saving an ordered pair from a list of numbers using python

Question

I have an array of numbers:

q1a = [1,2,2,2,4,3,1,3,3,4,0,0]

I want to save these in an array where it will be stored in as (number, proportion of the number) using PYTHON.

Such as : [[0 0.1667], [1 0.1667], [2 0.25], [3 0.25], [4 0.167]].

This is essential to calculate the distribution of the numbers. How can I do this?

Although I wrote the code to save the numbers as : (number, number of times it occurred in the list) but I cant figure it out how I can find the proportion of each number. Thanks.

sorted_sample_values_of_x = unique, counts = np.unique(q1a, return_counts=True)
np.asarray((unique, counts)).T
np.put(q1a, [0], [0])

sorted_x = np.matrix(sorted_sample_values_of_x)
sorted_x = np.transpose(sorted_x)
print('\n' 'Values of x (sorted):' '\n')
print(sorted_x)

possible duplicate of [item frequency count in python](http://stackoverflow.com/questions/893417/item-frequency-count-in-python) — maxymoo, Jul 21 '15 at 04:03

score 1 · Answer 1 · answered Jul 21 '15 at 04:03

>>> q1a = [1,2,2,2,4,3,1,3,3,4,0,0]
>>> from collections import Counter
>>> sorted([[x, float(y)/len(q1a)] for (x, y) in Counter(q1a).items()],
...        key=lambda x: x[0])
[[0, 0.16666666666666666],
 [1, 0.16666666666666666],
 [2, 0.25],
 [3, 0.25],
 [4, 0.16666666666666666]]

Anand S Kumar · Accepted Answer · 2015-07-21T05:14:04.107

1

You will need to do two things.

Convert sorted_x array as a float array.
And then divide it by sum of counts array.

Example -

In [34]: sorted_x = np.matrix(sorted_sample_values_of_x)

In [35]: sorted_x = np.transpose(sorted_x).astype(float)

In [36]: sorted_x
Out[36]:
matrix([[ 0.,  2.],
        [ 1.,  2.],
        [ 2.,  3.],
        [ 3.,  3.],
        [ 4.,  2.]])

In [37]: sorted_x[:,1] = sorted_x[:,1]/counts.sum()

In [38]: sorted_x
Out[38]:
matrix([[ 0.        ,  0.16666667],
        [ 1.        ,  0.16666667],
        [ 2.        ,  0.25      ],
        [ 3.        ,  0.25      ],
        [ 4.        ,  0.16666667]])

To store the numbers with the propertions in a new array, do -

In [41]: sorted_x = np.matrix(sorted_sample_values_of_x)

In [42]: sorted_x = np.transpose(sorted_x).astype(float)

In [43]: ns = sorted_x/np.array([1,counts.sum()])

In [44]: ns
Out[44]:
matrix([[ 0.        ,  0.16666667],
        [ 1.        ,  0.16666667],
        [ 2.        ,  0.25      ],
        [ 3.        ,  0.25      ],
        [ 4.        ,  0.16666667]])

edited Jul 21 '15 at 05:14

answered Jul 21 '15 at 04:06

Anand S Kumar

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thanks for your answer. can u plz tell me, How can I save the proportions into a different array? – jhon_wick Jul 21 '15 at 05:10
Give a different name on the left side - `newsorted_x = sorted_x[:,1]/counts.sum()` , this would give just the propertions though, is that what you want? Or do you want both the numbers as well as propertions? – Anand S Kumar Jul 21 '15 at 05:11
Added the code for storing both numbers and proportions in a new array in the answer. – Anand S Kumar Jul 21 '15 at 05:14
actually, i am still working on my program. and I figured out that I need the proportions only. thanks for your immediate reply. – jhon_wick Jul 21 '15 at 05:18
Glad I could be of help. – Anand S Kumar Jul 21 '15 at 05:18
Hi dont_give_up , I believe for new questions you should ask new question, that would help keep SO clean , we should not clutter multiple questions to same question. – Anand S Kumar Jul 21 '15 at 05:44
Also you can checkout `np.where(ns<0.02)` , this is what you need I believe. – Anand S Kumar Jul 21 '15 at 05:45

score 0 · Answer 3 · answered Jul 21 '15 at 04:07

In [12]: from collections import Counter

In [13]: a = [1,2,2,2,4,3,1,3,3,4,0,0]

In [14]: counter = Counter(a)

In [15]: sorted( [ [key, float(counter[key])/len(a)]  for key in counter ] )
Out[15]:
[[0, 0.16666666666666666],
 [1, 0.16666666666666666],
 [2, 0.25],
 [3, 0.25],
 [4, 0.16666666666666666]]

score 0 · Answer 4 · answered Jul 21 '15 at 04:19

#!/usr/bin/env python
import numpy as np
q1a = [1,2,2,2,4,3,1,3,3,4,0,0]

unique, counts = np.unique(q1a, return_counts=True)
counts = counts.astype(float) # convert to float
counts /= counts.sum()        # counts -> proportion
print(np.c_[unique, counts])

Output

[[ 0.          0.16666667]
 [ 1.          0.16666667]
 [ 2.          0.25      ]
 [ 3.          0.25      ]
 [ 4.          0.16666667]]

score 0 · Answer 5 · answered Jul 21 '15 at 04:20

As an alternative to collections.Counter, try collections.defaultdict. This allows you to accumulate the total frequency as you proceed through the input (i.e should be more efficient) and it's more readable (IMO).

from collections import defaultdict

q1a = [1,2,2,2,4,3,1,3,3,4,0,0]
n = float(len(q1a))
frequencies = defaultdict(int)
for i in q1a:
    frequencies[i] += 1/n

print frequencies.items()
[(0, 0.16666666666666666), (1, 0.16666666666666666), (2, 0.25), (3, 0.25), (4, 0.16666666666666666)]

dermen · Answer 6 · 2015-07-21T05:03:09.990

0

An fun alternative using numpy

print [(val, 1.*np.sum(q1a==val)/len(q1a) ) for val in np.unique(q1a) ]
#[(0, 0.16666666666666666),
#(1, 0.16666666666666666),
#(2, 0.25),
#(3, 0.25),
#(4, 0.16666666666666666)]

The 1. is to force float division

edited Jul 21 '15 at 05:03

answered Jul 21 '15 at 04:41

dermen

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saving an ordered pair from a list of numbers using python

6 Answers6

Output