I try to plot normalized histogram using example from numpy.random.normal documentation. For this purpose I generate normally distributed random sample.
mu_true = 0
sigma_true = 0.1
s = np.random.normal(mu_true, sigma_true, 2000)
Then I fitt normal distribution to the data and calculate pdf.
mu, sigma = stats.norm.fit(s)
points = np.linspace(stats.norm.ppf(0.01,loc=mu,scale=sigma),
stats.norm.ppf(0.9999,loc=mu,scale=sigma),100)
pdf = stats.norm.pdf(points,loc=mu,scale=sigma)
Display fitted pdf and data histogram.
plt.hist(s, 30, density=True);
plt.plot(points, pdf, color='r')
plt.show()
I use density=True, but it is obviously, that pdf and histogram are not normalized.
What can one suggests to plot truly normalized histogram and pdf?
Seaborn distplot also doesn't solve the problem.
import seaborn as sns
ax = sns.distplot(s)
What makes you think it is not normalised? At a guess, it's probably because the heights of each column extend to values greater than 1. However, this thinking is flawed because in a normalised histogram/pdf, the total area under it should sum to one (not the heights). When you are dealing with small steps in x (as you are), that are less than one, then it is not surprising that the column heights are greater than one!
You can see this clearly in the scipy example you link: the x-values are much greater (by an order of magnitude) so it follows that their y-values are also smaller. You will see the same effect if you change your distribution to cover a wider range of values. Try a sigma of 10 instead of 0.1, see what happens!
import numpy as np
from numpy.random import seed, randn
from scipy.stats import norm
import matplotlib.pyplot as plt
import seaborn as sns
sns.set_theme()
"Try this, for 𝜇 = 0"
seed(0)
points = np.linspace(-5,5,100)
pdf = norm.pdf(points,0,1)
plt.plot(points, pdf, color='r')
plt.hist(randn(50), density=True);
plt.show()
"or this, for 𝜇 = 10"
seed(0)
points = np.linspace(5,15,100)
pdf = norm.pdf(points,10,1)
plt.plot(points, pdf, color='r')
plt.hist(10+randn(50), density=True);
plt.show()
Related
I want to make a histogram from 30 csv files, and then fit a gaussian function to see if my data is optimal. After that, I need to find the mean and standard deviation of those peaks. The file data size are too large, I do not know if I extract individual column and organize their value range into number of bins correctly.
I know it is a bit long and too many questions, please answer as much as you want, thank you very much!
> this is the links of the data
Below so far I have done (actually not much, coz I am beginner to data visualization.)
Firstly, I import the packages, savgol_filter to make the bin transparent, it seems better.
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy.signal import savgol_filter
And then I convert the dimension and set limit.
def cm2inch(value):
return value/2.54
width = 9
height = 6.75
sliceMin, sliceMax = 300, 1002
Next I load all the data jupyter notebook by iteration 30 times, where I set up two arrays "times" and "voltages" to store the values.
times, voltages = [], []
for i in range(30):
time, ch1 = np.loadtxt(f"{i+1}.txt", delimiter=',', skiprows=5,unpack=True)
times.append(time)
voltages.append(ch1)
t = (np.array(times[0]) * 1e5)[sliceMin:sliceMax]
voltages = (np.array(voltages))[:, sliceMin:sliceMax]
1. I think I should need a hist function to plot the graph. Although I have the plot, but I am not sure if it is the proper way to generate the histogram.
hist, bin_edges = np.histogram(voltages, bins=500, density=True)
hist = savgol_filter(hist, 51, 3)
bin_centres = (bin_edges[:-1] + bin_edges[1:])/2
That is so far I have reached. the amplitude of the 3rd peak is too low, which is not what I expected. But please correct me if my expectation is wrong.
This is my histogram plot
I have updated my plot with the following code
labels = "hist"
if showGraph:
plt.title("Datapoints Distribution over Voltage [mV]", )
plt.xlabel("Voltage [mV]")
plt.ylabel("Data Points")
plt.plot(hist, label=labels)
plt.show()
2.(edited) I am not sure why my label cannot display, could you please correct me?
3.(edited) Besides, I want to make a fit curve by using gaussian function to the histogram. But there are three peaks, so how should I fit the function to them?
def gauss(x, *p):
A, mu, sigma = p
return A*np.exp(-(x-mu)**2/(2.*sigma**2))
4. (edited) I realised that I have not mentioned the mean value yet.
