Finding maximum value in a dictionary containing mixed items in Python

Question

I have a dictionary with either a integer or a tuple of integers as value. How do I find the maximum integer present in dicts' values?

Example:

x1 = {0:2, 2:1, 3:(1, 2), 20:3}

should return 3 and

x2 = {0:2, 2:1, 3:(1, 5), 20:3}

should return 5

Answer 1

You could try this aproach:

• create a set for storing integers
• loop through the values of the dictionary
  • add integer values to set
  • add each integer value of tuple values to set
• find max of set

Something like this:

def maxofdict(x):
   s = set()
   for v in x.values():
      if hasattr(v, '__len__'):
         s.update(v)
      else:
         s.add(v)
   return max(s)

Answer 2

You need a generic flatten() function. The Python standard library oddly enough doesn't provide one -- not even in itertools -- but googling around should get you an implementation. If you don't mind being potentially backwards incompatible, you can import a "private" implementation from tkinter:

from _tkinter import _flatten as flatten

def mixed_max(d):
    return max(flatten(d.items()))

mixed_max({0: 2, 2: 1, 3: (1,2), 4: 0}) # => 4
mixed_max({0: 2, 2: 1, 3: (1,5), 4: 0}) # => 5

Answer 3

Assuming the correct result for x1 = 4;

def maxOfMixedDict(x):
    max = 0
    for key, value in x.items():
        if(key > max):
            max = key
        try:
            for v2 in value:
                if(v2 > max):
                    max = v2
        except TypeError, e:
            pass

    return max

Answer 4

A one-liner:

max(max(v) if isinstance(v, collections.Iterable) else v for v in d.itervalues())

Needs at least Python 2.6 due to collections.Iterable ABC.

Answer 5

max(max(k,max(v) if isinstance(v,collections.Iterable) else v) for k,v in x1.items())

The other one-liner does not take account of the keys.

This is icky because it is not the designed use of a dictionary: the keys are meant to be keys, not themselves stores of data. I think you should reconsider your data structure.

EDIT: The above was nonsense. Thanks to @SilentGhost for pointing it out.

Answer 6

This is my version of one liner not needing 2.6:

x1 = {0:2, 2:1, 3:(1, 2), 20:3}
x2 = {0:2, 2:1, 3:(1, 5), 20:3}
print max(max(values) if hasattr(values,'__iter__') else values for values in x1.values()) 
print max(max(values) if hasattr(values,'__iter__') else values for values in x2.values()) 

Output:

3
5

HOWEVER I strongly suggest to go to origin of these values and change the storing of integers to singleton tuples.Then you can use cleaner code:

x1 = {0:(2,), 2:(1,), 3:(1, 2), 20:(3,)}
x2 = {0:(2,), 2:(1,), 3:(1, 5), 20:(3,)}
for x in (x1,x2):
    print max(max(values) for values in x.values())