Re: Arrays/List, filters, Pytho, Ruby
On 12/02/2011 04:49, Terry Reedy wrote:
On 2/11/2011 4:24 PM, LL.Snark wrote:
Hi,
I'm looking for a pythonic way to translate this short Ruby code :
t=[6,7,8,6,7,9,8,4,3,6,7]
i=t.index {|x| x
What does Ruby do if there is no such element?
For Python, the answer should be either None or ValueError.
Ruby code :
# What is the index, in array t,
# of the first element x such that x 8
# Raises an exception if no value found
==
The following solutions raise a StopIteration exception if no value is
found. See below how to use the second argument of next
next(filter(lambda x: x[1] (8,4)
from itertools import dropwhile
next(dropwhile(lambda x: x[1]>=t[0],enumerate(t)))
=> (8,4)
Or if you don't like enumerate :
next(filter(lambda x: x 4
t.index(next(filter(lambda x: x 8
next(filter(t[0].__gt__, t))
=> 4
Generators and enumerate :
next(i for i,v in enumerate(t) if v 8
If no values exists, this raises an exception. Second argument of next
can help you :
next((i for i,v in enumerate(t) if v 8 # returns None if no value found
===
If you use lists instead of generators, it works, but you have to go
across the whole array :
[x 8
[i for i,v in enumerate(t) if v < t[0]]
=> [ 8, 9 ] (indices of all array values less than t[0]
If you are only interested in the first value :
[i for i,v in enumerate(t) if v < t[0]][0]
=> 8 (but raises an exception if the list if no value is found)
[i for i,v in enumerate(t) if v < t[0]][0:1]
=> [8] (return an empty list if no value is found)
Keep in mind that, even if you just select the first value [0] or [0:1],
you go across the whole array t
=
I found it very interesting to read all these short lines of code.
Thanks to : Chris, Ian, Paul, Jon, Therry, Bruno, Arnaud and people that
have no name (like me...)
next(((i,v) for i,v in enumerate(t) if vhttp://mail.python.org/mailman/listinfo/python-list
Re: Arrays/List, filters, Pytho, Ruby
Arnaud Delobelle writes:
> "LL.Snark" writes:
>
>> Hi,
>>
>> I'm looking for a pythonic way to translate this short Ruby code :
>> t=[6,7,8,6,7,9,8,4,3,6,7]
>> i=t.index {|x| x>
>
> In Python3:
>
t = [6,7,8,6,7,9,8,4,3,6,7]
next(filter(t[0].__gt__, t))
> 4
Oops! I realised my mistake the moment I sent the reply. You can do this
to get the index:
>>> next(i for i, x in enumerate(t) if x < t[0])
7
--
Arnaud
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Re: Arrays/List, filters, Pytho, Ruby
"LL.Snark" writes:
> Hi,
>
> I'm looking for a pythonic way to translate this short Ruby code :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=t.index {|x| x
In Python3:
>>> t = [6,7,8,6,7,9,8,4,3,6,7]
>>> next(filter(t[0].__gt__, t))
4
--
Arnaud
--
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Re: Arrays/List, filters, Pytho, Ruby
On Feb 11, 1:24 pm, "LL.Snark" wrote:
> Hi,
>
> I'm looking for a pythonic way to translate this short Ruby code :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=t.index {|x| x
> If you don't know Ruby, the second line means :
> What is the index, in array t, of the first element x such that x
> If can write it in python several ways :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=0
> while t[i]>=t[0] : i+=1
>
> ... not pythonic I think...
>
> Or :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=[j for j in range(len(t)) if t[j]
> ...too cryptic...
>
> I'm using Python 3.
>
> Thx
[i for i,v in enumerate(t) if v < t[0]][0:1]
empty list if no match.
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Re: Arrays/List, filters, Pytho, Ruby
On 11 fév, 22:24, "LL.Snark" wrote:
> Hi,
>
> I'm looking for a pythonic way to translate this short Ruby code :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=t.index {|x| x1000), "not found")
or
next((i for i, x in enumerate(t) if x>1000), None)
or whatever matches your needs.
Note that if you use the second optionnal argument of next, you'll
need an additional pair of parentheses around
the generator expression which is the first argument of next.
Bruno.
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Re: Arrays/List, filters, Pytho, Ruby
On 2/11/2011 4:24 PM, LL.Snark wrote:
Hi,
I'm looking for a pythonic way to translate this short Ruby code :
t=[6,7,8,6,7,9,8,4,3,6,7]
i=t.index {|x| x
What does Ruby do if there is no such element?
