I would like to scale an array of size [192,4000] to a specific range. I would like each row (1:192) to be rescaled to a specific range e.g. (-840,840). I run a very simple code:
import numpy as np
from sklearn import preprocessing as sp
sample_mat = np.random.randint(-840,840, size=(192, 4000))
scaler = sp.MinMaxScaler(feature_range=(-840,840))
scaler = scaler.fit(sample_mat)
scaled_mat= scaler.transform(sample_mat)
This messes up my matrix range, even when max and min of my original matrix is exactly the same. I can't figure out what is wrong, any idea?
You can do this manually.
It is a linear transformation of the minmax normalized data.
interval_min = -840
interval_max = 840
scaled_mat = (sample_mat - np.min(sample_mat) / (np.max(sample_mat) - np.min(sample_mat)) * (interval_max - interval_min) + interval_min
MinMaxScaler support feature_range argument on initialization that can produce the output in a certain range.
scaler = MinMaxScaler(feature_range=(1, 2)) will yield output in the (1,2) range
Related
I am working with healthcare insurance claims data and would like to identify fraudulent claims. Have been reading online to try and find a better method. I came across the following code on scikit-learn.org
Does anyone know how to select the outliers? the code plot them in a graph but I would like to select those outliers if possible.
I have tried appending the y_predictions to the x dataframe but that has not worked.
print(__doc__)
import numpy as np
import matplotlib.pyplot as plt
from sklearn.neighbors import LocalOutlierFactor
np.random.seed(42)
# Generate train data
X = 0.3 * np.random.randn(100, 2)
# Generate some abnormal novel observations
X_outliers = np.random.uniform(low=-4, high=4, size=(20, 2))
X = np.r_[X + 2, X - 2, X_outliers]
# fit the model
clf = LocalOutlierFactor(n_neighbors=20)
y_pred = clf.fit_predict(X)
y_pred_outliers = y_pred[200:]
Below is the code i tried.
X['outliers'] = y_pred
The first 200 data are inliers while the last 20 are outliers. When you did fit_predict on X, you will get either outlier (-1) or inlier(1) in y_pred. So to get the predicted outliers, you need to get those y_pred = -1 and get the corresponding value in X. Below script will give you the outliers in X.
X_pred_outliers = [each[1] for each in list(zip(y_pred, X.tolist())) if each[0] == -1]
I combine y_pred and X into an array and check if y=-1, if yes then collect X values.
However, there are eight errors on the predictions (8 out of 220). These errors are -1 values in y_pred[:200] and 1 in y_pred[201:220]. Please be aware of the errors as well.
I'm trying to plot a simple moving averages function but the resulting array is a few numbers short of the full sample size. How do I plot such a line alongside a more standard line that extends for the full sample size? The code below results in this error message:
ValueError: x and y must have same first dimension, but have shapes (96,) and (100,)
This is using standard matplotlib.pyplot. I've tried just deleting X values using remove and del as well as switching all arrays to numpy arrays (since that's the output format of my moving averages function) then tried adding an if condition to the append in the while loop but neither has worked.
import random
import matplotlib
import matplotlib.pyplot as plt
import numpy as np
def movingaverage(values, window):
weights = np.repeat(1.0, window) / window
smas = np.convolve(values, weights, 'valid')
return smas
sampleSize = 100
min = -10
max = 10
window = 5
vX = np.array([])
vY = np.array([])
x = 0
val = 0
while x < sampleSize:
val += (random.randint(min, max))
vY = np.append(vY, val)
vX = np.append(vX, x)
x += 1
plt.plot(vX, vY)
plt.plot(vX, movingaverage(vY, window))
plt.show()
Expected results would be two lines on the same graph - one a simple moving average of the other.
Just change this line to the following:
smas = np.convolve(values, weights,'same')
The 'valid' option, only convolves if the window completely covers the values array. What you want is 'same', which does what you are looking for.
Edit: This, however, also comes with its own issues as it acts like there are extra bits of data with value 0 when your window does not fully sit on top of the data. This can be ignored if chosen, as is done in this solution, but another approach is to pad the array with specific values of your choosing instead (see Mike Sperry's answer).