I suppose that if I can locate the maximum value of the peak, then I can find the mean value of the specific peak. Do I need to fit the Gaussian first to find the peak, or I can find the straight ahead? Is it to find the local maximum so I can find it? If yes, how can I proceed it?
5. (edited) I know how to find the standard deviation from a single list, if I want to do similar logic, how to implement the code?
sample = [1,2,3,4,5,5,5,5,10]
standard_deviation = np.std(sample, ddof=1)
print(standard_deviation)
Feedback to suggestions:
I try to implement the gaussian fit, below are the packages I import.
from sklearn.mixture import GaussianMixture
import numpy as np
import matplotlib.pyplot as plt
Here isthe gaussian function, I put my 30 datasets voltages as the parameter of the Gaussian Mixture fit, which print our lots of values regarding mu and variance.
gmm = GaussianMixture(n_components=1)
gmm.fit(voltages)
print(gmm.means_, gmm.covariances_)
mu = gmm.means_[0][0]
variance = gmm.covariances_[0][0][0]
print(mu, variance)
I process the code one by one. There is an error on the second line:
fig, ax = plt.subplots(figsize=(6,6))
Xs = np.arange(min(voltages), max(voltages), 0.05)
The truth value of an array with more than one element is ambiguous.
Use a.any() or a.all()
I search from the web that, to use this is to indicate there is only one value, like if there are[T,T,F,F,T], you can have 4 possibilities.
I edit my code to:
Xs = np.arange(min(np.all(voltages)), max(np.all(voltages)), 0.05)
which gives me this:
'numpy.bool_' object is not iterable
I understand it is not a boolean object. At this stage, I do not know how to proceed the gaussian curve fit. Can anyone provides me an alternate way to do it?
To plot a histogram, the most vanilla matplotlib function, hist, is my go-to. Basically, if I have a list of samples, then I can plot a histogram of them with 100 bins via:
import matplotlib.pyplot as plt
plt.hist(samples, bins=100)
plt.show()
If you'd like to fit normal distribution(s) to your data, the best model for that is a Gaussian Mixture Model, which you can find more info about via scikit-learn's GMM page. That said, this is the code I use to fit a singular gaussian distribution to a dataset. If I wanted to fit k normal distributions, I'd need to use n_components=k. I've also included the resulting plot:
from sklearn.mixture import GaussianMixture
import numpy as np
import matplotlib.pyplot as plt
data = np.random.uniform(-1,1, size=(800,1))
data += np.random.uniform(-1,1, size=(800,1))
gmm = GaussianMixture(n_components=1)
gmm.fit(data)
print(gmm.means_, gmm.covariances_)
mu = gmm.means_[0][0]
variance = gmm.covariances_[0][0][0]
print(mu, variance)
fig, ax = plt.subplots(figsize=(6,6))
Xs = np.arange(min(data), max(data), 0.05)
ys = 1.0/np.sqrt(2*np.pi*variance) * np.exp(-0.5/variance * (Xs + mu)**2)
ax.hist(data, bins=100, label='data')
px = ax.twinx()
px.plot(Xs, ys, c='r', linestyle='dotted', label='fit')
ax.legend()
px.legend(loc='upper left')
plt.show()
As for question 3, I'm not sure which axis you'd like to capture the standard deviations of. If you'd like to get the standard deviation of columns, you can use np.std(data, axis=1), and use axis=0 for row-by-row standard deviation.
When I draw displot for discrete variables, the distribution might not be as what I think. For example.
We can find that there are crevices in the barplot so that the curve in kdeplot is "lower" in y axis.
In my work, it was even worse:
I think it may because the "width" or "weight" was not 1 for each bar. But I didn't find any parameter that can justify it.
I'd like to draw such curve (It should be more smooth)
One way to deal with this problem might be to adjust the "bandwidth" of the KDE (see the documentation for seaborn.kdeplot())
n = np.round(np.random.normal(5,2,size=(10000,)))
sns.distplot(n, kde_kws={'bw':1})
EDIT Here is an alternative with a different scale for the bars and the KDE
n = np.round(np.random.normal(5,2,size=(10000,)))
fig, ax1 = plt.subplots()
ax2 = ax1.twinx()
sns.distplot(n, kde=False, ax=ax1)
sns.distplot(n, hist=False, ax=ax2, kde_kws={'bw':1})
If the problem is that there are some emptry bins in the histogram, it probably makes sense to specify the bins to match the data. In this case, use bins=np.arange(0,16) to get the bins for all integers in the data.
import numpy as np; np.random.seed(1)
import matplotlib.pyplot as plt
import seaborn as sns
n = np.random.randint(0,15,10000)
sns.distplot(n, bins=np.arange(0,16), hist_kws=dict(ec="k"))
plt.show()
It seems sns.distplot (or displot https://seaborn.pydata.org/generated/seaborn.displot.html) is for plotting histograms and no barplots. Both Histogram and KDE (which is an approximation of the probability density function) make sense only with continuous random variables.
