Find minimum element in a dictionary of dictionaries
I need to find what element of apple
has the minimum size
.
Tnx for all answers. But there is one problem: I use Python 2.4.2 (I can't change it) and function min
haven't key
arg. 开发者_如何学PythonYes, I need key of apple
apple = {1:{'size':12,'color':'red'},2:{'size':10,'color':'green'}}
import operator
min(apple.values(), key=operator.itemgetter('size'))
will return you
{'color': 'green', 'size': 10}
UPDATE: to get the index:
min(apple, key=lambda k: apple[k]['size'])
Python has a very nice parameter for the min
function that allows using an arbitrary function to be minified instead of just using comparison on the elements:
result = min(apple.values(), key=lambda x:x['size'])
The key
parameter replaced in most cases the older idiom of decorate-process-undecorate that could have been applied here:
result = min((x['size'], x) for x in apple.values())[1]
If instead you want to know the number (key) of the apple (it's not clear in the question) then:
result = min(apple.keys(), key=lambda x:apples[x]['size'])
or (old style)
result = min((apples[x]['size'], x) for x in apple.keys())[1]
Use min
with a custom key
function that returns the size of each item.
apple = {1:{'size':12,'color':'red'},2:{'size':10,'color':'green'}}
print min(apple.keys(), key=lambda k, a=apple: a[k]['size'])
Which prints:
2
P.S. Since apple
is a collection I would make it plural -- apples
.
Don't know if it's the fastest way to do it, but anyway:
>>> apple = [ {'size':12, 'color': 'red' }, { 'size':10, 'color':'green'} ]
>>> a = dict(map(lambda apple: (apple['size'], apple), apple))
>>> a
{10: {'color': 'green', 'size': 10}, 12: {'color': 'red', 'size': 12}}
>>> min = a[min(a.keys())]
>>> min
{'color': 'green', 'size': 10}
def get_min(apple):
L = apple.values()
m = L[0]
for item in L:
if item['size'] < m['size']:
m = item
return m
P.S. Not very pythonic but linear time
min(map(lambda a:[apple[a]['size'],a], apple))[1]
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