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I used the seaborn.regplot to plot data, but not quite understand how the error bar in regplot was calculated. I have compared the results with the mean and standard deviation derived from mannual calculation. Here is my testing script.
import numpy as np
import pandas as pd
import seaborn as sn
def get_data_XYE(p):
x_list = []
lower_list = []
upper_list = []
for line in p.lines:
x_list.append(line.get_xdata()[0])
lower_list.append(line.get_ydata()[0])
upper_list.append(line.get_ydata()[1])
y = 0.5 * (np.asarray(lower_list) + np.asarray(upper_list))
y_error = np.asarray(upper_list) - y
x = np.asarray(x_list)
return x, y, y_error
x = [37.3448,36.6026,42.7795,34.7072,75.4027,226.2615,192.7984,140.8045,242.9952,458.451,640.6542,726.1024,231.7347,107.5605,200.2254,190.0006,314.1349,146.8131,152.4497,175.9096,284.9926,116.9681,118.2953,312.3787,815.8389,458.0146,409.5797,595.5373,188.9955,15.7716,36.1839,244.8689,57.4579,94.8717,112.2237,87.0687,72.79,22.3457,24.1728,29.505,80.8765,252.7454,280.6002,252.9573,348.246,112.705,98.7545,317.0541,300.9573,402.8411,406.6884,56.1286,30.1385,32.9909,497.556,19.3606,20.8409,95.2324,108.6074,15.7753,54.5511,45.5623,64.564,101.1934,81.8459,88.286,58.2642,56.1225,51.2943,38.0649,63.5882,63.6847,120.495,102.4097,49.3255,111.3309,171.6028,58.9526,28.7698,144.6884,180.0661,116.6028,146.2594,199.8702,128.9378,423.2363,119.8537,124.6508,518.8625,306.3023,79.5213,121.0309,116.9346,170.8863,930.361,48.9983,55.039,47.1092,72.0548,75.4045,103.521,83.4134,142.3253,146.6215,121.4467,101.4252,68.4812,291.4275,143.9475,142.647,78.9826,47.094,204.2196,89.0208,82.792,27.1346,142.4764,83.7874,67.3216,112.9531,138.2549,133.3446,86.2659,45.3464,56.1604,43.5882,54.3623,86.296,115.7272,96.5498,111.8081,36.1756,40.2947,34.2532,89.1452,53.9062,36.458,113.9297,176.9962,77.3125,77.8891,64.807,64.1515,127.7242,119.6876,976.2324,322.8454,434.2883,168.6923,250.0284,234.7329,131.0793,152.335,118.8838,243.1772,24.1776,168.6327,170.7541,167.8444,75.9315,110.1045,113.4417,60.5464,66.8956,79.7606,71.6659,72.5251,77.513,207.8019,21.8592,35.2787,169.7698,146.5012,412.9934,248.0708,318.5489,104.1278,184.7592,108.0581,175.2646,169.7698,340.3732,570.3396,23.9853,69.0405,66.7391,67.9435,294.6085,68.0537,77.6344,433.2713,104.3178,229.4615,187.8587,78.1399,121.4737,122.5451,384.5935,38.5232,117.6835,50.3308,318.2513,103.6695,20.7181,321.9601,510.3248,13.4754,16.1188,44.8082,37.7291,733.4587,446.6241,21.1822,287.9603,327.2367,274.1109,195.4713,158.2114,64.4537,26.9857,172.8503]
y = [37,40,30,29,24,23,27,12,21,20,29,28,27,32,23,29,28,22,28,23,24,29,32,18,22,12,12,14,29,31,34,31,22,40,25,36,27,27,29,35,33,25,25,27,27,19,35,26,18,24,25,37,52,47,34,39,40,48,41,44,35,36,53,46,38,44,23,26,26,28,27,21,25,21,20,27,35,24,46,34,22,30,30,30,31,26,25,28,21,31,24,27,33,21,31,33,29,33,32,21,25,22,39,31,34,26,23,18,20,18,34,25,20,12,23,25,21,21,25,31,17,27,28,29,25,24,25,21,24,27,23,22,23,22,22,26,22,19,26,35,33,35,29,26,26,30,22,32,33,33,28,32,26,29,36,37,37,28,24,30,25,20,29,24,33,35,30,32,31,33,40,35,37,24,34,29,27,24,36,26,26,26,27,27,20,17,28,34,18,20,20,18,19,23,20,22,25,32,44,41,39,41,40,44,36,42,31,32,26,29,23,29,29,28,31,22,29,24,28,28,25]
xbreaks = [13.4754, 27.1346, 43.5882, 58.9526, 72.79, 89.1452, 110.1045, 131.0793, 158.2114, 180.0661, 207.8019, 234.7329, 252.9573, 300.9573, 327.2367, 348.246, 412.9934, 434.2883, 458.451, 518.8625, 595.5373, 640.6542, 733.4587, 815.8389, 930.361, 976.2324]
df = pd.DataFrame([x,y]).T
df.columns = ['x','y']
# Check the bin average and std using agge
bins = pd.cut(df.x,xbreaks,right=False)
t = df[['x','y']].groupby(bins).agg({"x": "mean", "y": ["mean","std"]})
t.reset_index(inplace=True)
t.columns = ['range_cut','x_avg_cut','y_avg_cut','y_std_cut']
t.index.name ='id'
# Get the bin average from
g = sns.regplot(x='x',y='y',data=df,fit_reg=False,x_bins=xbreaks,seed=seed)
xye = pd.DataFrame(get_data_XYE(g)).T
xye.columns = ['x_regplot','y_regplot','e_regplot']
xye.index.name = 'id'
t2 = xye.merge(t,on='id',how='left')
t2
You can see the y and e from the two ways are different. I understand that the default x_ci or x_estimator may afect the result of regplot, but I still can not the these values in excel by removing some lowest and/or highest values in each bin.
