspond to the columns of A.
The other cells correspond to the covariance between two column vectors from A. For instance, the covariance between the first and the third column is located in the covariance matrix as the column 1 and the row 3 (or the column 3 and the row 1):
The position in the covariance matrix. Column corresponds to the first variable and row to the second (or the opposite). The covariance between the first and the third column vector of A is the element in column 1 and row 3 (or the opposite = same value).
Let’s check that the covariance between the first and the third column vector of A is equal to -2.67. The formula of the covariance between two variables Xand Y is:
The variables X and Y are the first and the third column vectors in the last example. Let’s split this formula to be sure that it is crystal clear:
The sum symbol (Σ) means that we will iterate on the elements of the vectors. We will start with the first element (i=1) and calculate the first element of X minus the mean of the vector X:
Multiply the result with the first element of Y minus the mean of the vector Y:
Reiterate the process for each element of the vectors and calculate the sum of all results:
Divide by the number of elements in the vector.
EXAMPLE - Let’s start with the matrix A:
We will calculate the covariance between the first and the third column vectors:
and
Which is x̄=3, ȳ=4, and n=3 so we have:
Code example -
Using NumPy, the covariance matrix can be calculated with the function np.cov.
It is worth noting that if you want NumPy to use the columns as vectors, the parameter rowvar=False has to be used. Also, bias=True divides by n and not by n-1.
Let’s create the array first:
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
A = np.array([[1, 3, 5], [5, 4, 1], [3, 8, 6]])
Now we will calculate the covariance with the NumPy function:
np.cov(A, rowvar=False, bias=True)
Finding the covariance matrix with the dot product
There is another way to compute the covariance matrix of A. You can center A around 0. The mean of the vector is subtracted from each element of the vector to have a vector with mean equal to 0. It is multiplied with its own transpose, and divided by the number of observations.
Let’s start with an implementation and then we’ll try to understand the link with the previous equation:
def calculateCovariance(X):
meanX = np.mean(X, axis = 0)
lenX = X.shape[0]
X = X - meanX
covariance = X.T.dot(X)/lenX
return covariance
print(calculateCovariance(A))
Output:
array([[ 2.66666667, 0.66666667, -2.66666667],
[ 0.66666667, 4.66666667, 2.33333333],
[-2.66666667, 2.33333333, 4.66666667]])
The dot product between two vectors can be expressed:
It is the sum of the products of each element of the vectors:
If we have a matrix A, the dot product between A and its transpose will give you a new matrix:
Visualize data and covariance matrices
In order to get more insights about the covariance matrix and how it can be useful, we will create a function to visualize it along with 2D data. You will be able to see the link between the covariance matrix and the data.
This function will calculate the covariance matrix as we have seen above. It will create two subplots — one for the covariance matrix and one for the data. The heatmap() function from Seaborn is used to create gradients of colour — small values will be coloured in light green and large values in dark blue. We chose one of our palette colours, but you may prefer other colours. The data is represented as a scatterplot.
def plotDataAndCov(data):
ACov = np.cov(data, rowvar=False, bias=True)
print 'Covariance matrix:\n', ACov
fig, ax = plt.subplots(nrows=1, ncols=2)
fig.set_size_inches(10, 10)
ax0 = plt.subplot(2, 2, 1)
# Choosing the colors
cmap = sns.color_palette("GnBu", 10)
sns.heatmap(ACov, cmap=cmap, vmin=0)
ax1 = plt.subplot(2, 2, 2)
# data can include the colors
if data.shape[1]==3:
c=data[:,2]
else:
c="#0A98BE"
ax1.scatter(data[:,0], data[:,1], c=c, s=40)
# Remove the top and right axes from the data plot
ax1.spines['right'].set_visible(False)
ax1.spines['top'].set_visible(False)
Uncorrelated data
Now that we have the plot function, we will generate some random data to visualize what the covariance matrix can tell us. We will start with some data drawn from a normal distribution with the NumPy function np.random.normal().
This function needs the mean, the standard deviation and the number of observations of the distribution as input. We will create two random variables of 300 observations with a standard deviation of 1. The first will have a mean of 1 and the second a mean of 2. If we randomly draw two sets of 300 observations from a normal distribution, both vectors will be uncorrelated.
np.random.seed(1234)
a1 = np.random.normal(2, 1, 300)
a2 = np.random.normal(1, 1, 300)
A = np.array([a1, a2]).T
A.shape
Note 1: We transpose the data with .T because the original shape is (2, 300) and we want the number of observations as rows (so with shape (300, 2)).
Note 2: We use np.random.seed function for reproducibility. The same random number will be used the next time we run the cell. Let’s check how the data looks like:
A[:10,:]
array([[ 2.47143516, 1.52704645],
[ 0.80902431, 1.7111124 ],
[ 3.43270697, 0.78245452],
[ 1.6873481 , 3.63779121],
[ 1.27941127, -0.74213763],
[ 2.88716294, 0.90556519],
[ 2.85958841, 2.43118375],
[ 1.3634765 , 1.59275845],
[ 2.01569637, 1.1702969 ],
[-0.24268495, -0.75170595]])
Nice, we have two column vectors; Now, we can check that the distributions are normal:
sns.distplot(A[:,0], color="#53BB04")
sns.distplot(A[:,1], color="#0A98BE")
plt.show()
plt.close()
We can see that the distributions have equivalent standard deviations but different means (1 and 2). So that’s exactly what we have asked for.
Now we can plot our dataset and its covariance matrix with our function:
plotDataAndCov(A)
plt.show()
plt.close()
Covariance matrix:
[[ 0.95171641 -0.0447816 ]
[-0.0447816 0.87959853]]
We can see on the scatterplot that the two dimensions are uncorrelated. Note that we have one dimension with a mean of 1 (y-axis) and the other with the mean of 2 (x-axis).
Also, the covariance matrix shows that the variance of each variable is very large (around 1) and the covariance of columns 1 and 2 is very small (around 0). Since we ensured that the two vectors are independent this is coherent. The opposite is not necessarily true: a covariance of 0 doesn’t guarantee independence.
Correlated data
Now, let’s construct dependent data by specifying one column from the other one.
np.random.seed(1234)
b1 = np.random.normal(3, 1, 300)
b2 = b1 + np.random.normal(7, 1, 300)/2.
B = np.array([b1, b2]).T
plotDataAndCov(B)
plt.show()
plt.close()
Covariance matrix:
[[ 0.95171641 0.92932561]
[ 0.92932561 1.12683445]]
The correlation between the two dimensions is visible on the scatter plot. We can see that a line could be drawn and used to predict y from x and vice versa. The covariance matrix is not diagonal (there are non-zero cells outside of the diagonal). That means that the covariance between dimensions is non-zero.
From this point with Covariance Matrcies, you can research further on the following:
Mean normalization
Standardization or normalization
Whitening
Zero-centering
Decorrelate
Rescaling