Difference between multiple elements in list with same string . Python 2.7

Question

This is a little confusing so I will try my best to explain my goal. In a nutshell i'm trying to look at a sublist within a list. In those sublists, some have the same starting element (sublist[0]) and i want to record the differences between that sublist with other sublists starting with the same element

data = [['o1415', '1', '0', '1'], ['o1415', '0', '0', '0'], ['o1414', '0', '0', '0'], ['o1414', '1', '0', '0'], ['o1414', '0', '0', '0'], ['o1408', '0', '0', '1'], ['o1406', '0', '0', '0']]
D_changes = {}

here is a list with 4 elements . . the first of which has a name, 2nd/3rd/4th elements have digits .

i'm trying to generate a dictionary that has the {name:[then,the,differences])}

for example data[0] and data[1] both have 'o1415' as their first element . since they have the same string for the first element i want to compare the rest of the lists with each other . so data[0] differs in data[0][1] and data[0][2] from data[1] . . . so i want to add 'o1415':['first','third'] to the empty dictionary D_changes.

another example would be 'o1414' which is in data[2],data[3],data[4] and for these lists, one element is different in the [1] position so i'd like to add 'o1414' : ['first'] to the empty dictionary above

in the end i want to obtain a dictionary with this type of content

desired_changes = {'o1415':['first','third'],'o1414':['first'],'o1408':[],'o1406':[]}

Questions asking for code must demonstrate a minimal understanding of the problem being solved. Include attempted solutions, why they didn't work, and the expected results. See also: Stack Overflow question checklist — Marcin, Oct 08 '13 at 21:30

roippi · Accepted Answer · 2013-10-08T21:51:41.047

1

I'll give you a direction more than a full answer.

First, load up a dict to group like items for further processing; I'll use a defaultdict:

d = defaultdict(list)

data = [['o1415', '1', '0', '1'], ['o1415', '0', '0', '0'], ['o1414', '0', '0', '0'], ['o1414', '1', '0', '0'], ['o1414', '0', '0', '0'], ['o1408', '0', '0', '1'], ['o1406', '0', '0', '0']]

for sub in data:
    d[sub[0]].append([int(x) for x in sub[1:]])

Then, for a given key, simply look at the zip of its values. i.e. for 'o1414':

d['o1414']
Out[58]: [[0, 0, 0], [1, 0, 0], [0, 0, 0]]

list(zip(*d['o1414']))
Out[59]: [(0, 1, 0), (0, 0, 0), (0, 0, 0)]

We know if they're all equal if it's all 1, or all 0; otherwise it's different. So just do:

[any(x) and not all(x) for x in zip(*d['o1414'])]
Out[60]: [True, False, False]

I particularly like the aesthetics of that - any(x) and not all(x). Python can be beautiful sometimes.

Anyway, True means that you have a differing value in that slot. I'll leave it up to you do do that for all your keys and to get it into the format that you want.

edited Oct 08 '13 at 21:51

answered Oct 08 '13 at 21:46

roippi

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[[1, 5, 0], [1, 6, 0]] [(1, 1), (5, 6), (0, 0)] [False, False, False] while using the same lines ? – O.rka Oct 15 '13 at 19:10
1

@draconisthe0ry `any` and `all` check for the truthiness of the collection - i.e. the value of the elements in a boolean context. Your original question was about a bitmap - where the elements could only be 1 or 0. That's what I answered. – roippi Oct 15 '13 at 19:42
oh ok i see the difference. you are right . thank you . what should i research to check if tuples have all the same elements within them ? – O.rka Oct 15 '13 at 19:46
@draconisthe0ry I answered that in your other thread. Any reason you've unaccepted this an an answer? – roippi Oct 15 '13 at 20:59
sorry about that! i just noticed that i didn't accept it – O.rka Oct 30 '13 at 18:18

score 0 · Answer 2 · answered Oct 08 '13 at 21:39

i figured it out. not sure that deserved a -1 vote . this probably isn't the most efficient way but it works

data = [['o1415', '1', '0', '1'], ['o1415', '0', '0', '0'], ['o1414', '0', '0', '0'], ['o1414', '1', '0', '0'], ['o1414', '0', '0', '0'], ['o1408', '0', '0', '1'], ['o1406', '0', '0', '0']]

D = {}   
for name in data:    
    while name:
        for k in data:
            temp = []
            if name[0] == k[0]:
                if name[1] != k[1]:
                    temp.append('first')
                if name[2] != k[2]:
                    temp.append('second')
                if name[3] != k[3]:
                    temp.append('third')
            for k in temp:
                if len(k) != 0:
                    D[name[0]] = temp
                    break
                else:
                    pass

        break

Difference between multiple elements in list with same string . Python 2.7

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