In seaborn.regplot, the x_bins are the center of each bin, and the original x values are assigned to the nearest bin value. Whereas in pandas.cut, the breaks define the bin edges.
I recently started to use Python, and I can't understand how to plot a confidence interval for a given datum (or set of data).
I already have a function that computes, given a set of measurements, a higher and lower bound depending on the confidence level that I pass to it, but how can I use those two values to plot a confidence interval?
There are several ways to accomplish what you asking for:
Using only matplotlib
from matplotlib import pyplot as plt
import numpy as np
#some example data
x = np.linspace(0.1, 9.9, 20)
y = 3.0 * x
#some confidence interval
ci = 1.96 * np.std(y)/np.sqrt(len(x))
fig, ax = plt.subplots()
ax.plot(x,y)
ax.fill_between(x, (y-ci), (y+ci), color='b', alpha=.1)
fill_between does what you are looking for. For more information on how to use this function, see: https://matplotlib.org/3.1.1/api/_as_gen/matplotlib.pyplot.fill_between.html
Output
Alternatively, go for seaborn, which supports this using lineplot or regplot,
see: https://seaborn.pydata.org/generated/seaborn.lineplot.html
Let's assume that we have three categories and lower and upper bounds of confidence intervals of a certain estimator across these three categories:
data_dict = {}
data_dict['category'] = ['category 1','category 2','category 3']
data_dict['lower'] = [0.1,0.2,0.15]
data_dict['upper'] = [0.22,0.3,0.21]
dataset = pd.DataFrame(data_dict)
You can plot the confidence interval for each of these categories using the following code:
for lower,upper,y in zip(dataset['lower'],dataset['upper'],range(len(dataset))):
plt.plot((lower,upper),(y,y),'ro-',color='orange')
plt.yticks(range(len(dataset)),list(dataset['category']))
Resulting with the following graph:
import matplotlib.pyplot as plt
import statistics
from math import sqrt
def plot_confidence_interval(x, values, z=1.96, color='#2187bb', horizontal_line_width=0.25):
mean = statistics.mean(values)
stdev = statistics.stdev(values)
confidence_interval = z * stdev / sqrt(len(values))
left = x - horizontal_line_width / 2
top = mean - confidence_interval
right = x + horizontal_line_width / 2
bottom = mean + confidence_interval
plt.plot([x, x], [top, bottom], color=color)
plt.plot([left, right], [top, top], color=color)
plt.plot([left, right], [bottom, bottom], color=color)
plt.plot(x, mean, 'o', color='#f44336')
return mean, confidence_interval
plt.xticks([1, 2, 3, 4], ['FF', 'BF', 'FFD', 'BFD'])
plt.title('Confidence Interval')
plot_confidence_interval(1, [10, 11, 42, 45, 44])
plot_confidence_interval(2, [10, 21, 42, 45, 44])
plot_confidence_interval(3, [20, 2, 4, 45, 44])
plot_confidence_interval(4, [30, 31, 42, 45, 44])
plt.show()
x: The x value of the input.
values: An array containing the repeated values (usually measured values) of y corresponding to the value of x.
z: The critical value of the z-distribution. Using 1.96 corresponds to the critical value of 95%.
