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I've got a list of Python objects that I'd like to sort by an attribute of the objects themselves. The list looks like:
>>> ut [<Tag: 128>, <Tag: 2008>, <Tag: <>, <Tag: actionscript>, <Tag: addresses>, <Tag: aes>, <Tag: ajax> ...] Each object has a count:
>>> ut[1].count 1L I need to sort the list by number of counts descending.
I've seen several methods for this, but I'm looking for best practice in Python.
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sort() method to sort a list of objects using some examples. By default, the sort() method sorts a given list into ascending order (or natural order). We can use Collections. reverseOrder() method, which returns a Comparator, for reverse sorting.
A Pythonic solution to in-place sort a list of objects using multiple attributes is to use the list. sort() function. It accepts two optional keyword-only arguments: key and reverse and produces a stable sort.
Python List sort() - Sorts Ascending or Descending List. The list. sort() method sorts the elements of a list in ascending or descending order using the default < comparisons operator between items. Use the key parameter to pass the function name to be used for comparison instead of the default < operator.
# To sort the list in place... ut.sort(key=lambda x: x.count, reverse=True) # To return a new list, use the sorted() built-in function... newlist = sorted(ut, key=lambda x: x.count, reverse=True) More on sorting by keys.
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A way that can be fastest, especially if your list has a lot of records, is to use operator.attrgetter("count"). However, this might run on an pre-operator version of Python, so it would be nice to have a fallback mechanism. You might want to do the following, then:
try: import operator except ImportError: keyfun= lambda x: x.count # use a lambda if no operator module else: keyfun= operator.attrgetter("count") # use operator since it's faster than lambda ut.sort(key=keyfun, reverse=True) # sort in-place If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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