How does Python sort a list of tuples?
Question:
Empirically, it seems that Python’s default list sorter, when passed a list of tuples, will sort by the first element in each tuple. Is that correct? If not, what’s the right way to sort a list of tuples by their first elements?
Answers:
Yes, this is the default. In fact, this is the basis of the classic “DSU” (Decorate-Sort-Undecorate) idiom in Python. See Code Like a Pythonista.
It automatically sorts a list of tuples by the first elements in the tuples, then by the second elements and so on tuple([1,2,3]) will go before tuple([1,2,4]). If you want to override this behaviour pass a callable as the second argument to the sort method. This callable should return 1, -1, 0.
No, tuples are sequence types just like strings. They are sorted the same, by comparing each element in turn:
>>> import random
>>> sorted([(0,0,0,int(random.getrandbits(4))) for x in xrange(10)])
[(0, 0, 0, 0), (0, 0, 0, 4), (0, 0, 0, 5), (0, 0, 0, 7), (0, 0, 0, 8),
(0, 0, 0, 9), (0, 0, 0, 12), (0, 0, 0, 12), (0, 0, 0, 12), (0, 0, 0, 14)]
The three zeroes are only there to show that something other than the first element must be getting inspected.
Try using the internal list sort method and pass a lambda. If your tuples first element is a integer, this should work.
# l is the list of tuples
l.sort(lambda x,y: x-y)
You can use any callable for the compare function, not necessarily a lambda. However it needs to return -1 (less than), 0 (equal) or 1 (greater than).
Check out “Devin Jeanpierre” answer to this question sort-a-dictionary-in-python-by-the-value where he says to use a tuple and shows how to sort by the second value
Empirically, it seems that Python’s default list sorter, when passed a list of tuples, will sort by the first element in each tuple. Is that correct? If not, what’s the right way to sort a list of tuples by their first elements?
Yes, this is the default. In fact, this is the basis of the classic “DSU” (Decorate-Sort-Undecorate) idiom in Python. See Code Like a Pythonista.
It automatically sorts a list of tuples by the first elements in the tuples, then by the second elements and so on tuple([1,2,3]) will go before tuple([1,2,4]). If you want to override this behaviour pass a callable as the second argument to the sort method. This callable should return 1, -1, 0.
No, tuples are sequence types just like strings. They are sorted the same, by comparing each element in turn:
>>> import random
>>> sorted([(0,0,0,int(random.getrandbits(4))) for x in xrange(10)])
[(0, 0, 0, 0), (0, 0, 0, 4), (0, 0, 0, 5), (0, 0, 0, 7), (0, 0, 0, 8),
(0, 0, 0, 9), (0, 0, 0, 12), (0, 0, 0, 12), (0, 0, 0, 12), (0, 0, 0, 14)]
The three zeroes are only there to show that something other than the first element must be getting inspected.
Try using the internal list sort method and pass a lambda. If your tuples first element is a integer, this should work.
# l is the list of tuples
l.sort(lambda x,y: x-y)
You can use any callable for the compare function, not necessarily a lambda. However it needs to return -1 (less than), 0 (equal) or 1 (greater than).
Check out “Devin Jeanpierre” answer to this question sort-a-dictionary-in-python-by-the-value where he says to use a tuple and shows how to sort by the second value