Result:
For a confidence interval across categories, building on what omer sagi suggested, let's say if we have a Pandas data frame with a column that contains categories (like category 1, category 2, and category 3) and another that has continuous data (like some kind of rating), here's a function using pd.groupby() and scipy.stats to plot difference in means across groups with confidence intervals:
import pandas as pd
import numpy as np
import scipy.stats as st
def plot_diff_in_means(data: pd.DataFrame, col1: str, col2: str):
"""
Given data, plots difference in means with confidence intervals across groups
col1: categorical data with groups
col2: continuous data for the means
"""
n = data.groupby(col1)[col2].count()
# n contains a pd.Series with sample size for each category
cat = list(data.groupby(col1, as_index=False)[col2].count()[col1])
# 'cat' has the names of the categories, like 'category 1', 'category 2'
mean = data.groupby(col1)[col2].agg('mean')
# The average value of col2 across the categories
std = data.groupby(col1)[col2].agg(np.std)
se = std / np.sqrt(n)
# Standard deviation and standard error
lower = st.t.interval(alpha = 0.95, df=n-1, loc = mean, scale = se)[0]
upper = st.t.interval(alpha = 0.95, df =n-1, loc = mean, scale = se)[1]
# Calculates the upper and lower bounds using SciPy
for upper, mean, lower, y in zip(upper, mean, lower, cat):
plt.plot((lower, mean, upper), (y, y, y), 'b.-')
# for 'b.-': 'b' means 'blue', '.' means dot, '-' means solid line
plt.yticks(
range(len(n)),
list(data.groupby(col1, as_index = False)[col2].count()[col1])
)
Given hypothetical data:
cat = ['a'] * 10 + ['b'] * 10 + ['c'] * 10
a = np.linspace(0.1, 5.0, 10)
b = np.linspace(0.5, 7.0, 10)
c = np.linspace(7.5, 20.0, 10)
rating = np.concatenate([a, b, c])
dat_dict = dict()
dat_dict['cat'] = cat
dat_dict['rating'] = rating
test_dat = pd.DataFrame(dat_dict)
which would look like this (but with more rows of course):
cat
rating
a
0.10000
a
0.64444
b
0.50000
b
0.12222
c
7.50000
c
8.88889
We can use the function to plot a difference in means with a confidence interval:
plot_diff_in_means(data = test_dat, col1 = 'cat', col2 = 'rating')
which gives us the following graph:
The Matplotlib or Seaborn box plot gives the interquartile range between the 25th percentile and 75th percentile. Is there a way to give custom interquartile range for the Boxplot ? I need to get the box plot such that the interquartile range is between 10th percentile and 90th percentile. Looked up on google and other sources, came to know about getting custom whiskers on the box plot but not custom interquartile range. Hoping would get some useful solutions here.
Yes, it is possible to plot a boxplot with box edges at any percentiles you desire.
Convention
With box and whisker plots it is convention to plot the 25th and 75th percentiles of the data. Thus, you should be aware that departing from this convention puts you at risk of misleading readers. You should also carefully consider what altering the box percentiles means to outlier classification and the whiskers of the boxplot.
Quick solution
A quick fix (ignoring any implications for whisker locations) is to compute the boxplot statistics we desire, alter the locations of q1 and q3, and then plot with ax.bxp:
import matplotlib.cbook as cbook
import matplotlib.pyplot as plt
import numpy as np
# Generate some random data to visualise
np.random.seed(2019)
data = np.random.normal(size=100)
stats = {}
# Compute the boxplot stats (as in the default matplotlib implementation)
stats['A'] = cbook.boxplot_stats(data, labels='A')[0]
stats['B'] = cbook.boxplot_stats(data, labels='B')[0]
stats['C'] = cbook.boxplot_stats(data, labels='C')[0]
# For box A compute the 1st and 99th percentiles
stats['A']['q1'], stats['A']['q3'] = np.percentile(data, [1, 99])
# For box B compute the 10th and 90th percentiles
stats['B']['q1'], stats['B']['q3'] = np.percentile(data, [10, 90])
# For box C compute the 25th and 75th percentiles (matplotlib default)
stats['C']['q1'], stats['C']['q3'] = np.percentile(data, [25, 75])
fig, ax = plt.subplots(1, 1)
# Plot boxplots from our computed statistics
ax.bxp([stats['A'], stats['B'], stats['C']], positions=range(3))
However, viewing the plot produced we see that altering q1 and q3 whilst leaving the whiskers unchanged may not be a sensible idea. You could counter this by recomputing eg. stats['A']['iqr'] and the whisker locations stats['A']['whishi'] and stats['A']['whislo'].
A more complete solution
Looking through matplotlib's source code we find that matplotlib uses matplotlib.cbook.boxplot_stats to compute the statistics used in the boxplot.
Within boxplot_stats we find the code q1, med, q3 = np.percentile(x, [25, 50, 75]). This is the line we can alter to change the plotted percentiles.
