How to take the first N items from a generator or list

Question

With linq I would  var top5   array Take 5     How to do this with Python

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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  N   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 element  generator next   for i in range N     Otherwise   first N element list generator   N    Where N is the number of elements you want to take  e g  N 5 for the first five elements

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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

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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

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Shaikovsky s answer is excellent     and heavily edited since I posted this answer   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

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This should work  top5   array  5

User · Answer

The answer for how to do this can be found here   gt  gt  gt  generator    i for i in xrange 10    gt  gt  gt  list next generator  for   in range 4    0  1  2  3   gt  gt  gt  list next generator  for   in range 4    4  5  6  7   gt  gt  gt  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

User · Answer

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

User · Answer

import itertools  top5   itertools islice array  5

[python] How to take the first N items from a generator or list?

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