Python join first n elements of list

With linq I would

var top5 = array.Take[5];

How to do this with Python?

martineau

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asked Mar 8, 2011 at 14:53

Jader DiasJader Dias

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Slicing a list

top5 = array[:5]

• To slice a list, there's a simple syntax: array[start:stop:step]
• You can omit any parameter. These are all valid: array[start:], array[:stop], array[::step]

Slicing a generator

import itertools
top5 = itertools.islice[my_list, 5] # grab the first five elements

• You can't slice a generator directly in Python. itertools.islice[] will wrap an object in a new slicing generator using the syntax itertools.islice[generator, start, stop, step]

• Remember, slicing a generator will exhaust it partially. If you want to keep the entire generator intact, perhaps turn it into a tuple or list first, like: result = tuple[generator]

Nico Schlömer

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answered Mar 8, 2011 at 15:00

5

import itertools

top5 = itertools.islice[array, 5]

answered Mar 8, 2011 at 14:56

Jader DiasJader Dias

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@Shaikovsky's answer is excellent, but I wanted to clarify a couple of points.

[next[generator] for _ in range[n]]

This is the most simple approach, but throws StopIteration if the generator is prematurely exhausted.

On the other hand, the following approaches return up to n items which is preferable in many circumstances:

List: [x for _, x in zip[range[n], records]]

Generator: [x for _, x in zip[range[n], records]]

Neuron

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answered Aug 17, 2016 at 11:29

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In my taste, it's also very concise to combine zip[] with xrange[n] [or range[n] in Python3], which works nice on generators as well and seems to be more flexible for changes in general.

# Option #1: taking the first n elements as a list
[x for _, x in zip[xrange[n], generator]]

# Option #2, using 'next[]' and taking care for 'StopIteration'
[next[generator] for _ in xrange[n]]

# Option #3: taking the first n elements as a new generator
[x for _, x in zip[xrange[n], generator]]

# Option #4: yielding them by simply preparing a function
# [but take care for 'StopIteration']
def top_n[n, generator]:
    for _ in xrange[n]:
        yield next[generator]

Neuron

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answered Oct 3, 2014 at 20:21

ShaikovskyShaikovsky

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The answer for how to do this can be found here

>>> generator = [i for i in xrange[10]]
>>> list[next[generator] for _ in range[4]]
[0, 1, 2, 3]
>>> list[next[generator] for _ in range[4]]
[4, 5, 6, 7]
>>> list[next[generator] for _ in range[4]]
[8, 9]

Notice that the last call asks for the next 4 when only 2 are remaining. The use of the list[] instead of [] is what gets the comprehension to terminate on the StopIteration exception that is thrown by next[].

answered Oct 15, 2015 at 19:37

ebergersonebergerson

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Do you mean the first N items, or the N largest items?

If you want the first:

top5 = sequence[:5]

This also works for the largest N items, assuming that your sequence is sorted in descending order. [Your LINQ example seems to assume this as well.]

If you want the largest, and it isn't sorted, the most obvious solution is to sort it first:

l = list[sequence]
l.sort[reverse=True]
top5 = l[:5]

For a more performant solution, use a min-heap [thanks Thijs]:

import heapq
top5 = heapq.nlargest[5, sequence]

answered Mar 8, 2011 at 14:57

ThomasThomas

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With itertools you will obtain another generator object so in most of the cases you will need another step the take the first n elements. There are at least two simpler solutions [a little bit less efficient in terms of performance but very handy] to get the elements ready to use from a generator:

Using list comprehension:

first_n_elements = [generator.next[] for i in range[n]]

Otherwise:

first_n_elements = list[generator][:n]

Where n is the number of elements you want to take [e.g. n=5 for the first five elements].

Neuron

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answered Feb 7, 2015 at 11:17

G MG M

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This should work

top5 = array[:5] 

answered Mar 8, 2011 at 14:57

Bala RBala R

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