CODEWITHSUNDEEP
pythondictionarytuples
halny
halny

Reputation: 63

Find key, value pair in dict according to second value of tuples using itemgetter() method

I have the following dictonary:

inventory = {'apple':(3, 133), 'banana':(9, 407), 'pear':(17, 205), 'orange':(1, 91)}

I would like to find key, value pair according to max value of the second element in the tuple so in this case it should return: banana':(9, 407) I know I can get this result using lambda:

key_max = max(inventory.keys(), key=(lambda k: inventory[k][1]))    
print key_max, inventory[key_max]

However I try to receive same result using itemgetter() method as I read it is faster but I can only get data according to the value of first element of the tuple:

from operator import itemgetter 
print max(inventory.iteritems(), key = itemgetter(1))

Is there a way I could get same result using itemgetter() method?

Upvotes: 1

Views: 283

Answers (2)

DirtyBit
DirtyBit

Reputation: 16782

Before I or someone else comes up with something fascinating:

inventory = {'apple':(3, 133), 'banana':(9, 407), 'pear':(17, 205), 'orange':(1, 91)}

maxVal = 0                   # to get the max of 2nd elem from tuple
for k,v in inventory.items():
    maxVal = max(maxVal, v[1])   

for k,v in inventory.items():
    if v[1] == maxVal:     # if the 2nd elem of tuple eqs to the maxVal
        print(k, v)

OUTPUT:

banana (9, 407)

Note: I recommend the Edited version.

EDIT

Using lambda:

print(max(inventory.items(), key=lambda x: x[1][1]))  # ('banana', (9, 407))

Upvotes: 2

meowgoesthedog
meowgoesthedog

Reputation: 15035

Although I don't see the point of this exercise...

Here's a solution using only itemgetter and no lambdas, as was the original question:

key = max(zip(inventory.keys(), map(itemgetter(1), inventory.values())), key=itemgetter(1))[0]

print(key, inventory[key])
# banana (9, 407)

Note that the above will only be efficient in Python 3.x where the built-in functions return iterators / view objects instead of lists.

An equally efficient Python 2.x equivalent requires the use of library functions such as itertools.izip:

# from itertools import izip, imap

key = max(izip(inventory.iterkeys(), imap(itemgetter(1), inventory.itervalues())), key=itemgetter(1))[0]

print key, inventory[key]

Upvotes: 2

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