Python – limit output from a sort method

djangoitertoolspython

if my views code is:

arttags = sorted(arttags, key=operator.attrgetter('date_added'), reverse=True)

what is the argument that will limit the result to 50 tags?

I'm assuming this:

.... limit=50)

is incorrect.

more complete code follows:

videoarttags = Media.objects.order_by('date_added'),filter(topic__exact='art') 
audioarttags = Audio.objects.order_by('date_added'),filter(topic__exact='art') 
conarttags = Concert.objects.order_by('date_added'),filter(topic__exact='art') 
arttags = list(chain(videoarttags, audioarttags, conarttags)) 
arttags = sorted(arttags, key=operator.attrgetter('date_added'), reverse=True)

how do incorporate –

itertools.islice(sorted(...),50)

Best Solution

what about heapq.nlargest:
Return a list with the n largest elements from the dataset defined by iterable.key, if provided, specifies a function of one argument that is used to extract a comparison key from each element in the iterable: key=str.lower Equivalent to: sorted(iterable, key=key, reverse=True)[:n]

>>> from heapq import nlargest
>>> data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0]
>>> nlargest(3, data)
[9, 8, 7]

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Python – How to sort a list of dictionaries by a value of the dictionary

The sorted() function takes a key= parameter

newlist = sorted(list_to_be_sorted, key=lambda d: d['name'])

Alternatively, you can use operator.itemgetter instead of defining the function yourself

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For completeness, add reverse=True to sort in descending order

newlist = sorted(list_to_be_sorted, key=itemgetter('name'), reverse=True)

Python – How to execute a program or call a system command

Use the subprocess module in the standard library:

import subprocess
subprocess.run(["ls", "-l"])

The advantage of subprocess.run over os.system is that it is more flexible (you can get the stdout, stderr, the "real" status code, better error handling, etc...).

Even the documentation for os.system recommends using subprocess instead:

The subprocess module provides more powerful facilities for spawning new processes and retrieving their results; using that module is preferable to using this function. See the Replacing Older Functions with the subprocess Module section in the subprocess documentation for some helpful recipes.

On Python 3.4 and earlier, use subprocess.call instead of .run:

subprocess.call(["ls", "-l"])

Best Solution

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