Geek Answers Handbook

What is the standard way to make this look in Python?

Suppose I have a list of tuples in this format:

(1, 2, 3)
(1, 0, 2)
(3, 9 , 11)
(0, 2, 8)
(2, 3, 4)
(2, 4, 5)
(2, 7, 8)
....

How can I sort the list by the first element of tuples and then by the second? I would like to get on this list:

(0, 2, 8)
(1, 0, 2)
(1, 2, 3)
(2, 3, 4)
(2, 4, 5)
(2, 7, 8)
(3, 9 , 11)

I was thinking of pretending to look for the first element, and then go through the list and build a hash with subarrays. I probably confuse things :) and that is why I asked for other ways to do this.

+5
source share
5 answers

Why not just let python sort the list for you?

my_list = [
(1, 2, 3),
(1, 0, 2),
(3, 9 , 11),
(0, 2, 8),
(2, 3, 4),
(2, 4, 5),
(2, 7, 8),
]

print sorted(my_list)
>>>[(0, 2, 8), (1, 0, 2), (1, 2, 3), (2, 3, 4), (2, 4, 5), (2, 7, 8), (3, 9, 11)]
+8
source

Python automatically does the right thing:

>>> a = [(1, 2, 3), (1, 0, 2), (3, 9, 11), (0, 2, 8), (2, 3, 4), (2, 4, 5), (2, 7, 8)]
>>> a.sort()
>>> a
[(0, 2, 8), (1, 0, 2), (1, 2, 3), (2, 3, 4), (2, 4, 5), (2, 7, 8), (3, 9, 11)]
+3
source

.

:

#!/usr/bin/python2
l = [
  (1, 2, 3),
  (1, 0, 2),
  (3, 9 , 11),
  (0, 2, 8),
  (2, 3, 4),
  (2, 4, 5),
  (2, 7, 8),
]

l.sort()
print l
+1

, :

>>> l = [(1, 2, 3), (1, 0, 2), (3, 9, 11), (0, 2, 8), (2, 3, 4), (2, 4, 5), (2, 7, 8)]
>>> l.sort()
>>> l
[(0, 2, 8), (1, 0, 2), (1, 2, 3), (2, 3, 4), (2, 4, 5), (2, 7, 8), (3, 9, 11)]
0
>>> x = [
...   (1, 2, 3),
...   (1, 0, 2),
...   (3, 9 , 11),
...   (0, 2, 8),
...   (2, 3, 4),
...   (2, 4, 5),
...   (2, 7, 8),
... ]
>>> x.sort()
>>> x
[(0, 2, 8), (1, 0, 2), (1, 2, 3), (2, 3, 4), (2, 4, 5), (2, 7, 8), (3, 9, 11)]
0

More articles:


All Articles