Here is how you would pad a numpy array out to the desired length with 'nan's (replace 'nan' with other values, or replace 'constant' with another mode depending on desired results)
https://docs.scipy.org/doc/numpy/reference/generated/numpy.pad.html
import numpy as np
bob = np.asarray([1,2,3])
alice = np.pad(bob,(0,100-len(bob)),'constant',constant_values=('nan','nan'))
So in your code it would look something like this:
import random
import matplotlib
import matplotlib.pyplot as plt
import numpy as np
def movingaverage(values,window):
weights = np.repeat(1.0,window)/window
smas = np.convolve(values,weights,'valid')
shorted = int((100-len(smas))/2)
print(shorted)
smas = np.pad(smas,(shorted,shorted),'constant',constant_values=('nan','nan'))
return smas
sampleSize = 100
min = -10
max = 10
window = 5
vX = np.array([])
vY = np.array([])
x = 0
val = 0
while x < sampleSize:
val += (random.randint(min,max))
vY = np.append(vY,val)
vX = np.append(vX,x)
x += 1
plt.plot(vX,vY)
plt.plot(vX,(movingaverage(vY,window)))
plt.show()
To answer your basic question, the key is to take a slice of the x-axis appropriate to the data of the moving average. Since you have a convolution of 100 data elements with a window of size 5, the result is valid for the last 96 elements. You would plot it like this:
plt.plot(vX[window - 1:], movingaverage(vY, window))
That being said, your code could stand to have some optimization done on it. For example, numpy arrays are stored in fixed size static buffers. Any time you do append or delete on them, the entire thing gets reallocated, unlike Python lists, which have amortization built in. It is always better to preallocate if you know the array size ahead of time (which you do).
Secondly, running an explicit loop is rarely necessary. You are generally better off using the under-the-hood loops implemented at the lowest level in the numpy functions instead. This is called vectorization. Random number generation, cumulative sums and incremental arrays are all fully vectorized in numpy. In a more general sense, it's usually not very effective to mix Python and numpy computational functions, including random.
Finally, you may want to consider a different convolution method. I would suggest something based on numpy.lib.stride_tricks.as_strided. This is a somewhat arcane, but very effective way to implement a sliding window with numpy arrays. I will show it here as an alternative to the convolution method you used, but feel free to ignore this part.
All in all:
import matplotlib
import matplotlib.pyplot as plt
import numpy as np
def movingaverage(values, window):
# this step creates a view into the same buffer
values = np.lib.stride_tricks.as_strided(values, shape=(window, values.size - window + 1), strides=values.strides * 2)
smas = values.sum(axis=0)
smas /= window # in-place to avoid temp array
return smas
sampleSize = 100
min = -10
max = 10
window = 5
v_x = np.arange(sampleSize)
v_y = np.cumsum(np.random.random_integers(min, max, sampleSize))
plt.plot(v_x, v_y)
plt.plot(v_x[window - 1:], movingaverage(v_y, window))
plt.show()
A note on names: in Python, variable and function names are conventionally name_with_underscore. CamelCase is reserved for class names. np.random.random_integers uses inclusive bounds just like random.randint, but allows you to specify the number of samples to generate. Confusingly, np.random.randint has an exclusive upper bound, more like random.randrange.
I'm trying to normalize data with missing (i.e. nan) values before processing it, using scikit-learn preprocessing.
Apparently, some scalers (e.g. StandardScaler) handle the missing values the way I want - by which I mean normalize the existing values while keeping the nans - while other (e.g. Normalizer) just raise an error.
I've looked around and haven't found - how can I use the Normalizer with missing values, or replicate its behavior (with norm='l1' and norm='l2'; I need to test several normalization options) some other way?
from sklearn.preprocessing import Normalizer, StandardScaler
import numpy as np
data = np.array([0,1,2,np.nan, 3,4])
scaler = StandardScaler(with_mean=True, with_std=True)
scaler.fit_transform(data.reshape(-1,1))
normalizer = Normalizer(norm='l2')
normalizer.fit_transform(data.reshape(-1,1))
The problem with your request is that Normalizer operates in this fashion, accordingly to documentation:
Normalize samples individually to unit norm.
Each sample (i.e. each row of the data matrix) with at least one non
zero component is rescaled independently of other samples so that its
norm (l1 or l2) equals one (source here)
That means that each row have to sum to unit norm. How to deal with a missing value? Ideally it seems you don't want it to count in the sum and you want the row to normalize regardless of it, but the internal function check_array prevents from it by throwing an error.
You need to circumvent such a situation. The most reasonable way to do it is to:
first create a mask in order to record which elements were missing in your array
create a response array filled with missing values
apply the Normalizer to your array after selecting only the valid entries
record on your response array the normalized values based on their original position
here some code detailing the process, based on your example:
from sklearn.preprocessing import Normalizer, StandardScaler
import numpy as np
data = np.array([0,1,2,np.nan, 3,4])
# set valid mask
nan_mask = np.isnan(data)
valid_mask = ~nan_mask
normalizer = Normalizer(norm='l2')
# create a result array
result = np.full(data.shape, np.nan)
# assign only valid cases to
result[valid_mask] = normalizer.fit_transform(data[valid_mask].reshape(-1,1)).reshape(data[valid_mask].shape)
I'd like to normalize my training set before passing it to my NN so instead of doing it manually (subtract mean and divide by std), I tried keras.utils.normalize() and I am amazed about the results I got.
