current community

your communities

more stack exchange communities

Stack Exchange
tour help
careers 2.0
Take the 2-minute tour ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free, no registration required.
up vote 2 down vote favorite

I have a list of tuples as follows:

[(x,{'y':'1,3','z':'2'}),
(y,{'a':'4'}),
(z,{'b':'2,3'})]

I need to convert this to a numpy array format as follows:

    x   y   z   a   b
x   0   1,3 2   0   0
y   1,3 0   0   4   0
z   2   0   0   0   2,3
a   0   4   0   0   0
b   0   0   2,3 0   0

To support this,store the node-name as a list to give them mapping indices.

[x,y,z,a,b]

Given the indices - what is the most efficient to way to create the numpy array from this structure?
Also - as new entries come into the original list of tuples,it will add into the index list and the numpy array as appropriate.

Edit of an existing element will not happen.

Help is appreciated.

share|improve this question
2  
is your comma a "decimal" comma as in the German and French locales? –  Oz123 Jul 1 '13 at 10:54
    
No - its just a simple english separator.I may even may it a list but the problem statement would remain the same.The intersection value would become the matrix element/entry. –  IUnknown Jul 1 '13 at 10:57
add comment

2 Answers

up vote 1 down vote accepted

If you use object dtypes you can build your array in the approach below. Since you need a 2D symmetry, it is easier to create a 2D array first and from this build the structured array:

import numpy as np
o = ['x','y','z','a','b']
a = np.zeros((len(o),len(o)),dtype=object)
s  =[('x',{'y':'1,3','z':'2'}), ('y',{'a':'4'}), ('z',{'b':'2,3'})]
for vi in s:
    i = o.index(vi[0])
    for vj in vi[1].items():
        j = o.index(vj[0])
        a[i,j] = vj[1]
        a[j,i] = a[i,j]

# building the structured array
b = np.zeros((len(o),), dtype=[(i,object) for i in o])
for i,vi in enumerate(o):
    b[vi] = a[i,:]

# building a dictionary to access the values
d = dict(( (vi, dict(( (vj, a[i,j]) for j,vj in enumerate(o) ))) for i,vi in enumerate(o) ))
share|improve this answer
    
Thanks - needed help in understanding a syntax b = np.zeros((len(o),), dtype=[(i,object) for i in o]) Wouldn't this be same as - b = np.zeros((len(o),len(o)), dtype=object) –  IUnknown Jul 2 '13 at 4:47
    
The first is creating the structured array to receive the values and the second is creating a 2D array to receive the values.... I thing the structured array is what you mean by using mapping indices... From a you access the values like a[i,j] and from b like b['x'][i] –  Saullo Castro Jul 2 '13 at 5:50
1  
Is there a way to create the structured array such that it is accessible as b['x']['y'],and thus bypass indices completely? –  IUnknown Jul 2 '13 at 10:43
    
then you should build a dictionary: I've updated the answer with this third option... –  Saullo Castro Jul 2 '13 at 10:56
add comment
up vote 0 down vote

A more numpythonic version... The values are stored as strings. That can be changed, but you will probably need to better define the syntax of your input list of dicts:

import numpy as np
import operator as op

data = [('x', {'y' : '1,3', 'z' : '2'}),
        ('y', {'a' : '4'}),
        ('z', {'b' : '2,3'})]

keys = np.array(['x', 'y', 'z', 'a', 'b'])
keys_sort = np.argsort(keys)

rows = [(item[0], item[1].keys(), item[1].values()) for item in data]


rows = np.array(reduce(op.add, ([item[0]]*len(item[1]) for item in data)))
cols = np.array(reduce(op.add, (item[1].keys() for item in data)))
vals = np.array(reduce(op.add, (item[1].values() for item in data)))

row_idx = keys_sort[np.searchsorted(keys, rows, sorter=keys_sort)]
col_idx = keys_sort[np.searchsorted(keys, cols, sorter=keys_sort)]

out_arr = np.empty((len(keys), len(keys)), dtype=vals.dtype)
out_arr[:] = '0'
out_arr[row_idx, col_idx] = vals
out_arr[col_idx, row_idx] = vals

>>> out_arr
array([['0', '1,3', '2', '0', '0'],
       ['1,3', '0', '0', '4', '0'],
       ['2', '0', '0', '0', '2,3'],
       ['0', '4', '0', '0', '0'],
       ['0', '0', '2,3', '0', '0']], 
      dtype='|S3')
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.