So a potential solution would be to make a copy of matplotlib.cbook.boxplot_stats and alter it as we desire. Here I call the function my_boxplot_stats and add an argument percents to make it easy to alter the locations of q1 and q3.
import itertools
from matplotlib.cbook import _reshape_2D
import matplotlib.pyplot as plt
import numpy as np
# Function adapted from matplotlib.cbook
def my_boxplot_stats(X, whis=1.5, bootstrap=None, labels=None,
autorange=False, percents=[25, 75]):
def _bootstrap_median(data, N=5000):
# determine 95% confidence intervals of the median
M = len(data)
percentiles = [2.5, 97.5]
bs_index = np.random.randint(M, size=(N, M))
bsData = data[bs_index]
estimate = np.median(bsData, axis=1, overwrite_input=True)
CI = np.percentile(estimate, percentiles)
return CI
def _compute_conf_interval(data, med, iqr, bootstrap):
if bootstrap is not None:
# Do a bootstrap estimate of notch locations.
# get conf. intervals around median
CI = _bootstrap_median(data, N=bootstrap)
notch_min = CI[0]
notch_max = CI[1]
else:
N = len(data)
notch_min = med - 1.57 * iqr / np.sqrt(N)
notch_max = med + 1.57 * iqr / np.sqrt(N)
return notch_min, notch_max
# output is a list of dicts
bxpstats = []
# convert X to a list of lists
X = _reshape_2D(X, "X")
ncols = len(X)
if labels is None:
labels = itertools.repeat(None)
elif len(labels) != ncols:
raise ValueError("Dimensions of labels and X must be compatible")
input_whis = whis
for ii, (x, label) in enumerate(zip(X, labels)):
# empty dict
stats = {}
if label is not None:
stats['label'] = label
# restore whis to the input values in case it got changed in the loop
whis = input_whis
# note tricksyness, append up here and then mutate below
bxpstats.append(stats)
# if empty, bail
if len(x) == 0:
stats['fliers'] = np.array([])
stats['mean'] = np.nan
stats['med'] = np.nan
stats['q1'] = np.nan
stats['q3'] = np.nan
stats['cilo'] = np.nan
stats['cihi'] = np.nan
stats['whislo'] = np.nan
stats['whishi'] = np.nan
stats['med'] = np.nan
continue
# up-convert to an array, just to be safe
x = np.asarray(x)
# arithmetic mean
stats['mean'] = np.mean(x)
# median
med = np.percentile(x, 50)
## Altered line
q1, q3 = np.percentile(x, (percents[0], percents[1]))
# interquartile range
stats['iqr'] = q3 - q1
if stats['iqr'] == 0 and autorange:
whis = 'range'
# conf. interval around median
stats['cilo'], stats['cihi'] = _compute_conf_interval(
x, med, stats['iqr'], bootstrap
)
# lowest/highest non-outliers
if np.isscalar(whis):
if np.isreal(whis):
loval = q1 - whis * stats['iqr']
hival = q3 + whis * stats['iqr']
elif whis in ['range', 'limit', 'limits', 'min/max']:
loval = np.min(x)
hival = np.max(x)
else:
raise ValueError('whis must be a float, valid string, or list '
'of percentiles')
else:
loval = np.percentile(x, whis[0])
hival = np.percentile(x, whis[1])
# get high extreme
wiskhi = np.compress(x <= hival, x)
if len(wiskhi) == 0 or np.max(wiskhi) < q3:
stats['whishi'] = q3
else:
stats['whishi'] = np.max(wiskhi)
# get low extreme
wisklo = np.compress(x >= loval, x)
if len(wisklo) == 0 or np.min(wisklo) > q1:
stats['whislo'] = q1
else:
stats['whislo'] = np.min(wisklo)
# compute a single array of outliers
stats['fliers'] = np.hstack([
np.compress(x < stats['whislo'], x),
np.compress(x > stats['whishi'], x)
])
# add in the remaining stats
stats['q1'], stats['med'], stats['q3'] = q1, med, q3
return bxpstats
With this in place we can compute our statistics and then plot with plt.bxp.
# Generate some random data to visualise
np.random.seed(2019)
data = np.random.normal(size=100)
stats = {}
# Compute the boxplot stats with our desired percentiles
stats['A'] = my_boxplot_stats(data, labels='A', percents=[1, 99])[0]
stats['B'] = my_boxplot_stats(data, labels='B', percents=[10, 90])[0]
stats['C'] = my_boxplot_stats(data, labels='C', percents=[25, 75])[0]
fig, ax = plt.subplots(1, 1)
# Plot boxplots from our computed statistics
ax.bxp([stats['A'], stats['B'], stats['C']], positions=range(3))
See that with this solution the whiskers are adjusted in our function based on our selected percentiles.:
Explanation:
I have two numpy arrays: dataX and dataY, and I am trying to filter each array to reduce the noise. The image shown below shows the actual input data (blue dots) and an example of what I want it to be like(red dots). I do not need the filtered data to be as perfect as in the example but I do want it to be as straight as possible. I have provided sample data in the code.
