Sort a list numerically in Python

Question

So I have this list, we\'ll call it listA. I\'m trying to get the [3] item in each list e.g. [\'5.01\',\'5.88\',\'2.10\',\'9.45\',\'17.58\',\'2.76\'] in sorted orde

Answer 1

If you want to return the complete list in sorted order, this may work. This takes your input list and runs sorted on top of it. The reverse argument set to True sorts the list in reverse (descending) order, and the key argument specifies the method by which to sort, which in this case is the float of the third argument of each list:

In [5]: l = [['John Doe', u'25.78', u'20.77', '5.01'] # continues...    
In [6]: sorted(l, reverse=True, key=lambda x: float(x[3]))
Out[6]:
[['Santa', u'31.13', u'13.55', '17.58'],
 ['Michael Jordan', u'28.50', u'19.05', '9.45'],
 ['Jane Doe', u'21.08', u'15.20', '5.88'],
 ['John Doe', u'25.78', u'20.77', '5.01'],
 ['Easter Bunny', u'17.20', u'14.44', '2.76'],
 ['James Bond', u'20.57', u'18.47', '2.10']]

If you only need the values in sorted order, the other answers provide viable ways of doing so.

Answer 2

from operator import itemgetter
lstA.sort(reverse = True, key = itemgetter(3))

And you are good to go. Using itemgetter() within a sort is extremely helpful when you have multiple indexs to sort on. Lets say you wanted to sort alphabetically on the first value in case of a tie you could do the following

from operator import itemgetter
lstA.sort(key = itemgetter(0))
lstA.sort(reverse = True, key = itemgetter(3))

And Voilà!

Answer 3

An anonymous lambda function is not necessary for this task. You can use operator.itemgetter, which may be more intuitive:

from operator import itemgetter

res1 = sorted(map(itemgetter(3), listA), reverse=True, key=float)

['17.58', '9.45', '5.88', '5.01', '2.76', '2.10']

Answer 4

Try this:

>>> listA=[['John Doe', u'25.78', u'20.77', '5.01'], ['Jane Doe', u'21.08', u'15.20', '5.88'], ['James Bond', u'20.57', u'18.47', '2.10'], ['Michael Jordan', u'28.50', u'19.05', '9.45'], ['Santa', u'31.13', u'13.55', '17.58'], ['Easter Bunny', u'17.20', u'14.44', '2.76']]

>>> [x[3] for x in sorted(listA, reverse = True, key = lambda i : float(i[3]))]
['17.58', '9.45', '5.88', '5.01', '2.76', '2.10']