So in your case, as you'd like to plot a distribution of a discrete random variable, you must go for a bar plot and plotting the Probability Mass Function (PMF) instead.
import numpy as np
import matplotlib.pyplot as plt
array = np.random.randint(15, size=10000)
unique, counts = np.unique(array, return_counts=True)
freq =counts/10000 # to change into frequency, no count
# plotting the points
plt.bar(unique, freq)
# naming the x axis
plt.xlabel('Value')
# naming the y axis
plt.ylabel('Frequency')
#Title
plt.title("Discrete uniform distribution")
# function to show the plot
plt.show()
I'd like to plot a normalized histogram from a vector using matplotlib. I tried the following:
plt.hist(myarray, normed=True)
as well as:
plt.hist(myarray, normed=1)
but neither option produces a y-axis from [0, 1] such that the bar heights of the histogram sum to 1.
If you want the sum of all bars to be equal unity, weight each bin by the total number of values:
weights = np.ones_like(myarray) / len(myarray)
plt.hist(myarray, weights=weights)
Note for Python 2.x: add casting to float() for one of the operators of the division as otherwise you would end up with zeros due to integer division
It would be more helpful if you posed a more complete working (or in this case non-working) example.
I tried the following:
import numpy as np
import matplotlib.pyplot as plt
x = np.random.randn(1000)
fig = plt.figure()
ax = fig.add_subplot(111)
n, bins, rectangles = ax.hist(x, 50, density=True)
fig.canvas.draw()
plt.show()
This will indeed produce a bar-chart histogram with a y-axis that goes from [0,1].
Further, as per the hist documentation (i.e. ax.hist? from ipython), I think the sum is fine too:
*normed*:
If *True*, the first element of the return tuple will
be the counts normalized to form a probability density, i.e.,
``n/(len(x)*dbin)``. In a probability density, the integral of
the histogram should be 1; you can verify that with a
trapezoidal integration of the probability density function::
pdf, bins, patches = ax.hist(...)
print np.sum(pdf * np.diff(bins))
Giving this a try after the commands above:
np.sum(n * np.diff(bins))
I get a return value of 1.0 as expected. Remember that normed=True doesn't mean that the sum of the value at each bar will be unity, but rather than the integral over the bars is unity. In my case np.sum(n) returned approx 7.2767.
I know this answer is too late considering the question is dated 2010 but I came across this question as I was facing a similar problem myself. As already stated in the answer, normed=True means that the total area under the histogram is equal to 1 but the sum of heights is not equal to 1. However, I wanted to, for convenience of physical interpretation of a histogram, make one with sum of heights equal to 1.
I found a hint in the following question - Python: Histogram with area normalized to something other than 1
But I was not able to find a way of making bars mimic the histtype="step" feature hist(). This diverted me to : Matplotlib - Stepped histogram with already binned data
If the community finds it acceptable I should like to put forth a solution which synthesises ideas from both the above posts.
import matplotlib.pyplot as plt
# Let X be the array whose histogram needs to be plotted.
nx, xbins, ptchs = plt.hist(X, bins=20)
plt.clf() # Get rid of this histogram since not the one we want.
nx_frac = nx/float(len(nx)) # Each bin divided by total number of objects.
width = xbins[1] - xbins[0] # Width of each bin.
x = np.ravel(zip(xbins[:-1], xbins[:-1]+width))
y = np.ravel(zip(nx_frac,nx_frac))
plt.plot(x,y,linestyle="dashed",label="MyLabel")
#... Further formatting.
This has worked wonderfully for me though in some cases I have noticed that the left most "bar" or the right most "bar" of the histogram does not close down by touching the lowest point of the Y-axis. In such a case adding an element 0 at the begging or the end of y achieved the necessary result.
Just thought I'd share my experience. Thank you.