What I have tried:
Firstly, you can see that the data isn't 'continuous', so I first divided them into individual 'segments' ( 4 of them in this example), and then applied a filter to each 'segment'. Someone suggested that I use a Savitzky-Golay filter. The full, run-able code is below:
import scipy as sc
import scipy.signal
import numpy as np
import matplotlib.pyplot as plt
# Sample Data
ydata = np.array([1,0,1,2,1,2,1,0,1,1,2,2,0,0,1,0,1,0,1,2,7,6,8,6,8,6,6,8,6,6,8,6,6,7,6,5,5,6,6, 10,11,12,13,12,11,10,10,11,10,12,11,10,10,10,10,12,12,10,10,17,16,15,17,16, 17,16,18,19,18,17,16,16,16,16,16,15,16])
xdata = np.array([1,2,3,1,5,4,7,8,6,10,11,12,13,10,12,13,17,16,19,18,21,19,23,21,25,20,26,27,28,26,26,26,29,30,30,29,30,32,33, 1,2,3,1,5,4,7,8,6,10,11,12,13,10,12,13,17,16,19,18,21,19,23,21,25,20,26,27,28,26,26,26,29,30,30,29,30,32])
# Used a diff array to find where there is a big change in Y.
# If there's a big change in Y, then there must be a change of 'segment'.
diffy = np.diff(ydata)
# Create empty numpy arrays to append values into
filteredX = np.array([])
filteredY = np.array([])
# Chose 3 to be the value indicating the change in Y
index = np.where(diffy >3)
# Loop through the array
start = 0
for i in range (0, (index[0].size +1) ):
# Check if last segment is reached
if i == index[0].size:
print xdata[start:]
partSize = xdata[start:].size
# Window length must be an odd integer
if partSize % 2 == 0:
partSize = partSize - 1
filteredDataX = sc.signal.savgol_filter(xdata[start:], partSize, 3)
filteredDataY = sc.signal.savgol_filter(ydata[start:], partSize, 3)
filteredX = np.append(filteredX, filteredDataX)
filteredY = np.append(filteredY, filteredDataY)
else:
print xdata[start:index[0][i]]
partSize = xdata[start:index[0][i]].size
if partSize % 2 == 0:
partSize = partSize - 1
filteredDataX = sc.signal.savgol_filter(xdata[start:index[0][i]], partSize, 3)
filteredDataY = sc.signal.savgol_filter(ydata[start:index[0][i]], partSize, 3)
start = index[0][i]
filteredX = np.append(filteredX, filteredDataX)
filteredY = np.append(filteredY, filteredDataY)
# Plots
plt.plot(xdata,ydata, 'bo', label = 'Input Data')
plt.plot(filteredX, filteredY, 'ro', label = 'Filtered Data')
plt.xlabel('X')
plt.ylabel('Y')
plt.title('Result')
plt.legend()
plt.show()
This is my result:
When each point is connected, the result looks as follows.
I have played around with the order, but it seems like a third order gave the best result.
I have also tried these filters, among a few others:
scipy.signal.medfilt
scipy.ndimage.filters.uniform_filter1d
But so far none of the filters I have tried were close to what I really wanted. What is the best way to filter data such as this? Looking forward to your help.
One way to get something looking close to your ideal would be clustering + linear regression.
Note that you have to provide the number of clusters and I also cheated a bit in scaling up y before clustering.
import numpy as np
from scipy import cluster, stats
ydata = np.array([1,0,1,2,1,2,1,0,1,1,2,2,0,0,1,0,1,0,1,2,7,6,8,6,8,6,6,8,6,6,8,6,6,7,6,5,5,6,6, 10,11,12,13,12,11,10,10,11,10,12,11,10,10,10,10,12,12,10,10,17,16,15,17,16, 17,16,18,19,18,17,16,16,16,16,16,15,16])
xdata = np.array([1,2,3,1,5,4,7,8,6,10,11,12,13,10,12,13,17,16,19,18,21,19,23,21,25,20,26,27,28,26,26,26,29,30,30,29,30,32,33, 1,2,3,1,5,4,7,8,6,10,11,12,13,10,12,13,17,16,19,18,21,19,23,21,25,20,26,27,28,26,26,26,29,30,30,29,30,32])
def split_to_lines(x, y, k):
yo = np.empty_like(y, dtype=float)
# get the cluster centers and the labels for each point
centers, map_ = cluster.vq.kmeans2(np.array((x, y * 2)).T.astype(float), k)
# for each cluster, use the labels to select the points belonging to
# the cluster and do a linear regression
for i in range(k):
slope, interc, *_ = stats.linregress(x[map_==i], y[map_==i])
# use the regression parameters to construct y values on the
# best fit line
yo[map_==i] = x[map_==i] * slope + interc
return yo
import pylab
pylab.plot(xdata, ydata, 'or')
pylab.plot(xdata, split_to_lines(xdata, ydata, 4), 'ob')
pylab.show()
Is there a numpy builtin to do something like the following? That is, take a list d and return a list filtered_d with any outlying elements removed based on some assumed distribution of the points in d.
