I have a list a = ['a1', 'b1', 'c1', 'd1', 'a2', 'b2', 'c2', 'd2',]
How can I get list b = ['a1,', 'a2', 'b1', 'b2', 'c1', 'c2', 'd1', 'd2'] without using .sorted()?
Thanks!
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If you don't want to use sorted() then sort it yourself. :)Azad Salahli– Azad Salahli2011-03-06 08:12:25 +00:00Commented Mar 6, 2011 at 8:12
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And why would you not want to use sorted?KT100– KT1002013-04-29 18:28:30 +00:00Commented Apr 29, 2013 at 18:28
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6 Answers
There is no .sorted() method for lists, though there is the sorted() function, as S.Mark pointed out (which returns a new sorted list), and a .sort() method (which sorts a list in place and returns None). If you meant to not use the sorted() function, then:
a = ['a1', 'b1', 'c1', 'd1', 'a2', 'b2', 'c2', 'd2',]
a.sort()
b = a
otherwise, maybe you can clarify your question further.
3 Comments
Bob
Okey. But if I have a list of playing card, then .sorted() doesn't work correctly!
BenjaminGolder
@Bob, I don't understand what you're looking for. Are you looking for a way to sort things that are not letters or numbers? A way to sort strings by their meaning rather than their alphabetic order?
It seems a bit arbitrary, not to use sorted(). I think you mean, that you don't want to sort the list in the (default) alphanumerical order.
Here is how you define a key for sorting strings that represent playing cards (a1 through d13) by suit, then rank:
>>> def cardsortkey(card):
... return (card[0], int(card[1:]))
...
>>> cardsortkey('a1')
('a', 1)
>>> a = ['a1', 'b1', 'c1', 'd1',
... 'a2', 'b2', 'c2', 'd2',
... 'a11', 'b11', 'c11', 'd11']
>>> sorted(a, key=cardsortkey)
['a1', 'a2', 'a11', 'b1', 'b2', 'b11', 'c1', 'c2', 'c11', 'd1', 'd2', 'd11']
Is that what you need?
1 Comment
Maciej Ziarko
+1. I agree, your answers is much more Pythonic and much more elegant. But I don't think that it is more obvious for noob python coder.
without using sorted, but expensive way.
a = ['a1', 'b1', 'c1', 'd1', 'a2', 'b2', 'c2', 'd2',]
split it to 2 parts
['a1', 'b1', 'c1', 'd1',] ['a2', 'b2', 'c2', 'd2',]
zip it
[('a1', 'a2'), ('b1', 'b2'), ('c1', 'c2'), ('d1', 'd2')]
and flatten it (with itertools here)
import itertools
itertools.chain(*zip(a[:len(a)/2],a[len(a)/2:]))
itertools returns iterator, so If you need list, wrapped it with list(), and assigned it to b
b = list(itertools.chain(*zip(a[:len(a)/2],a[len(a)/2:])))
=> ['a1', 'a2', 'b1', 'b2', 'c1', 'c2', 'd1', 'd2']
Comments
l = ['a1', 'b1', 'c1', 'd1', 'a2', 'b2', 'c2', 'd2']
numbersPerLetter = 2
lsorted = []
for i in range(len(l) / numbersPerLetter):
lsorted.extend([l[x+i] for x in range(0, len(l), len(l) / numbersPerLetter)])
print(lsorted)
Output:
['a1', 'a2', 'b1', 'b2', 'c1', 'c2', 'd1', 'd2']
In Python 3.X you have to change / to // in order to make it work.
6 Comments
Bob
Could you please explain the loop inside the first loop?
Maciej Ziarko
It is called a list comprehension. More here: docs.python.org/tutorial/…. For each iteration it creates a little list with the same letters. I mean indexes 0,4 next 1,5 next 2,6 and finally 3,7. More on range with 3 arguments here: docs.python.org/library/functions.html#range
Maciej Ziarko
@hop: By the way, I've just noticed that your answer is not really answer to the question. Funny.
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You can also sort it this way
for i1, e1 in enumerate(a):
for i2, e2 in enumerate(a):
if e2 > e1:
e1 = a[i2]
a[i2] = a[i1]
a[i1] = e1