Here is another simple solution using np.histogram() method.
myarray = np.random.random(100)
results, edges = np.histogram(myarray, normed=True)
binWidth = edges[1] - edges[0]
plt.bar(edges[:-1], results*binWidth, binWidth)
You can indeed check that the total sums up to 1 with:
> print sum(results*binWidth)
1.0
The easiest solution is to use seaborn.histplot, or seaborn.displot with kind='hist', and specify stat='probability'
probability: or proportion: normalize such that bar heights sum to 1
density: normalize such that the total area of the histogram equals 1
data: pandas.DataFrame, numpy.ndarray, mapping, or sequence
seaborn is a high-level API for matplotlib
Tested in python 3.8.12, matplotlib 3.4.3, seaborn 0.11.2
Imports and Data
import seaborn as sns
import matplotlib.pyplot as plt
# load data
df = sns.load_dataset('penguins')
sns.histplot
axes-level plot
# create figure and axes
fig, ax = plt.subplots(figsize=(6, 5))
p = sns.histplot(data=df, x='flipper_length_mm', stat='probability', ax=ax)
sns.displot
figure-level plot
p = sns.displot(data=df, x='flipper_length_mm', stat='probability', height=4, aspect=1.5)
Since matplotlib 3.0.2, normed=True is deprecated. To get the desired output I had to do:
import numpy as np
data=np.random.randn(1000)
bins=np.arange(-3.0,3.0,51)
counts, _ = np.histogram(data,bins=bins)
if density: # equivalent of normed=True
counts_weighter=counts.sum()
else: # equivalent of normed=False
counts_weighter=1.0
plt.hist(bins[:-1],bins=bins,weights=counts/counts_weighter)
Trying to specify weights and density simultaneously as arguments to plt.hist() did not work for me. If anyone know of a way to get that working without having access to the normed keyword argument then please let me know in the comments and I will delete/modify this answer.
If you want bin centres then don't use bins[:-1] which are the bin edges - you need to choose a suitable scheme for how to calculate the centres (which may or may not be trivially derived).
I have two set of data with one containing around 11 million data points and the another around 5000. I would like to plot them both on one histogram. But because of the difference in size I need to normalise the frequency so I can plot them on the same figure. Below I have simulated what I have done with my data to be able to plot them. I have used the normed=True.
from numpy.random import randn
import matplotlib.pyplot as plt
import random
datalist1=[]
for x in range(1,50000):
datalist1.append(random.uniform(1,2))
datalist2=randn(5000000)
fig= plt.figure(1)
plt.hist(datalist1,bins=20,color='b',alpha=0.3,label='theoretical',histtype='stepfilled', normed=True)
plt.hist(datalist2,bins=20,alpha=0.5,color='g',label='experimental',histtype='stepfilled',normed=True)
plt.xlabel("Value")
plt.ylabel("Normalised Frequency")
plt.legend()
plt.show()
Can you please tell me if this is a good way to get around this issue? I would like to match the tallest hight between the two histogram frequencies to be 1 (or 100%).
The normed=True setting normalizes the histogram to an area of 1. That gives the histogram an interpretation as estimates of probability density functions.
In short, it actually makes sense not to normalize on the peak but on the area.
But if you really want to normalize by height you can modify the polygon data of the histogram:
h = plt.hist(datalist1,bins=20,color='b',alpha=0.3,label='theoretical',histtype='stepfilled', normed=True)
p = h[2][0]
p.xy[:,1] /= p.xy[:, 1].max()
h = plt.hist(datalist2,bins=20,alpha=0.5,color='g',label='experimental',histtype='stepfilled',normed=True)
p = h[2][0]
p.xy[:,1] /= p.xy[:, 1].max()
This solution feels a bit hackish, but at least it's quick and dirty :)
I'd like to plot a normalized histogram from a vector using matplotlib. I tried the following:
plt.hist(myarray, normed=True)
as well as:
plt.hist(myarray, normed=1)
but neither option produces a y-axis from [0, 1] such that the bar heights of the histogram sum to 1.
If you want the sum of all bars to be equal unity, weight each bin by the total number of values:
weights = np.ones_like(myarray) / len(myarray)
plt.hist(myarray, weights=weights)
Note for Python 2.x: add casting to float() for one of the operators of the division as otherwise you would end up with zeros due to integer division
It would be more helpful if you posed a more complete working (or in this case non-working) example.