import numpy as np
def reject_outliers(data):
m = 2
u = np.mean(data)
s = np.std(data)
filtered = [e for e in data if (u - 2 * s < e < u + 2 * s)]
return filtered
>>> d = [2,4,5,1,6,5,40]
>>> filtered_d = reject_outliers(d)
>>> print filtered_d
[2,4,5,1,6,5]
I say 'something like' because the function might allow for varying distributions (poisson, gaussian, etc.) and varying outlier thresholds within those distributions (like the m I've used here).
Something important when dealing with outliers is that one should try to use estimators as robust as possible. The mean of a distribution will be biased by outliers but e.g. the median will be much less.
Building on eumiro's answer:
def reject_outliers(data, m = 2.):
d = np.abs(data - np.median(data))
mdev = np.median(d)
s = d/mdev if mdev else np.zero(len(d))
return data[s<m]
Here I have replace the mean with the more robust median and the standard deviation with the median absolute distance to the median. I then scaled the distances by their (again) median value so that m is on a reasonable relative scale.
Note that for the data[s<m] syntax to work, data must be a numpy array.
This method is almost identical to yours, just more numpyst (also working on numpy arrays only):
def reject_outliers(data, m=2):
return data[abs(data - np.mean(data)) < m * np.std(data)]
Benjamin Bannier's answer yields a pass-through when the median of distances from the median is 0, so I found this modified version a bit more helpful for cases as given in the example below.
def reject_outliers_2(data, m=2.):
d = np.abs(data - np.median(data))
mdev = np.median(d)
s = d / (mdev if mdev else 1.)
return data[s < m]
Example:
data_points = np.array([10, 10, 10, 17, 10, 10])
print(reject_outliers(data_points))
print(reject_outliers_2(data_points))
Gives:
[[10, 10, 10, 17, 10, 10]] # 17 is not filtered
[10, 10, 10, 10, 10] # 17 is filtered (it's distance, 7, is greater than m)
Building on Benjamin's, using pandas.Series, and replacing MAD with IQR:
def reject_outliers(sr, iq_range=0.5):
pcnt = (1 - iq_range) / 2
qlow, median, qhigh = sr.dropna().quantile([pcnt, 0.50, 1-pcnt])
iqr = qhigh - qlow
return sr[ (sr - median).abs() <= iqr]
For instance, if you set iq_range=0.6, the percentiles of the interquartile-range would become: 0.20 <--> 0.80, so more outliers will be included.
An alternative is to make a robust estimation of the standard deviation (assuming Gaussian statistics). Looking up online calculators, I see that the 90% percentile corresponds to 1.2815σ and the 95% is 1.645σ (http://vassarstats.net/tabs.html?#z)
As a simple example:
import numpy as np
# Create some random numbers
x = np.random.normal(5, 2, 1000)
# Calculate the statistics
print("Mean= ", np.mean(x))
print("Median= ", np.median(x))
print("Max/Min=", x.max(), " ", x.min())
print("StdDev=", np.std(x))
print("90th Percentile", np.percentile(x, 90))
# Add a few large points
x[10] += 1000
x[20] += 2000
x[30] += 1500
# Recalculate the statistics
print()
print("Mean= ", np.mean(x))
print("Median= ", np.median(x))
print("Max/Min=", x.max(), " ", x.min())
print("StdDev=", np.std(x))
print("90th Percentile", np.percentile(x, 90))
# Measure the percentile intervals and then estimate Standard Deviation of the distribution, both from median to the 90th percentile and from the 10th to 90th percentile
p90 = np.percentile(x, 90)
p10 = np.percentile(x, 10)
p50 = np.median(x)
# p50 to p90 is 1.2815 sigma
rSig = (p90-p50)/1.2815
print("Robust Sigma=", rSig)
rSig = (p90-p10)/(2*1.2815)
print("Robust Sigma=", rSig)
The output I get is:
Mean= 4.99760520022
Median= 4.95395274981
Max/Min= 11.1226494654 -2.15388472011
Sigma= 1.976629928
90th Percentile 7.52065379649
Mean= 9.64760520022
Median= 4.95667658782
Max/Min= 2205.43861943 -2.15388472011
Sigma= 88.6263902244
90th Percentile 7.60646688694
Robust Sigma= 2.06772555531
Robust Sigma= 1.99878292462
Which is close to the expected value of 2.
