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ipfn

=======================

Iterative proportional fitting is an algorithm used is many different fields such as economics or social sciences, to alter results in such a way that aggregates along one or several dimensions match known marginals (or aggregates along these same dimensions).

The algorithm exists in 2 versions:

* numpy version, which the fastest by far

* pandas version, which is much slower but easier to use than the numpy version

The algorithm recognizes the input variable type and and uses the appropriate version to solve the problem. To install the package:

* pip install ipfn

For more information and examples, please visit:

* `wikipedia page on ipf <https: en.wikipedia.org="" wiki="" iterative_proportional_fitting="">`_

* `slides explaining the methodology and links to specific examples <http: www.demog.berkeley.edu="" ~eddieh="" ipfdescription="" akdolwdipftwod.pdf="">`_

* https://github.com/Dirguis/ipfn

----

The project is similar to the ipfp package available for R and tests have been run to ensure same results.

----

Example with the numpy version of the algorithm:

------------------------------------------------

Please, follow the example below to run the package. Several additional examples in addition to the one listed below, are listed in the ipfn.py script. This example is taken from `<http: www.demog.berkeley.edu="" ~eddieh="" ipfdescription="" akdolwdipfthreed.pdf="">`_

First, let us define a matrix of N=3 dimensions, the matrix being of specific size 2*4*3 and populate that matrix with some values ::

from ipfn import *

import numpy as np

import pandas as pd

m = np.zeros((2,4,3))

m[0,0,0] = 1

m[0,0,1] = 2

m[0,0,2] = 1

m[0,1,0] = 3

m[0,1,1] = 5

m[0,1,2] = 5

m[0,2,0] = 6

m[0,2,1] = 2

m[0,2,2] = 2

m[0,3,0] = 1

m[0,3,1] = 7

m[0,3,2] = 2

m[1,0,0] = 5

m[1,0,1] = 4

m[1,0,2] = 2

m[1,1,0] = 5

m[1,1,1] = 5

m[1,1,2] = 5

m[1,2,0] = 3

m[1,2,1] = 8

m[1,2,2] = 7

m[1,3,0] = 2

m[1,3,1] = 7

m[1,3,2] = 6

Now, let us define some marginals. They all have to be less than N=3 dimensions and be consistent with the dimensions of contingency table m. For example, the marginal along the first dimension will be made of 2 elements. We want the sum of elements in m for dimensions 2 and 3 to equal the marginal::

m[0,:,:].sum() == marginal[0]

m[1,:,:].sum() == marginal[1]

The marginals are::

xipp = np.array([52, 48])

xpjp = np.array([20, 30, 35, 15])

xppk = np.array([35, 40, 25])

xijp = np.array([[9, 17, 19, 7], [11, 13, 16, 8]])

xpjk = np.array([[7, 9, 4], [8, 12, 10], [15, 12, 8], [5, 7, 3]])

I used the letter p to denote the dimension(s) being summed over

Define the aggregates list and the corresponding list of dimension to indicate the algorithm which dimension(s) to sum over for each aggregate::

aggregates = [xipp, xpjp, xppk, xijp, xpjk]

dimensions = [[0], [1], [2], [0, 1], [1, 2]]

Finally, run the algorithm::

IPF = ipfn(m, aggregates, dimensions)

m = IPF.iteration()

print xijp[0,0]

print m[0, 0, :].sum()

Example with the pandas version of the algorithm:

------------------------------------------------

In the same fashion, we can run a similar example, but using a dataframe::

from ipfn import *

import numpy as np

import pandas as pd

m = np.array([1., 2., 1., 3., 5., 5., 6., 2., 2., 1., 7., 2.,

5., 4., 2., 5., 5., 5., 3., 8., 7., 2., 7., 6.], )

dma_l = [501, 501, 501, 501, 501, 501, 501, 501, 501, 501, 501, 501,

502, 502, 502, 502, 502, 502, 502, 502, 502, 502, 502, 502]

size_l = [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4,

1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

age_l = ['20-25','30-35','40-45',

'20-25','30-35','40-45',

'20-25','30-35','40-45',

'20-25','30-35','40-45',

'20-25','30-35','40-45',

'20-25','30-35','40-45',

'20-25','30-35','40-45',

'20-25','30-35','40-45']

df = pd.DataFrame()

df['dma'] = dma_l

df['size'] = size_l

df['age'] = age_l

df['total'] = m

xipp = df.groupby('dma')['total'].sum()

xpjp = df.groupby('size')['total'].sum()

xppk = df.groupby('age')['total'].sum()

xijp = df.groupby(['dma', 'size'])['total'].sum()

xpjk = df.groupby(['size', 'age'])['total'].sum()

