How to sort over dictionaries

[David Raymond]

Remember to check what the res["date"] types actually are. If they're just text, then it looked like they were in M/D/Y format, which won't sort correctly as text, hence you might want to include using datetime.strptime() to turn them into sortable datetimes.


-----Original Message-----
From: Python-list [mailto:python-list-bounces+david.raymond=tomtom.com at python.org] On Behalf Of harish at moonshots.co.in
Sent: Thursday, August 30, 2018 4:31 AM
To: python-list at python.org
Subject: Re: How to sort over dictionaries



> > sort = sorted(results, key=lambda res:itemgetter('date'))
> > print(sort)
> > 
> > 
> > I have tried the above code peter but it was showing error like ....
> > TypeError: '<' not supported between instances of 'operator.itemgetter'
> > and 'operator.itemgetter'
> 
> lambda res: itemgetter('date')
> 
> is short for
> 
> def keyfunc(res):
>     return itemgetter('date')
> 
> i. e. it indeed returns the itemgetter instance. But you want to return 
> res["date"]. For that you can either use a custom function or the 
> itemgetter, but not both.
> 
> (1) With regular function:
> 
> def keyfunc(res):
>     return res["date"]
> sorted_results = sorted(results, key=keyfunc)
> 
> (1a) With lambda:
> 
> keyfunc = lambda res: res["date"]
> sorted_results = sorted(results, key=keyfunc)
> 
> (2) With itemgetter:
> 
> keyfunc = itemgetter("date")
> sorted_results = sorted(results, key=keyfunc)
> 
> Variants 1a and 2 can also be written as one-liners.



Thanks Peter....No 2 has worked.
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