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I have the following data structure (a list of lists)
[ ['4', '21', '1', '14', '2008-10-24 15:42:58'], ['3', '22', '4', '2somename', '2008-10-24 15:22:03'], ['5', '21', '3', '19', '2008-10-24 15:45:45'], ['6', '21', '1', '1somename', '2008-10-24 15:45:49'], ['7', '22', '3', '2somename', '2008-10-24 15:45:51'] ] I would like to be able to
Use a function to reorder the list so that I can group by each item in the list. For example I'd like to be able to group by the second column (so that all the 21's are together)
Use a function to only display certain values from each inner list. For example i'd like to reduce this list to only contain the 4th field value of '2somename'
so the list would look like this
[ ['3', '22', '4', '2somename', '2008-10-24 15:22:03'], ['7', '22', '3', '2somename', '2008-10-24 15:45:51'] ] 
905
There will be three distinct ways to sort the nested lists. The first is to use Bubble Sort, the second is to use the sort() method, and the third is to use the sorted() method.
Use the sort() Function to Sort a List of Lists in Python. The sort() method sorts the list of lists in Python according to the first element of each inner list. This method makes changes in the original list itself. We use the reverse parameter to sort in descending order.
@SilentGhost A comparison function is called on pairs of items and "should return a negative integer if self < other, zero if self == other, a positive integer if self > other." Whereas a key function is called once for each item and returns the value(s) you want an item sorted on.
For the first question, the first thing you should do is sort the list by the second field using itemgetter from the operator module:
x = [ ['4', '21', '1', '14', '2008-10-24 15:42:58'], ['3', '22', '4', '2somename', '2008-10-24 15:22:03'], ['5', '21', '3', '19', '2008-10-24 15:45:45'], ['6', '21', '1', '1somename', '2008-10-24 15:45:49'], ['7', '22', '3', '2somename', '2008-10-24 15:45:51'] ] from operator import itemgetter x.sort(key=itemgetter(1)) Then you can use itertools' groupby function:
from itertools import groupby y = groupby(x, itemgetter(1)) Now y is an iterator containing tuples of (element, item iterator). It's more confusing to explain these tuples than it is to show code:
for elt, items in groupby(x, itemgetter(1)): print(elt, items) for i in items: print(i) Which prints:
21 <itertools._grouper object at 0x511a0> ['4', '21', '1', '14', '2008-10-24 15:42:58'] ['5', '21', '3', '19', '2008-10-24 15:45:45'] ['6', '21', '1', '1somename', '2008-10-24 15:45:49'] 22 <itertools._grouper object at 0x51170> ['3', '22', '4', '2somename', '2008-10-24 15:22:03'] ['7', '22', '3', '2somename', '2008-10-24 15:45:51'] For the second part, you should use list comprehensions as mentioned already here:
from pprint import pprint as pp pp([y for y in x if y[3] == '2somename']) Which prints:
[['3', '22', '4', '2somename', '2008-10-24 15:22:03'], ['7', '22', '3', '2somename', '2008-10-24 15:45:51']] 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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