# xppk = df.groupby('age')['total'].sum()

xipp.loc[501] = 52

xipp.loc[502] = 48

xpjp.loc[1] = 20

xpjp.loc[2] = 30

xpjp.loc[3] = 35

xpjp.loc[4] = 15

xppk.loc['20-25'] = 35

xppk.loc['30-35'] = 40

xppk.loc['40-45'] = 25

xijp.loc[501] = [9, 17, 19, 7]

xijp.loc[502] = [11, 13, 16, 8]

xpjk.loc[1] = [7, 9, 4]

xpjk.loc[2] = [8, 12, 10]

xpjk.loc[3] = [15, 12, 8]

xpjk.loc[4] = [5, 7, 3]

aggregates = [xipp, xpjp, xppk, xijp, xpjk]

dimensions = [['dma'], ['size'], ['age'], ['dma', 'size'], ['size', 'age']]

IPF = ipfn(df, aggregates, dimensions)

df = IPF.iteration()

print df

print df.groupby('size')['total'].sum(), xpjp

Added notes:

------------

Several examples, using the numpy or pandas version of the algorithm are listed in the script `ipfn.py <https: github.com="" dirguis="" ipfn.git="">`_. Comment, uncomment to parts of interests and run the script::

python ipfn.py

To call the algorithm in a program, execute::

import ipfn

=======================

Iterative proportional fitting is an algorithm used is many different fields such as economics or social sciences, to alter results in such a way that aggregates along one or several dimensions match known marginals (or aggregates along these same dimensions).

The algorithm exists in 2 versions:

* numpy version, which the fastest by far

* pandas version, which is much slower but easier to use than the numpy version

The algorithm recognizes the input variable type and and uses the appropriate version to solve the problem. To install the package:

* pip install ipfn

For more information and examples, please visit:

* `wikipedia page on ipf <https: en.wikipedia.org="" wiki="" iterative_proportional_fitting="">`_

* `slides explaining the methodology and links to specific examples <http: www.demog.berkeley.edu="" ~eddieh="" ipfdescription="" akdolwdipftwod.pdf="">`_

* https://github.com/Dirguis/ipfn

----

The project is similar to the ipfp package available for R and tests have been run to ensure same results.

----

Example with the numpy version of the algorithm:

------------------------------------------------

Please, follow the example below to run the package. Several additional examples in addition to the one listed below, are listed in the ipfn.py script. This example is taken from `<http: www.demog.berkeley.edu="" ~eddieh="" ipfdescription="" akdolwdipfthreed.pdf="">`_

First, let us define a matrix of N=3 dimensions, the matrix being of specific size 2*4*3 and populate that matrix with some values ::

from ipfn import *

import numpy as np

import pandas as pd

m = np.zeros((2,4,3))

m[0,0,0] = 1

m[0,0,1] = 2

m[0,0,2] = 1

m[0,1,0] = 3

m[0,1,1] = 5

m[0,1,2] = 5

m[0,2,0] = 6

m[0,2,1] = 2

m[0,2,2] = 2

m[0,3,0] = 1

m[0,3,1] = 7

m[0,3,2] = 2

m[1,0,0] = 5

m[1,0,1] = 4

m[1,0,2] = 2

m[1,1,0] = 5

m[1,1,1] = 5

m[1,1,2] = 5

m[1,2,0] = 3

m[1,2,1] = 8

m[1,2,2] = 7

m[1,3,0] = 2

m[1,3,1] = 7

m[1,3,2] = 6

Now, let us define some marginals. They all have to be less than N=3 dimensions and be consistent with the dimensions of contingency table m. For example, the marginal along the first dimension will be made of 2 elements. We want the sum of elements in m for dimensions 2 and 3 to equal the marginal::

m[0,:,:].sum() == marginal[0]

m[1,:,:].sum() == marginal[1]

