[(u'we', 'PRP'), (u'saw', 'VBD'), (u'you', 'PRP'), (u'bruh', 'VBP'), (u'.', '.')]
I want to order this alphabetically, by "PRP, VBD, PRP, and VBP" It's not the traditional sort, right?
views:
141answers:
4
+15
A:
Use itemgetter:
>>> a = [(u'we', 'PRP'), (u'saw', 'VBD'), (u'you', 'PRP'), (u'bruh', 'VBP'), (u'.', '.')]
>>> import operator
>>> a.sort(key = operator.itemgetter(1))
>>> a
[(u'.', '.'), (u'we', 'PRP'), (u'you', 'PRP'), (u'saw', 'VBD'), (u'bruh', 'VBP')]
slurdge
2009-11-23 10:31:20
+1 for the use of `itemgetter`, didn't know it existed :)
abyx
2009-11-23 10:33:20
+1 for using key instead of cmp.
SpoonMeiser
2009-11-23 10:55:02
Acceptable solution ... but I don't think it's worth an import. :) I prefer Dave Webb's way.
jellybean
2009-11-23 10:57:48
+1
A:
You can pass a comparison function in the sort method.
Example:
l = [(u'we', 'PRP'), (u'saw', 'VBD'), (u'you', 'PRP'), (u'bruh', 'VBP'), (u'.', '.')]
l.sort(lambda x, y: cmp(x[1], y[1])) # compare tuple's 2nd elements
Output:
>>> l
[(u'.', '.'),
(u'we', 'PRP'),
(u'you', 'PRP'),
(u'saw', 'VBD'),
(u'bruh', 'VBP')]
>>>
Nick D
2009-11-23 10:33:56
Using a key with sort is preferable to using cmp, as the key only gets called N times where cmp will be called NlogN times. You have no choice in python3, support for cmp has been removed
gnibbler
2009-11-23 10:49:06
@gnibbler, thanks for the tip. Why the key will be called only N times? I didn't know that.
Nick D
2009-11-23 10:59:14
@Nick, because with `key` Python will go through the list once to calculate the keys, then sort using those keys. With `cmp`, on the other hand, it will call the function every time it needs to compare two values.
Daniel Roseman
2009-11-23 11:10:54
hmm, so it collects the keys prior sorting. Thanks. But won't this step use extra memory? Is it really faster than using the `cmp`, for large lists? I have to test the 2 approaches ;-)
Nick D
2009-11-23 11:26:39
+7
A:
The sort method takes a key argument to extract a comparison key from each argument, i.e. key is a function which transforms the list item into the value you wish to sort on.
In this case it's very easy to use a lambda to extract the second item from each tuple:
>>> mylist = [(u'we', 'PRP'), (u'saw', 'VBD'), (u'you', 'PRP'), (u'bruh', 'VBP')
, (u'.', '.')]
>>> mylist.sort(key = lambda x: x[1])
>>> mylist
[(u'.', '.'), (u'we', 'PRP'), (u'you', 'PRP'), (u'saw', 'VBD'), (u'bruh', 'VBP')]
Dave Webb
2009-11-23 10:48:45
Actualy, it is the same as itemgetter but at least you see what's it is doing.
Natim
2009-11-23 10:51:39
A:
Or use the Built-in Function sorted (similar to Nick_D's answer):
ls = [(u'we', 'PRP'), (u'saw', 'VBD'), (u'you', 'PRP'), (u'bruh', 'VBP'), (u'.', '.')]
sorted(ls, cmp=lambda t1, t2: cmp(t1[1], t2[1]))
Jogusa
2009-11-23 10:50:18