Running this:
r = np.random.rand(3000) * 1000
nr = normalize(r)
print(np.mean(r))
print(np.mean(nr))
print(np.std(r))
print(np.std(nr))
print(np.min(r))
print(np.min(nr))
print(np.max(r))
print(np.max(nr))
Results in that:
495.60440066771866
0.015737914577213984
291.4440194021
0.009254802974329002
0.20755517410064872
6.590913227674956e-06
999.7631481267636
0.03174747238214018
Unfortunately, the docs don't explain what's happening under the hood. Can you please explain what it does and if I should use keras.utils.normalize instead of what I would have done manually?
It is not the kind of normalization you expect. Actually, it uses np.linalg.norm() under the hood to normalize the given data using Lp-norms:
def normalize(x, axis=-1, order=2):
"""Normalizes a Numpy array.
# Arguments
x: Numpy array to normalize.
axis: axis along which to normalize.
order: Normalization order (e.g. 2 for L2 norm).
# Returns
A normalized copy of the array.
"""
l2 = np.atleast_1d(np.linalg.norm(x, order, axis))
l2[l2 == 0] = 1
return x / np.expand_dims(l2, axis)
For example, in the default case, it would normalize the data using L2-normalization (i.e. the sum of squared of elements would be equal to one).
You can either use this function, or if you don't want to do mean and std normalization manually, you can use StandardScaler() from sklearn or even MinMaxScaler().
I am currently moving my data analysis from R to Python. When scaling a dataset in R i would use R.scale(), which in my understanding would do the following: (x-mean(x))/sd(x)
To replace that function I tried to use sklearn.preprocessing.scale(). From my understanding of the description it does the same thing. Nonetheless I ran a little test-file and found out, that both of these methods have different return-values. Obviously the standard deviations are not the same... Is someone able to explain why the standard deviations "deviate" from one another?
MWE:
# import packages
from sklearn import preprocessing
import numpy
import rpy2.robjects.numpy2ri
from rpy2.robjects.packages import importr
rpy2.robjects.numpy2ri.activate()
# Set up R namespaces
R = rpy2.robjects.r
np1 = numpy.array([[1.0,2.0],[3.0,1.0]])
print "Numpy-array:"
print np1
print "Scaled numpy array through R.scale()"
print R.scale(np1)
print "-------"
print "Scaled numpy array through preprocessing.scale()"
print preprocessing.scale(np1, axis = 0, with_mean = True, with_std = True)
scaler = preprocessing.StandardScaler()
scaler.fit(np1)
print "Mean of preprocessing.scale():"
print scaler.mean_
print "Std of preprocessing.scale():"
print scaler.std_
Output:
It seems to have to do with how standard deviation is calculated.
>>> import numpy as np
>>> a = np.array([[1, 2],[3, 1]])
>>> np.std(a, axis=0)
array([ 1. , 0.5])
>>> np.std(a, axis=0, ddof=1)
array([ 1.41421356, 0.70710678])
From numpy.std documentation,
ddof : int, optional
Means Delta Degrees of Freedom. The divisor used in calculations is N - ddof, where N represents the number of elements. By default ddof is zero.
Apparently, R.scale() uses ddof=1, but sklearn.preprocessing.StandardScaler() uses ddof=0.
EDIT: (To explain how to use alternate ddof)
There doesn't seem to be a straightforward way to calculate std with alternate ddof, without accessing the variables of the StandardScaler() object itself.
sc = StandardScaler()
sc.fit(data)
# Now, sc.mean_ and sc.std_ are the mean and standard deviation of the data
# Replace the sc.std_ value using std calculated using numpy
sc.std_ = numpy.std(data, axis=0, ddof=1)
The current answers are good, but sklearn has changed a bit meanwhile. The new syntax that makes sklearn behave exactly like R.scale() now is:
from sklearn.preprocessing import StandardScaler
import numpy as np
sc = StandardScaler()
sc.fit(data)
sc.scale_ = np.std(data, axis=0, ddof=1).to_list()
sc.transform(data)
Feature request:
https://github.com/scikit-learn/scikit-learn/issues/23758
R.scale documentation says:
The root-mean-square for a (possibly centered) column is defined as sqrt(sum(x^2)/(n-1)), where x is a vector of the non-missing values and n is the number of non-missing values. In the case center = TRUE, this is the same as the standard deviation, but in general it is not. (To scale by the standard deviations without centering, use scale(x, center = FALSE, scale = apply(x, 2, sd, na.rm = TRUE)).)
However, sklearn.preprocessing.StandardScale always scale with standard deviation.
In my case, I want to replicate R.scale in Python without centered,I followed #Sid advice in a slightly different way:
import numpy as np
def get_scale_1d(v):
# I copy this function from R source code haha
v = v[~np.isnan(v)]
std = np.sqrt(
np.sum(v ** 2) / np.max([1, len(v) - 1])
)
return std
sc = StandardScaler()
sc.fit(data)
sc.std_ = np.apply_along_axis(func1d=get_scale_1d, axis=0, arr=x)
sc.transform(data)