If we want to remove points above/below 5 standard deviations (with 1000 points we would expect 1 value > 3 standard deviations):
y = x[abs(x - p50) < rSig*5]
# Print the statistics again
print("Mean= ", np.mean(y))
print("Median= ", np.median(y))
print("Max/Min=", y.max(), " ", y.min())
print("StdDev=", np.std(y))
Which gives:
Mean= 4.99755359935
Median= 4.95213030447
Max/Min= 11.1226494654 -2.15388472011
StdDev= 1.97692712883
I have no idea which approach is the more efficent/robust
I wanted to do something similar, except setting the number to NaN rather than removing it from the data, since if you remove it you change the length which can mess up plotting (i.e. if you're only removing outliers from one column in a table, but you need it to remain the same as the other columns so you can plot them against each other).
To do so I used numpy's masking functions:
def reject_outliers(data, m=2):
stdev = np.std(data)
mean = np.mean(data)
maskMin = mean - stdev * m
maskMax = mean + stdev * m
mask = np.ma.masked_outside(data, maskMin, maskMax)
print('Masking values outside of {} and {}'.format(maskMin, maskMax))
return mask
I would like to provide two methods in this answer, solution based on "z score" and solution based on "IQR".
The code provided in this answer works on both single dim numpy array and multiple numpy array.
Let's import some modules firstly.
import collections
import numpy as np
import scipy.stats as stat
from scipy.stats import iqr
z score based method
This method will test if the number falls outside the three standard deviations. Based on this rule, if the value is outlier, the method will return true, if not, return false.
def sd_outlier(x, axis = None, bar = 3, side = 'both'):
assert side in ['gt', 'lt', 'both'], 'Side should be `gt`, `lt` or `both`.'
d_z = stat.zscore(x, axis = axis)
if side == 'gt':
return d_z > bar
elif side == 'lt':
return d_z < -bar
elif side == 'both':
return np.abs(d_z) > bar
IQR based method
This method will test if the value is less than q1 - 1.5 * iqr or greater than q3 + 1.5 * iqr, which is similar to SPSS's plot method.
def q1(x, axis = None):
return np.percentile(x, 25, axis = axis)
def q3(x, axis = None):
return np.percentile(x, 75, axis = axis)
def iqr_outlier(x, axis = None, bar = 1.5, side = 'both'):
assert side in ['gt', 'lt', 'both'], 'Side should be `gt`, `lt` or `both`.'
d_iqr = iqr(x, axis = axis)
d_q1 = q1(x, axis = axis)
d_q3 = q3(x, axis = axis)
iqr_distance = np.multiply(d_iqr, bar)
stat_shape = list(x.shape)
if isinstance(axis, collections.Iterable):
for single_axis in axis:
stat_shape[single_axis] = 1
else:
stat_shape[axis] = 1
if side in ['gt', 'both']:
upper_range = d_q3 + iqr_distance
upper_outlier = np.greater(x - upper_range.reshape(stat_shape), 0)
if side in ['lt', 'both']:
lower_range = d_q1 - iqr_distance
lower_outlier = np.less(x - lower_range.reshape(stat_shape), 0)
if side == 'gt':
return upper_outlier
if side == 'lt':
return lower_outlier
if side == 'both':
return np.logical_or(upper_outlier, lower_outlier)
Finally, if you want to filter out the outliers, use a numpy selector.
Have a nice day.
Consider that all the above methods fail when your standard deviation gets very large due to huge outliers.
(Simalar as the average caluclation fails and should rather caluclate the median. Though, the average is "more prone to such an error as the stdDv".)
You could try to iteratively apply your algorithm or you filter using the interquartile range:
(here "factor" relates to a n*sigma range, yet only when your data follows a Gaussian distribution)
import numpy as np
def sortoutOutliers(dataIn,factor):
quant3, quant1 = np.percentile(dataIn, [75 ,25])
iqr = quant3 - quant1
iqrSigma = iqr/1.34896
medData = np.median(dataIn)
dataOut = [ x for x in dataIn if ( (x > medData - factor* iqrSigma) and (x < medData + factor* iqrSigma) ) ]
return(dataOut)
So many answers, but I'm adding a new one that can be useful for the author or even for other users.
You could use the Hampel filter. But you need to work with Series.
Hampel filter returns the Outliers indices, then you can delete them from the Series, and then convert it back to a List.
