Python: nested list and dictionary comparison

Question

I'm having trouble comparing a nested list with a dictionary of multiple values. The dictionary and nested list are like so:

list = [[['a']], [['b']], [['c'], ['d']], [['e'], ['f'], ['g']]]

dict = {'adv0' : ('a', 'b'), 'adv1' : ('f', 'c'), 'adv2' : ('d', 'e', 'q')}

I want to create an array where each sublist ('a', 'b', 'c & d', 'e & f & g') is compared to each value and if any item in the sublist is a member of that value it creates a 0 entry, otherwise it creates a 1 entry.

[0, 1, 1] since 'a' is only in adv0, [0, 1, 1] since 'b' is only in adv0, [1, 0, 0] since adv1 and adv2 contain 'c' or 'd', [1, 0, 0] since adv1 and adv2 contain one of 'e', 'f' or 'g'. Hence we get the array [0,1,1,0,1,1,1,0,0,1,0,0].

The following code is my horrible attempt at a solution which doesn't work:

l = []
for sublist in list:
      for items in sublist:
        for x in items:
          for key in dict:
            if x in dict[key]:
              l.extend('0')
            elif x not in dict[key]:
              l.extend('1')
print l

Note that dictionaries are not ordered. It sounds like you want the resulting list ordered by a natural sort on the dict keys -- but it would be good to mention that explicitly. — Jesse W at Z - Given up on SE
– Jesse W at Z - Given up on SE, Commented Oct 6, 2014 at 21:03

Mark Tolonen · Accepted Answer · 2014-10-06 21:15:13Z

1

One problem is dictionaries are unordered, so you may not iterate through the keys as you expect. use an OrderedDict or iterate specifically through the keys you want. This works:

L = [[['a']], [['b']], [['c'], ['d']], [['e'], ['f'], ['g']]]
D = {'adv0' : ('a', 'b'), 'adv1' : ('f', 'c'), 'adv2' : ('d', 'e', 'q')}

l = []
for sublist in L:
    for key in ('adv0','adv1','adv2'):
        if any(item[0] in D[key] for item in sublist):
            l.append(0) # changed to give explicit output you listed.
        else:
            l.append(1) # ditto
print l

Output:

[0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0]

answered Oct 6, 2014 at 21:15

Mark Tolonen

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Jesse W at Z - Given up on SE · Accepted Answer · 2014-10-06 21:18:14Z

1

Here you go: (I renamed the input to lst and dct to avoid conflicting with the built-in list and dict objects)

[0 if any(v[0] in dct[k] for v in sublst) else 1 
   for sublst in lst 
     for k in sorted(dct.keys())]

That was fun. (But, sadly, 3 minutes too late...)

answered Oct 6, 2014 at 21:18

Jesse W at Z - Given up on SE

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1 Comment

Jesse W at Z - Given up on SE Over a year ago

I was quite surprised how similar out solutions were.

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