Finding the minimum/maximum of a sequence of sequences is possible:
list_of_tuples = [(0, 10), (1, 15), (2, 8)]
min(list_of_tuples)
# Output: (0, 10)
but if you want to sort by a specific element in each sequence use the key-argument:
min(list_of_tuples, key=lambda x: x[0]) # Sorting ...
You can't pass an empty sequence into max or min:
min([])
ValueError: min() arg is an empty sequence
However, with Python 3, you can pass in the keyword argument default with a value that will be returned if the sequence is empty, instead of raising an exception:
max([], default=42) ...
Getting the minimum or maximum or using sorted depends on iterations over the object. In the case of dict, the iteration is only over the keys:
adict = {'a': 3, 'b': 5, 'c': 1}
min(adict)
# Output: 'a'
max(adict)
# Output: 'c'
sorted(adict)
# Output: ['a', 'b', 'c']
To keep the dictionary ...
Getting the minimum of a sequence (iterable) is equivalent of accessing the first element of a sorted sequence:
min([2, 7, 5])
# Output: 2
sorted([2, 7, 5])[0]
# Output: 2
The maximum is a bit more complicated, because sorted keeps order and max returns the first encountered value. In case th...
min, max, and sorted all need the objects to be orderable. To be properly orderable, the class needs to define all of the 6 methods __lt__, __gt__, __ge__, __le__, __ne__ and __eq__:
class IntegerContainer(object):
def __init__(self, value):
self.value = value
def __rep...
To find some number (more than one) of largest or smallest values of an iterable, you can use the nlargest and nsmallest of the heapq module:
import heapq
# get 5 largest items from the range
heapq.nlargest(5, range(10))
# Output: [9, 8, 7, 6, 5]
heapq.nsmallest(5, range(10))
# Output: [...
