Python 2.7 creating a matrix from lists

Question

I have two lists:

a_list contains distinct tupled ids:

a_list = [('1'), ('2'), ('3'), ('4')...etc]

and lists within lists of tuples called check_list:

check_list = [[[('1'), ('2')]], [[('2'), ('3')], [('3'),('4')]], [[('1'),('3')]]...etc]

I've got a problem that is a little too complex for my python skill level... I'm attempting to form a matrix to be outputted to a csv file, with the following structure:

     1 2 3 4

1 1 1 0 0

2 0 1 2 1

3 1 0 1 0

Where each value is the count of the number from a_list in check_list.

I've looked at numpy and I have no knowledge. I've played around with this but I cant seem to get a full understanding to transfer it to my problem. I'm pretty limited with array knowledge, too. Thanks in advance.

Do you actually have tuples in your sublists because as posted the parens are doing nothing? — Padraic Cunningham, Feb 16 '16 at 12:24
Do you want to tell us anything with the parenthesis around the strings? Are these actually supposed to be tuples? What's the problem, counting or writing to a file? — timgeb, Feb 16 '16 at 12:25
yes, the parenthesis are supposed to be tuples and are supposed to be there. The problem is the counting. — user47467, Feb 16 '16 at 12:29
Then edit your question. As it stands, `a_list` is just a list of strings because `('1') == '1'`. — timgeb, Feb 16 '16 at 12:30

Padraic Cunningham · Accepted Answer · 2016-02-16T12:53:55.203

You can get the counts using a Counter dict, sort by key to get the counts in Order using an OrderedDict to maintain that order and use dict.viewkeys to find what keys are missing.

from __future__ import print_function
from collections import Counter, OrderedDict
from itertools import chain

check_list = [[[('1',), ('2',)]], [[('2',), ('3',)], [('3',), ('4',)]], [[('1',), ('3',)]]]

a_list = [('1',), ('2',), ('3',), ('4',)]

cn = OrderedDict(sorted(Counter(a_list).items()))

print(" ".join([str(t[0]) for t in cn]))
for chk in check_list:
    _cn = Counter(chain(*chk))
    # cn.keys() python 3
    diff = cn.viewkeys() - _cn
    for k  in cn:
        if k not in diff:
            print(_cn[k], end=" ")
        else:
            print(0, end=" ")
    print()

Output:

If you don't care about the order you can remove the sorted/Ordereddict logic:

from collections import Counter
from itertools import chain

check_list = [[[('1',), ('2',)]], [[('2',), ('3',)], [('3',), ('4',)]], [[('1',), ('3',)]]]

a_list = [('1',), ('2',), ('3',), ('4',)]

cn = Counter(a_list)

print(" ".join([str(t[0]) for t in cn]))
for chk in check_list:
    _cn = Counter(chain(*chk))
    diff = cn.viewkeys() - _cn
    for k  in cn:
        if k not in diff:
            print(_cn[k], end=" ")
        else:
            print(0, end=" ")
    print()

That will give you an arbitrary order:

To write to a csv:

from collections import Counter, OrderedDict
from itertools import chain
from csv import writer

check_list = [[[('1',), ('2',)]], [[('2',), ('3',)], [('3',), ('4',)]], [[('1',), ('3',)]]]

a_list = [('1',), ('2',), ('3',), ('4',)]

with open("out.csv","w") as out:
    wr = writer(out,delimiter=" ")
    cn = OrderedDict(sorted(Counter(a_list).items()))
    wr.writerow(list(chain(*cn)))
    for chk in check_list:
        _cn = Counter(chain(*chk))
        diff = cn.viewkeys() - _cn
        wr.writerow([_cn[k] if k not in diff else 0 for k in cn])

out.csv:

Perfect! And I understand it too! – user47467 Feb 16 '16 at 13:40 — user47467, Feb 16 '16 at 13:40

Python 2.7 creating a matrix from lists

1 Answers1