how to plot the empirical cdf of an array?

Community matplotlib-users
1

How can I plot the empirical CDF of an array of numbers in matplotlib
in Python? I'm looking for the cdf analog of pylab's "hist" function.

One thing I can think of is:

from scipy.stats import cumfreq
a = array([...]) # my array of numbers
num_bins = 20
b = cumfreq(a, num_bins)
plt.plot(b)

Is that correct though? Is there an easier/better way?

thanks.

2

Hi,

I would use pyplot.hist to produce a histogram with a bar for each bin. By use
of the keyword argument 'cumulative' you can select cumulative frequency
distribution instead of density.

Kind regards,
Matthias

···

On Friday July 9 2010 06:02:58 per freem wrote:

How can I plot the empirical CDF of an array of numbers in matplotlib
in Python? I'm looking for the cdf analog of pylab's "hist" function.

One thing I can think of is:

from scipy.stats import cumfreq
a = array([...]) # my array of numbers
num_bins = 20
b = cumfreq(a, num_bins)
plt.plot(b)

Is that correct though? Is there an easier/better way?

3

I recalled David Huard posted the below,
which apparently was once in the sandbox...
hth,
Alan Isaac

def empiricalcdf(data, method='Hazen'):
     """Return the empirical cdf.

     Methods available (here i goes from 1 to N)
         Hazen: (i-0.5)/N
         Weibull: i/(N+1)
         Chegodayev: (i-.3)/(N+.4)
         Cunnane: (i-.4)/(N+.2)
         Gringorten: (i-.44)/(N+.12)
         California: (i-1)/N

     :see:
http://svn.scipy.org/svn/scipy/trunk/scipy/sandbox/dhuard/stats.py
     :author: David Huard
     """
     i = np.argsort(np.argsort(data)) + 1.
     nobs = len(data)
     method = method.lower()
     if method == 'hazen':
         cdf = (i-0.5)/nobs
     elif method == 'weibull':
         cdf = i/(nobs+1.)
     elif method == 'california':
         cdf = (i-1.)/nobs
     elif method == 'chegodayev':
         cdf = (i-.3)/(nobs+.4)
     elif method == 'cunnane':
         cdf = (i-.4)/(nobs+.2)
     elif method == 'gringorten':
         cdf = (i-.44)/(nobs+.12)
     else:
         raise 'Unknown method. Choose among Weibull, Hazen, Chegodayev,
Cunnane, Gringorten and California.'
     return cdf

···

On 7/9/2010 12:02 AM, per freem wrote:

How can I plot the empirical CDF of an array of numbers in matplotlib
in Python?

4

I'd like to clarify: I want the empirical cdf, but I want it to be
normalized. There's a normed=True option to plt.hist but how can I do
the equivalent for CDFs?

···

On Fri, Jul 9, 2010 at 9:14 AM, Alan G Isaac <alan.isaac@...287...> wrote:

On 7/9/2010 12:02 AM, per freem wrote:

How can I plot the empirical CDF of an array of numbers in matplotlib
in Python?

I recalled David Huard posted the below,
which apparently was once in the sandbox...
hth,
Alan Isaac

def empiricalcdf(data, method='Hazen'):
"""Return the empirical cdf.

Methods available \(here i goes from 1 to N\)
    Hazen:       \(i\-0\.5\)/N
    Weibull:     i/\(N\+1\)
    Chegodayev:  \(i\-\.3\)/\(N\+\.4\)
    Cunnane:     \(i\-\.4\)/\(N\+\.2\)
    Gringorten:  \(i\-\.44\)/\(N\+\.12\)
    California:  \(i\-1\)/N

:see:

http://svn.scipy.org/svn/scipy/trunk/scipy/sandbox/dhuard/stats.py
:author: David Huard
"""
i = np.argsort(np.argsort(data)) + 1.
nobs = len(data)
method = method.lower()
if method == 'hazen':
cdf = (i-0.5)/nobs
elif method == 'weibull':
cdf = i/(nobs+1.)
elif method == 'california':
cdf = (i-1.)/nobs
elif method == 'chegodayev':
cdf = (i-.3)/(nobs+.4)
elif method == 'cunnane':
cdf = (i-.4)/(nobs+.2)
elif method == 'gringorten':
cdf = (i-.44)/(nobs+.12)
else:
raise 'Unknown method. Choose among Weibull, Hazen, Chegodayev,
Cunnane, Gringorten and California.'
return cdf

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5

There is no such thing as a normalized empirical CDF. Or rather, there is no such thing as an unnormalized empirical CDF.

Alan's code is good. Unless if you have a truly staggering number of points, there is no reason to bin the data first.

···

On 7/9/10 10:02 AM, per freem wrote:

I'd like to clarify: I want the empirical cdf, but I want it to be
normalized. There's a normed=True option to plt.hist but how can I do
the equivalent for CDFs?

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
  that is made terrible by our own mad attempt to interpret it as though it had
  an underlying truth."
   -- Umberto Eco

6

How does Alan's code compare with using cumfreq and then plotting its
result? Is the only difference that cumfreq bins the data?

···

On Fri, Jul 9, 2010 at 10:12 AM, Robert Kern <robert.kern@...287...> wrote:

On 7/9/10 10:02 AM, per freem wrote:

I'd like to clarify: I want the empirical cdf, but I want it to be
normalized. There's a normed=True option to plt.hist but how can I do
the equivalent for CDFs?

There is no such thing as a normalized empirical CDF. Or rather, there is no
such thing as an unnormalized empirical CDF.

Alan's code is good. Unless if you have a truly staggering number of points,
there is no reason to bin the data first.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

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7

Also, I am not sure how to use alan's code.

If I try:

ec = empirical_cdf(my_data)
plt.plot(ec)

it doesn't actually look like a cdf

···

On Fri, Jul 9, 2010 at 10:17 AM, per freem <perfreem@...287...> wrote:

How does Alan's code compare with using cumfreq and then plotting its
result? Is the only difference that cumfreq bins the data?

On Fri, Jul 9, 2010 at 10:12 AM, Robert Kern <robert.kern@...287...> wrote:

On 7/9/10 10:02 AM, per freem wrote:

I'd like to clarify: I want the empirical cdf, but I want it to be
normalized. There's a normed=True option to plt.hist but how can I do
the equivalent for CDFs?

There is no such thing as a normalized empirical CDF. Or rather, there is no
such thing as an unnormalized empirical CDF.

Alan's code is good. Unless if you have a truly staggering number of points,
there is no reason to bin the data first.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

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Visit sprint.com/first -- http://p.sf.net/sfu/sprint-com-first
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8

Make sure my_data is sorted first.

plt.plot(my_data, ec)

You probably want to use one of the "steps" linestyles; I'm not sure which one would be best. It probably doesn't matter much.

···

On 7/9/10 10:31 AM, per freem wrote:

Also, I am not sure how to use alan's code.

If I try:

ec = empirical_cdf(my_data)
plt.plot(ec)

it doesn't actually look like a cdf

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
  that is made terrible by our own mad attempt to interpret it as though it had
  an underlying truth."
   -- Umberto Eco