Thủ Thuật Hướng dẫn Python join first n elements of list Mới nhất Chi Tiết

Bạn đang tìm kiếm từ khóa Python join first n elements of list Mới nhất được Cập Nhật vào lúc : 2022-09-21 14:40:00 . Với phương châm chia sẻ Kinh Nghiệm Hướng dẫn trong nội dung bài viết một cách Chi Tiết Mới Nhất. Nếu sau khi đọc tài liệu vẫn ko hiểu thì hoàn toàn có thể lại Comment ở cuối bài để Mình lý giải và hướng dẫn lại nha.

Mẹo về Python join first n elements of list Chi Tiết

You đang tìm kiếm từ khóa Python join first n elements of list được Cập Nhật vào lúc : 2022-09-21 14:40:18 . Với phương châm chia sẻ Kinh Nghiệm Hướng dẫn trong nội dung nội dung bài viết một cách Chi Tiết Mới Nhất. Nếu sau khi đọc nội dung nội dung bài viết vẫn ko hiểu thì hoàn toàn hoàn toàn có thể lại Comments ở cuối bài để Mình lý giải và hướng dẫn lại nha.

With linq I would

Nội dung chính

    Slicing a list
    Slicing a generator
    How do I merge the first two elements in a list Python?
    How do you join elements in a list in Python?
    How do you print the top 5 elements of a list in Python?
    How do you make a list with N elements in Python?

var top5 = array.Take(5);

How to do this with Python?

martineau

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asked Mar 8, 2011 14:53

Jader DiasJader Dias

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1

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 15:00

5

import itertools

top5 = itertools.islice(array, 5)

answered Mar 8, 2011 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, 2022 11:29

6

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 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 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 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 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

11:17

G MG M

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This should work

top5 = array[:5]

answered Mar 8, 2011 14:57

Bala RBala R

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How do I merge the first two elements in a list Python?

To join specific list elements (e.g., with indices 0 , 2 , and 4 ) and return the joined string that’s the concatenation of all those, use the expression ”. join([lst[i] for i in [0, 2, 4]]) . The list comprehension statement creates a list consisting of elements lst[0] , lst[2] , and lst[4] .

How do you join elements in a list in Python?

To join specific list elements in Python:. Use list slicing to select the specific elements in the list.. Use the str. join() method to join the elements into a string.. Replace the list elements with the string..

How do you print the top 5 elements of a list in Python?

Method #1 : Using sorted() + lambda

The combination of above functionality can be used to perform this particular task. In this, we just employ sorted function with reverse flag true, and print the top N elements using list slicing.

How do you make a list with N elements in Python?

To create a list of n placeholder elements, multiply the list of a single placeholder element with n . For example, use [None] * 5 to create a list [None, None, None, None, None] with five elements None . You can then overwrite some elements with index assignments.

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