The marginals are::

xipp = np.array([52, 48])

xpjp = np.array([20, 30, 35, 15])

xppk = np.array([35, 40, 25])

xijp = np.array([[9, 17, 19, 7], [11, 13, 16, 8]])

xpjk = np.array([[7, 9, 4], [8, 12, 10], [15, 12, 8], [5, 7, 3]])

I used the letter p to denote the dimension(s) being summed over

Define the aggregates list and the corresponding list of dimension to indicate the algorithm which dimension(s) to sum over for each aggregate::

aggregates = [xipp, xpjp, xppk, xijp, xpjk]

dimensions = [[0], [1], [2], [0, 1], [1, 2]]

Finally, run the algorithm::

IPF = ipfn(m, aggregates, dimensions)

m = IPF.iteration()

print xijp[0,0]

print m[0, 0, :].sum()

Example with the pandas version of the algorithm:

------------------------------------------------

In the same fashion, we can run a similar example, but using a dataframe::

from ipfn import *

import numpy as np

import pandas as pd

m = np.array([1., 2., 1., 3., 5., 5., 6., 2., 2., 1., 7., 2.,

5., 4., 2., 5., 5., 5., 3., 8., 7., 2., 7., 6.], )

dma_l = [501, 501, 501, 501, 501, 501, 501, 501, 501, 501, 501, 501,

502, 502, 502, 502, 502, 502, 502, 502, 502, 502, 502, 502]

size_l = [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4,

1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

age_l = ['20-25','30-35','40-45',

'20-25','30-35','40-45',

'20-25','30-35','40-45',

'20-25','30-35','40-45',

'20-25','30-35','40-45',

'20-25','30-35','40-45',

'20-25','30-35','40-45',

'20-25','30-35','40-45']

df = pd.DataFrame()

df['dma'] = dma_l

df['size'] = size_l

df['age'] = age_l

df['total'] = m

xipp = df.groupby('dma')['total'].sum()

xpjp = df.groupby('size')['total'].sum()

xppk = df.groupby('age')['total'].sum()

xijp = df.groupby(['dma', 'size'])['total'].sum()

xpjk = df.groupby(['size', 'age'])['total'].sum()

# xppk = df.groupby('age')['total'].sum()

xipp.loc[501] = 52

xipp.loc[502] = 48

xpjp.loc[1] = 20

xpjp.loc[2] = 30

xpjp.loc[3] = 35

xpjp.loc[4] = 15

xppk.loc['20-25'] = 35

xppk.loc['30-35'] = 40

xppk.loc['40-45'] = 25

xijp.loc[501] = [9, 17, 19, 7]

xijp.loc[502] = [11, 13, 16, 8]

xpjk.loc[1] = [7, 9, 4]

xpjk.loc[2] = [8, 12, 10]

xpjk.loc[3] = [15, 12, 8]

xpjk.loc[4] = [5, 7, 3]

aggregates = [xipp, xpjp, xppk, xijp, xpjk]

dimensions = [['dma'], ['size'], ['age'], ['dma', 'size'], ['size', 'age']]

IPF = ipfn(df, aggregates, dimensions)

df = IPF.iteration()

print df

print df.groupby('size')['total'].sum(), xpjp

Added notes:

------------

Several examples, using the numpy or pandas version of the algorithm are listed in the script `ipfn.py <https: github.com="" dirguis="" ipfn.git="">`_. Comment, uncomment to parts of interests and run the script::

python ipfn.py

To call the algorithm in a program, execute::

import ipfn

TODO: Figure out how to actually get changelog content.

Changelog content for this version goes here.

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File Name & Checksum SHA256 Checksum Help | Version | File Type | Upload Date |
---|---|---|---|

ipfn-1.1.6-py2.py3-none-any.whl (10.3 kB) Copy SHA256 Checksum SHA256 | py2.py3 | Wheel | Jan 15, 2017 |