For Python, the answer should be either None or ValueError.
If can write it in python several ways :
t=[6,7,8,6,7,9,8,4,3,6,7]
i=0
while t[i]>=t[0] : i+=1
This will raise IndexError when i gets too big.
... not pythonic I think...
Or :
t=[6,7,8,6,7,9,8,4,3,6,7]
i=[j for j in range(len(t)) if t[j]
This will raise IndexError if the list is empty.
--
Terry Jan Reedy
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Re: Arrays/List, filters, Pytho, Ruby
On 11/02/2011 22:24, LL.Snark wrote:
Hi,
I'm looking for a pythonic way to translate this short Ruby code :
t=[6,7,8,6,7,9,8,4,3,6,7]
i=t.index {|x| x=t[0] : i+=1
... not pythonic I think...
Or :
t=[6,7,8,6,7,9,8,4,3,6,7]
i=[j for j in range(len(t)) if t[j]
Thx for your answers.
May I add some comments ?
=
t = [6,7,8,6,7,9,8,4,3,6,7]
i = -1
for index, item in enumerate(t):
if item < t[0]:
i = index
break
is OK for me, while a bit too long :)
==
from itertools import dropwhile
t=[6,7,8,6,7,9,8,4,3,6,7]
i = dropwhile(lambda k: t[k]>=t[0], t).__next__()
(I added the __ around next. Am i true ?)
This does not work for me.
i is effectively 7, but 7 is the second item of the array.
If you try :
t=[6,8,8,6,7,9,8,4,3,6,7]
i = dropwhile(lambda k: t[k]>=t[0], t).__next__()
You will get 8.
I guess this is because k is a value of the array, not an index.
In the first example, we compare t[6], then t[7], then t[8], t[6], t[7]
etc... to t[0].
(It happens that the first item of the list is dropped... then the first
element of the resulting list is 7... the answer... but it was just luck)
=
The javalicious one :
class ComparisonPredicate:
def __init__(self, func):
self.func = func
def __eq__(self, other):
return self.func(other)
t = [6, 7, 8, 6, 7, 9, 8, 4, 3, 6, 7]
print(t.index(ComparisonPredicate(lambda x: x < t[0])))
It took me some time to understand :)
It's a shame there is no built-in object like ComparisonPredicate,
because the line t.index(ComparisonPredicate(lambda x: x < t[0])) looks
good to me.
===
Finally, the other enumerate solution may be written like this :
t = [6,7,8,6,7,9,8,4,3,6,7]
for i, j in enumerate(t):
if j < t[0]: break
else : i=-1
Quite short.
=
Finally, with your solutions, I build another one. Here it is :
t=[6,7,8,6,7,9,8,4,3,6,7]
i,j=filter(lambda x: x[1]=t[0],enumerate(t)).__next__()
Or else :
t=[6,7,8,6,7,9,8,4,3,6,7]
t.index(filter(lambda x: xThe last one behaves like the Ruby one, if a value is found. If no walue
is found, the Python code raises a StopException. With the Ruby code, i
is nil.
Is there another way to access the first item of an iterator ?
( __next__() is ugly )
==
Paul, the Ruby version stops when it finds the first matching element
t=[6,7,8,6,7,9,8,4,3,6,7]
i=t.index {|x| x(I figured this out by making a 1 000 000 elements array. It was faster
with a matching value at the beginning of the array)
Many thanks for your answers
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Re: Arrays/List, filters, Pytho, Ruby
On Feb 11, 9:24 pm, "LL.Snark" wrote:
> Hi,
>
> I'm looking for a pythonic way to translate this short Ruby code :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=t.index {|x| x
> If you don't know Ruby, the second line means :
> What is the index, in array t, of the first element x such that x
> If can write it in python several ways :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=0
> while t[i]>=t[0] : i+=1
>
> ... not pythonic I think...
>
> Or :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=[j for j in range(len(t)) if t[j]
> ...too cryptic...
>
> I'm using Python 3.
>
> Thx
My take (but using Python 2.x):
import operator as op
from functools import partial
from itertools import islice
t = [6,7,8,6,7,9,8,4,3,6,7]
def first_conditional(seq, first=op.itemgetter(0), pred=op.gt):
f = first(seq)
cmpfunc = partial(pred, f)
for idx, val in enumerate(islice(seq, 1, None)):
if cmpfunc(val): return idx + 1
return -1 # or raise an exception?
Off top of head, so needs work, but is fairly generic.
Jon.
