Get the first item from an iterable that matches a condition

Question

I would like to get the first item from a list matching a condition  It s important that the resulting method not process the entire list  which could be quite large  For example  the following function is adequate   def first the iterable  condition   lambda x  True       for i in the iterable          if condition i               return i   This function could be used something like this    gt  gt  gt  first range 10   0  gt  gt  gt  first range 10   lambda i  i  gt  3  4   However  I can t think of a good built-in   one-liner to let me do this  I don t particularly want to copy this function around if I don t have to  Is there a built-in way to get the first item matching a condition

User · Answer

In Python 3   a    None  False  0  1  assert next filter None  a      1   In Python 2 6   a    None  False  0  1  assert next iter filter None  a       1   EDIT  I thought it was obvious  but apparently not  instead of None you can pass a function  or a lambda  with a check for the condition   a    2 3 4 5 6 7 8  assert next filter lambda x  x 2  a      3

User · Answer

Oneliner   thefirst    i for i in range 10  if i  gt  3  0    If youre not sure that any element will be valid according to the criteria  you should enclose this with try except since that  0  can raise an IndexError

User · Answer

By using    index for index  value in enumerate the iterable  if condition value     one can check the condition of the value of the first item in the iterable  and obtain its index without the need to evaluate all of the items in the iterable   The complete expression to use is  first index   next index for index  value in enumerate the iterable  if condition value     Here first index assumes the value of the first value identified in the expression discussed above

User · Answer

You could also use the argwhere function in Numpy  For example   i  Find the first  l  in  helloworld    import numpy as np l   list  helloworld     Create list i   np argwhere np array l    l     i   array   2   3   8    index of first   i min     ii  Find first random number   0 1  import numpy as np r   np random rand 50    Create random numbers i   np argwhere r gt 0 1  index of first   i min     iii  Find the last random number   0 1  import numpy as np r   np random rand 50    Create random numbers i   np argwhere r gt 0 1  index of last   i max

User · Answer

As a reusable  documented and tested function  def first iterable  condition   lambda x  True               Returns the first item in the  iterable  that     satisfies the  condition        If the condition is not given  returns the first item of     the iterable       Raises  StopIteration  if no item satysfing the condition is found        gt  gt  gt  first   1 2 3   condition lambda x  x   2    0      2      gt  gt  gt  first range 3  100       3      gt  gt  gt  first           Traceback  most recent call last               StopIteration              return next x for x in iterable if condition x     Version with default argument   zorf suggested a version of this function where you can have a predefined return value if the iterable is empty or has no items matching the condition   def first iterable  default   None  condition   lambda x  True               Returns the first item in the  iterable  that     satisfies the  condition        If the condition is not given  returns the first item of     the iterable       If the  default  argument is given and the iterable is empty      or if it has no items matching the condition  the  default  argument     is returned if it matches the condition       The  default  argument being None is the same as it not being given       Raises  StopIteration  if no item satisfying the condition is found     and default is not given or doesn t satisfy the condition        gt  gt  gt  first   1 2 3   condition lambda x  x   2    0      2      gt  gt  gt  first range 3  100       3      gt  gt  gt  first           Traceback  most recent call last               StopIteration      gt  gt  gt  first     default 1      1      gt  gt  gt  first     default 1  condition lambda x  x   2    0      Traceback  most recent call last               StopIteration      gt  gt  gt  first  1 3 5   default 1  condition lambda x  x   2    0      Traceback  most recent call last               StopIteration              try          return next x for x in iterable if condition x       except StopIteration          if default is not None and condition default               return default         else              raise

User · Answer

Similar to using ifilter  you could use a generator expression    gt  gt  gt   x for x in xrange 10  if x  gt  5  next   6   In either case  you probably want to catch StopIteration though  in case no elements satisfy your condition   Technically speaking  I suppose you could do something like this    gt  gt  gt  foo   None  gt  gt  gt  for foo in  x for x in xrange 10  if x  gt  5   break       gt  gt  gt  foo 6   It would avoid having to make a try except block  But that seems kind of obscure and abusive to the syntax

User · Answer

The itertools module contains a filter function for iterators  The first element of the filtered iterator can be obtained by calling next   on it   from itertools import ifilter  print ifilter  lambda i  i  gt  3   range 10   next

User · Answer

I would write this   next x for x in xrange 10  if x  gt  3

User · Answer

For older versions of Python where the next built-in doesn t exist    x for x in range 10  if x  gt  3  next

User · Answer

Damn Exceptions   I love this answer  However  since next   raise a StopIteration exception when there are no items  i would use the following snippet to avoid an exception   a      item   next  x for x in a   None      For example    a      item   next x for x in a    Will raise a StopIteration exception   Traceback  most recent call last     File   lt stdin gt    line 1  in  lt module gt  StopIteration

User · Answer

The most efficient way in Python 3 are one of the following  using a similar example    With  comprehension  style   next i for i in range 100000000  if i    1000    WARNING  The expression works also with Python 2  but in the example is used range that returns an iterable object in Python 3 instead of a list like Python 2  if you want to construct an iterable in Python 2 use xrange instead    Note that the expression avoid to construct a list in the comprehension expression next  i for        that would cause to create a list with all the elements before filter the elements  and would cause to process the entire options  instead of stop the iteration once i    1000   With  functional  style   next filter lambda i  i    1000  range 100000000      WARNING  This doesn t work in Python 2  even replacing range with xrange due that filter create a list instead of a iterator  inefficient   and the next function only works with iterators   Default value  As mentioned in other responses  you must add a extra-parameter to the function next if you want to avoid an exception raised when the condition is not fulfilled    functional  style   next filter lambda i  i    1000  range 100000000    False     comprehension  style   With this style you need to surround the comprehension expression with    to avoid a SyntaxError  Generator expression must be parenthesized if not sole argument   next  i for i in range 100000000  if i    1000   False

User · Answer

In Python 2 6 or newer   If you want StopIteration to be raised if no matching element is found   next x for x in the iterable if x  gt  3    If you want default value  e g  None  to be returned instead   next  x for x in the iterable if x  gt  3   default value    Note that you need an extra pair of parentheses around the generator expression in this case - they are needed whenever the generator expression isn t the only argument   I see most answers resolutely ignore the next built-in and so I assume that for some mysterious reason they re 100  focused on versions 2 5 and older -- without mentioning the Python-version issue  but then I don t see that mention in the answers that do mention the next built-in  which is why I thought it necessary to provide an answer myself -- at least the  correct version  issue gets on record this way -    In 2 5  the  next   method of iterators immediately raises StopIteration if the iterator immediately finishes -- i e   for your use case  if no item in the iterable satisfies the condition   If you don t care  i e   you know there must be at least one satisfactory item  then just use  next    best on a genexp  line for the next built-in in Python 2 6 and better    If you do care  wrapping things in a function as you had first indicated in your Q seems best  and while the function implementation you proposed is just fine  you could alternatively use itertools  a for     break loop  or a genexp  or a try except StopIteration as the function s body  as various answers suggested   There s not much added value in any of these alternatives so I d go for the starkly-simple version you first proposed

User · Answer

For anyone using Python 3 8 or newer I recommend using  quot Assignment Expressions quot  as described in PEP 572 -- Assignment Expressions  if any  match    i   gt  3 for i in range 10        print match

User · Answer

This question already has great answers  I m only adding my two cents because I landed here trying to find a solution to my own problem  which is very similar to the OP    If you want to find the INDEX of the first item matching a criteria using generators  you can simply do   next index for index  value in enumerate iterable  if condition

[python] Get the first item from an iterable that matches a condition

Examples related to python

Examples related to iterator