Difference between multiple elements in list with

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':[]}

Answer 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.

Answer 2

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