I tried the following:
import numpy as np
import matplotlib.pyplot as plt
x = np.random.randn(1000)
fig = plt.figure()
ax = fig.add_subplot(111)
n, bins, rectangles = ax.hist(x, 50, density=True)
fig.canvas.draw()
plt.show()
This will indeed produce a bar-chart histogram with a y-axis that goes from [0,1].
Further, as per the hist documentation (i.e. ax.hist? from ipython), I think the sum is fine too:
*normed*:
If *True*, the first element of the return tuple will
be the counts normalized to form a probability density, i.e.,
``n/(len(x)*dbin)``. In a probability density, the integral of
the histogram should be 1; you can verify that with a
trapezoidal integration of the probability density function::
pdf, bins, patches = ax.hist(...)
print np.sum(pdf * np.diff(bins))
Giving this a try after the commands above:
np.sum(n * np.diff(bins))
I get a return value of 1.0 as expected. Remember that normed=True doesn't mean that the sum of the value at each bar will be unity, but rather than the integral over the bars is unity. In my case np.sum(n) returned approx 7.2767.
I know this answer is too late considering the question is dated 2010 but I came across this question as I was facing a similar problem myself. As already stated in the answer, normed=True means that the total area under the histogram is equal to 1 but the sum of heights is not equal to 1. However, I wanted to, for convenience of physical interpretation of a histogram, make one with sum of heights equal to 1.
I found a hint in the following question - Python: Histogram with area normalized to something other than 1
But I was not able to find a way of making bars mimic the histtype="step" feature hist(). This diverted me to : Matplotlib - Stepped histogram with already binned data
If the community finds it acceptable I should like to put forth a solution which synthesises ideas from both the above posts.
import matplotlib.pyplot as plt
# Let X be the array whose histogram needs to be plotted.
nx, xbins, ptchs = plt.hist(X, bins=20)
plt.clf() # Get rid of this histogram since not the one we want.
nx_frac = nx/float(len(nx)) # Each bin divided by total number of objects.
width = xbins[1] - xbins[0] # Width of each bin.
x = np.ravel(zip(xbins[:-1], xbins[:-1]+width))
y = np.ravel(zip(nx_frac,nx_frac))
plt.plot(x,y,linestyle="dashed",label="MyLabel")
#... Further formatting.
This has worked wonderfully for me though in some cases I have noticed that the left most "bar" or the right most "bar" of the histogram does not close down by touching the lowest point of the Y-axis. In such a case adding an element 0 at the begging or the end of y achieved the necessary result.
Just thought I'd share my experience. Thank you.
Here is another simple solution using np.histogram() method.
myarray = np.random.random(100)
results, edges = np.histogram(myarray, normed=True)
binWidth = edges[1] - edges[0]
plt.bar(edges[:-1], results*binWidth, binWidth)
You can indeed check that the total sums up to 1 with:
> print sum(results*binWidth)
1.0
The easiest solution is to use seaborn.histplot, or seaborn.displot with kind='hist', and specify stat='probability'
probability: or proportion: normalize such that bar heights sum to 1
density: normalize such that the total area of the histogram equals 1
data: pandas.DataFrame, numpy.ndarray, mapping, or sequence
seaborn is a high-level API for matplotlib
Tested in python 3.8.12, matplotlib 3.4.3, seaborn 0.11.2
Imports and Data
import seaborn as sns
import matplotlib.pyplot as plt
# load data
df = sns.load_dataset('penguins')
sns.histplot
axes-level plot
# create figure and axes
fig, ax = plt.subplots(figsize=(6, 5))
p = sns.histplot(data=df, x='flipper_length_mm', stat='probability', ax=ax)
sns.displot
figure-level plot
p = sns.displot(data=df, x='flipper_length_mm', stat='probability', height=4, aspect=1.5)
Since matplotlib 3.0.2, normed=True is deprecated. To get the desired output I had to do:
import numpy as np
data=np.random.randn(1000)
bins=np.arange(-3.0,3.0,51)
counts, _ = np.histogram(data,bins=bins)
if density: # equivalent of normed=True
counts_weighter=counts.sum()
else: # equivalent of normed=False
counts_weighter=1.0
plt.hist(bins[:-1],bins=bins,weights=counts/counts_weighter)
Trying to specify weights and density simultaneously as arguments to plt.hist() did not work for me. If anyone know of a way to get that working without having access to the normed keyword argument then please let me know in the comments and I will delete/modify this answer.
If you want bin centres then don't use bins[:-1] which are the bin edges - you need to choose a suitable scheme for how to calculate the centres (which may or may not be trivially derived).