To use Hampel filter, you can easily install the package with pip:
pip install hampel
Usage:
# Imports
from hampel import hampel
import pandas as pd
list_d = [2, 4, 5, 1, 6, 5, 40]
# List to Series
time_series = pd.Series(list_d)
# Outlier detection with Hampel filter
# Returns the Outlier indices
outlier_indices = hampel(ts = time_series, window_size = 3)
# Drop Outliers indices from Series
filtered_d = time_series.drop(outlier_indices)
filtered_d.values.tolist()
print(f'filtered_d: {filtered_d.values.tolist()}')
And the output will be:
filtered_d: [2, 4, 5, 1, 6, 5]
Where, ts is a pandas Series object and window_size is a total window size will be computed as 2 * window_size + 1.
For this Series I set window_size with the value 3.
The cool thing about working with Series is being able to generate graphics:
# Imports
import matplotlib.pyplot as plt
plt.style.use('seaborn-darkgrid')
# Plot Original Series
time_series.plot(style = 'k-')
plt.title('Original Series')
plt.show()
# Plot Cleaned Series
filtered_d.plot(style = 'k-')
plt.title('Cleaned Series (Without detected Outliers)')
plt.show()
And the output will be:
To learn more about Hampel filter, I recommend the following readings:
Python implementation of the Hampel Filter
Outlier Detection with Hampel Filter
Clean-up your time series data with a Hampel Filter
if you want to get the index position of the outliers idx_list will return it.
def reject_outliers(data, m = 2.):
d = np.abs(data - np.median(data))
mdev = np.median(d)
s = d/mdev if mdev else 0.
data_range = np.arange(len(data))
idx_list = data_range[s>=m]
return data[s<m], idx_list
data_points = np.array([8, 10, 35, 17, 73, 77])
print(reject_outliers(data_points))
after rejection: [ 8 10 35 17], index positions of outliers: [4 5]
For a set of images (each image has 3 dimensions), where I wanted to reject outliers for each pixel I used:
mean = np.mean(imgs, axis=0)
std = np.std(imgs, axis=0)
mask = np.greater(0.5 * std + 1, np.abs(imgs - mean))
masked = np.multiply(imgs, mask)
Then it is possible to compute the mean:
masked_mean = np.divide(np.sum(masked, axis=0), np.sum(mask, axis=0))
(I use it for Background Subtraction)
Here I find the outliers in x and substitute them with the median of a window of points (win) around them (taking from Benjamin Bannier answer the median deviation)
def outlier_smoother(x, m=3, win=3, plots=False):
''' finds outliers in x, points > m*mdev(x) [mdev:median deviation]
and replaces them with the median of win points around them '''
x_corr = np.copy(x)
d = np.abs(x - np.median(x))
mdev = np.median(d)
idxs_outliers = np.nonzero(d > m*mdev)[0]
for i in idxs_outliers:
if i-win < 0:
x_corr[i] = np.median(np.append(x[0:i], x[i+1:i+win+1]))
elif i+win+1 > len(x):
x_corr[i] = np.median(np.append(x[i-win:i], x[i+1:len(x)]))
else:
x_corr[i] = np.median(np.append(x[i-win:i], x[i+1:i+win+1]))
if plots:
plt.figure('outlier_smoother', clear=True)
plt.plot(x, label='orig.', lw=5)
plt.plot(idxs_outliers, x[idxs_outliers], 'ro', label='outliers')
plt.plot(x_corr, '-o', label='corrected')
plt.legend()
return x_corr
Trim outliers in a numpy array along axis and replace them with min or max values along this axis, whichever is closer. The threshold is z-score:
def np_z_trim(x, threshold=10, axis=0):
""" Replace outliers in numpy ndarray along axis with min or max values
within the threshold along this axis, whichever is closer."""
mean = np.mean(x, axis=axis, keepdims=True)
std = np.std(x, axis=axis, keepdims=True)
masked = np.where(np.abs(x - mean) < threshold * std, x, np.nan)
min = np.nanmin(masked, axis=axis, keepdims=True)
max = np.nanmax(masked, axis=axis, keepdims=True)
repl = np.where(np.abs(x - max) < np.abs(x - min), max, min)
return np.where(np.isnan(masked), repl, masked)
My solution drops the top and bottom percentiles, keeping values that are equal to the boundary:
def remove_percentile_outliers(data, percent_to_drop=0.001):
low, high = data.quantile([percent_to_drop / 2, 1-percent_to_drop / 2])
return data[(data >= low )&(data <= high)]
My solution let the outliers equal to the previous value.
test_data = [2,4,5,1,6,5,40, 3]
def reject_outliers(data, m=2):
mean = np.mean(data)
std = np.std(data)
for i in range(len(data)) :
if np.abs(data[i] -mean) > m*std :
data[i] = data[i-1]
return data
reject_outliers(test_data)
Output:
[2, 4, 5, 1, 6, 5, 5, 3]