February 27th, 2005, 10:01 PM

Sorting 2d array/list
Simple question I'm sure someone can answer in a jiffy.
I have a 2d array like the following:
array = [['value', 2.454], ['2ndvalue', 6.252], ['3rdvalue', 1.252]...]
I now wish to sort this array by the 2nd numerical attribute. Is there some standard inbuilt function in python to do this or do I have to write my own code using a standard sorting algorithm???
February 28th, 2005, 01:01 AM

there are dictionarys!
theres are basicly like a list, except you can use it to find the a value instead of using the numerical index key.
lucky for you there is a way to build a dictionary from a nested list. the function is dict(). you can find more at pythons web site http://docs.python.org/tut/node7.html#SECTION007500000000000000000
Comments on this post
February 28th, 2005, 01:23 AM

Here is another function for making a dictionary from 2 lists
Code:
def mkDict(self,keyList, valueList, cond=None):
'''
Make a new dict out of two lists. The first list provides keys,
the 2nd provides values. The cond is a function taking 2 arguments:
key and value. Example: lambda k,v:v%2==0
A valid item means its cond(key, value) is true. Only valid items
are included in the returned dict.
>>> a
[0, 1, 2, 3, 4]
>>> b= [chr(x) for x in range(65,70)]
>>> b
['A', 'B', 'C', 'D', 'E']
>>> mkDict(b,a)
{'A': 0, 'C': 2, 'B': 1, 'E': 4, 'D': 3}
>>> mkDict(b,a,lambda x,y:y%2==0)
{'A': 0, 'C': 2, 'E': 4}
Note:
mkDict(a,b) (without the conditional check) is equivalent to
dict(zip(a,b)), which is much faster. If you know there's no
conditional check, better use dict(zip(a,b)) instead.
'''
if cond==None:
return dict(zip(keyList, valueList))
else:
return dict( [(k,v) for k,v in zip(keyList, valueList) if cond(k,v)] )
Here is another function for sorting a dictionary by value
Code:
def sort_by_value(self,d):
""" Returns the keys of dictionary d sorted by their values """
items=d.items()
backitems=[ [v[1],v[0]] for v in items]
backitems.sort()
return [ backitems[i][1] for i in range(0,len(backitems))]
I would convert the array to a dictionary ,sort it and convert it back to 2d array ( list or whatever)
something like this ( havent tested the code below)
Code:
keys=[]
values=[]
for item in length(array)
keys.append(item[0])
values.append(item[1])
dict = mkDict(keys,values)# defined above
dict = sort_by_value(dict)#defined above
sortedarray= []
for key in dict.keys()
arrayItem=[]
arrayItem.append(key)
arrayItem.append(dict[key])
sortedarray.append(arrayItem)
#your sorted 2d Array is sortedArray here
Might not be the best way to do this whole thing though
February 28th, 2005, 07:50 AM

Hi!
I think this is the easiest solution:
Code:
x = [['a', 2.454],['b', 6.252],['c', 1.252]]
def sort_second(a,b):
return cmp(a[1],b[1])
x.sort(sort_second)
print x
Regards, mawe
Comments on this post
February 28th, 2005, 12:41 PM

I agree totally
Originally Posted by mawe
Hi!
I think this is the easiest solution:
Code:
x = [['a', 2.454],['b', 6.252],['c', 1.252]]
def sort_second(a,b):
return cmp(a[1],b[1])
x.sort(sort_second)
print x
Regards, mawe
Above sol is the easiest way to do it, I agree.
February 28th, 2005, 04:35 PM

Originally Posted by mawe
Hi!
I think this is the easiest solution:
Code:
x = [['a', 2.454],['b', 6.252],['c', 1.252]]
def sort_second(a,b):
return cmp(a[1],b[1])
x.sort(sort_second)
print x
Regards, mawe
I agree totally