Hello NG,
I was wondering if there is a faster/nicer method (than a for loop)
that will allow me to find the elements (AND their indices) in a list that
verify a certain condition. For example, assuming that I have a list like:
mylist = [0, 1, 1, 1, 1, 5, 6, 7, 8, 1, 10]
I would like to find the indices of the elements in the list that are equal
to 1 (in this case, the 1,2,3,4,9 elements are equal to 1). I could easily
use a for loop but I was wondering if there is a faster method...
Thanks for every suggestion.
Andrea.
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<an***********@ agip.it> a écrit dans le message de news: ma************* *************** **...*@pyth on.org... For example, assuming that I have a list like:
mylist = [0, 1, 1, 1, 1, 5, 6, 7, 8, 1, 10]
I would like to find the indices of the elements in the list that are equal to 1 (in this case, the 1,2,3,4,9 elements are equal to 1).
List comprehension is your friend:
PythonWin 2.4 (#60, Nov 30 2004, 11:49:19) [MSC v.1310 32 bit (Intel)] on
win32.
Portions Copyright 1994-2004 Mark Hammond (mh******@skipp inet.com.au) - see
'Help/About PythonWin' for further copyright information. mylist = [0, 1, 1, 1, 1, 5, 6, 7, 8, 1, 10] [i for i, j in enumerate(mylis t) if j==1]
[1, 2, 3, 4, 9]
an***********@a gip.it wrote: Hello NG,
I was wondering if there is a faster/nicer method (than a for loop) that will allow me to find the elements (AND their indices) in a list that verify a certain condition. For example, assuming that I have a list like:
mylist = [0, 1, 1, 1, 1, 5, 6, 7, 8, 1, 10]
I would like to find the indices of the elements in the list that are equal to 1 (in this case, the 1,2,3,4,9 elements are equal to 1). I could easily use a for loop but I was wondering if there is a faster method...
Thanks for every suggestion.
Andrea. ------------------------------------------------------------------------------------------------------------------------------------------ Message for the recipient only, if received in error, please notify the sender and read http://www.eni.it/disclaimer/
You could do a list comprehension /generator expression. Like this:
[i for i in range(len(mylis t)) if mylist[i] == 1]
--
--------------------------------------
Ola Natvig <ol********@inf osense.no>
infoSense AS / development
On Fri, 10 Dec 2004 16:01:26 +0100, an***********@a gip.it wrote: Hello NG,
I was wondering if there is a faster/nicer method (than a for loop) that will allow me to find the elements (AND their indices) in a list that verify a certain condition. For example, assuming that I have a list like:
mylist = [0, 1, 1, 1, 1, 5, 6, 7, 8, 1, 10]
I would like to find the indices of the elements in the list that are equal to 1 (in this case, the 1,2,3,4,9 elements are equal to 1). I could easily use a for loop but I was wondering if there is a faster method...
Thanks for every suggestion.
One way: mylist = [0, 1, 1, 1, 1, 5, 6, 7, 8, 1, 10] [i for i,v in enumerate(mylis t) if v==1]
[1, 2, 3, 4, 9]
Regards,
Bengt Richter an***********@a gip.it wrote: Hello NG,
I was wondering if there is a faster/nicer method (than a for loop) that will allow me to find the elements (AND their indices) in a list that verify a certain condition. For example, assuming that I have a list like:
mylist = [0, 1, 1, 1, 1, 5, 6, 7, 8, 1, 10]
I would like to find the indices of the elements in the list that are equal to 1 (in this case, the 1,2,3,4,9 elements are equal to 1). I could easily use a for loop but I was wondering if there is a faster method...
Everyone has already given you the answer (enumerate in a LC or GE), I'd
just comment that it's easy enough to extend their answers to any given
condition: def getindices(sequ ence, predicate):
.... return [i for i, v in enumerate(seque nce) if predicate(v)]
.... getindices([0,1,1,1,1,5,6,7 ,8,1,10], bool)
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] def equalsone(v):
.... return v == 1
.... getindices([0,1,1,1,1,5,6,7 ,8,1,10], equalsone)
[1, 2, 3, 4, 9] def f(v):
.... return pow(v, 3, 4) == 3
.... getindices([0,1,1,1,1,5,6,7 ,8,1,10], f)
[7]
Steve
On Fri, 10 Dec 2004 16:27:29 GMT, Steven Bethard <st************ @gmail.com> wrote: an***********@ agip.it wrote: Hello NG,
I was wondering if there is a faster/nicer method (than a for loop) that will allow me to find the elements (AND their indices) in a list that verify a certain condition. For example, assuming that I have a list like:
mylist = [0, 1, 1, 1, 1, 5, 6, 7, 8, 1, 10]
I would like to find the indices of the elements in the list that are equal to 1 (in this case, the 1,2,3,4,9 elements are equal to 1). I could easily use a for loop but I was wondering if there is a faster method...
Everyone has already given you the answer (enumerate in a LC or GE), I'd just comment that it's easy enough to extend their answers to any given condition:
def getindices(sequ ence, predicate):... return [i for i, v in enumerate(seque nce) if predicate(v)] ... getindices([0,1,1,1,1,5,6,7 ,8,1,10], bool)[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] def equalsone(v):... return v == 1 ... getindices([0,1,1,1,1,5,6,7 ,8,1,10], equalsone)[1, 2, 3, 4, 9] def f(v):... return pow(v, 3, 4) == 3 ... getindices([0,1,1,1,1,5,6,7 ,8,1,10], f)
[7]
Conclusion:
Python is programmer's Lego ;-)
Regards,
Bengt Richter This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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