On 02/07/2014 11:06 PM, Igor Korot wrote:
Hi, ALL,
I'm trying to do a very easy task: sort python dictionary by value
where value is a datetime object.
When trying to do that in Python shell everthing works as expected.
C:\Documents and Settings\Igor.FORDANWORK>python
Python 2.7.5 (default, May 15 2013, 22:43:36) [MSC v.1500 32 bit
(Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
a = {}
import datetime
a['1'] = datetime.datetime(2012,12,28,12,15,30,100)
a['2'] = datetime.datetime(2012,12,28,12,17,29,100)
a['3'] = datetime.datetime(2012,12,28,12,16,44,100)
sorted(a.items(), key=a.get)
[('1', datetime.datetime(2012, 12, 28, 12, 15, 30, 100)), ('3', datetime.datetim
e(2012, 12, 28, 12, 16, 44, 100)), ('2', datetime.datetime(2012, 12, 28, 12, 17,
29, 100))]
However, trying to do the same thing from the script does not sort the
dictionary:
sorted(my_dict.items(), key=my_dict.get, reverse=False)
for key, value in my_dict.items():
print value, key
the dictionary prints with unsorted items.
I am trying to run it on Windows XP and the data are coming from the DB.
What am I missing?
Thank you.
You are missing the difference between sorted and sort.
The sorted function returns a new list, but leaves the original unchanged.
The sort method changes the list in place, and returns None.
In your first, interactive example, the returned sorted list is automatically displayed --
that's why you see the sorted list. In your script, however, you are sorting the list but not
saving the result -- it's immediately thrown away. In effect, your "sorted(...)" line is a
no-op. You follow this by displaying the original, unsorted dictionary.
Secondly, your sorting key is wrong. You are sorting a list not a dictionary. It is a list of
(key,value) tuples, which you want to sort on the second element of the tuple. You can do this
with a simple lambda expression: key=lambda t:t[1]
Finally, change your for loop to print the data from this sorted list of tuples, not from the
dictionary.
-=- Larry -=-
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