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Re: Arrays/List, filters, Pytho, Ruby
"LL.Snark" writes:
> I'm looking for a pythonic way to translate this short Ruby code :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=t.index {|x| x=t[0], t).next()
Note the above can throw an exception if no suitable element is found.
t=[6,7,8,6,7,9,8,4,3,6,7]
i=[j for j in range(len(t)) if t[j]http://mail.python.org/mailman/listinfo/python-list
Re: Arrays/List, filters, Pytho, Ruby
On Fri, Feb 11, 2011 at 1:51 PM, Dan Stromberg wrote:
> On Fri, Feb 11, 2011 at 1:43 PM, André Roberge
> wrote:
>> On Friday, February 11, 2011 5:24:15 PM UTC-4, LL.Snark wrote:
>>> Hi,
>>>
>>> I'm looking for a pythonic way to translate this short Ruby code :
>>> t=[6,7,8,6,7,9,8,4,3,6,7]
>>> i=t.index {|x| x>>
>>> If you don't know Ruby, the second line means :
>>> What is the index, in array t, of the first element x such that x>>
>>> If can write it in python several ways :
>>> t=[6,7,8,6,7,9,8,4,3,6,7]
>>> i=0
>>> while t[i]>=t[0] : i+=1
>>>
>>> ... not pythonic I think...
>>>
>>> Or :
>>> t=[6,7,8,6,7,9,8,4,3,6,7]
>>> i=[j for j in range(len(t)) if t[j]>>
>>> ...too cryptic...
>>>
>> You could go with something like (untested)
>> t = [6,7,8,6,7,9,8,4,3,6,7]
>> for i, j in enumerate(t):
>> if j < t[0]:
>> break
>> else:
>> i = 0
>>
>> ;-)
>>
>>
>>
>>> I'm using Python 3.
>>>
>>> Thx
>
t = [6,7,8,6,7,9,8,4,3,6,7]
generator = (element for element in t[1:] if element >= t[0])
print(next(generator))
>
Oops; a correction. Fast and concise - and if you decide you need the
2nd or 10th, they'll be on their way as soon as you request them (lazy
evaluation):
>>> generator = (ind+1 for ind, element in enumerate(t[1:]) if element >= t[0])
>>> print(next(generator))
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Re: Arrays/List, filters, Pytho, Ruby
On Fri, Feb 11, 2011 at 1:51 PM, Dan Stromberg wrote:
> On Fri, Feb 11, 2011 at 1:43 PM, André Roberge
> wrote:
> > On Friday, February 11, 2011 5:24:15 PM UTC-4, LL.Snark wrote:
> >> Hi,
> >>
> >> I'm looking for a pythonic way to translate this short Ruby code :
> >> t=[6,7,8,6,7,9,8,4,3,6,7]
> >> i=t.index {|x| x >>
> >> If you don't know Ruby, the second line means :
> >> What is the index, in array t, of the first element x such that x >>
> >> If can write it in python several ways :
> >> t=[6,7,8,6,7,9,8,4,3,6,7]
> >> i=0
> >> while t[i]>=t[0] : i+=1
> >>
> >> ... not pythonic I think...
> >>
> >> Or :
> >> t=[6,7,8,6,7,9,8,4,3,6,7]
> >> i=[j for j in range(len(t)) if t[j] >>
> >> ...too cryptic...
> >>
> > You could go with something like (untested)
> > t = [6,7,8,6,7,9,8,4,3,6,7]
> > for i, j in enumerate(t):
> >if j < t[0]:
> >break
> > else:
> >i = 0
> >
> > ;-)
> >
> >
> >
> >> I'm using Python 3.
> >>
> >> Thx
>
> >>> t = [6,7,8,6,7,9,8,4,3,6,7]
> >>> generator = (element for element in t[1:] if element >= t[0])
> >>> print(next(generator))
> --
> http://mail.python.org/mailman/listinfo/python-list
>
Since we only care about the 1st generated value ...
t = [6,7,8,6,7,9,8,4,3,6,7]
t.index(next((elem for elem in t[1:] if elem < t[0])))
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Re: Arrays/List, filters, Pytho, Ruby
On Feb 11, 2:24 pm, "LL.Snark" wrote:
> Hi,
>
> I'm looking for a pythonic way to translate this short Ruby code :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=t.index {|x| xhttp://mail.python.org/mailman/listinfo/python-list
Re: Arrays/List, filters, Pytho, Ruby
On Fri, Feb 11, 2011 at 1:43 PM, André Roberge wrote:
> On Friday, February 11, 2011 5:24:15 PM UTC-4, LL.Snark wrote:
>> Hi,
>>
>> I'm looking for a pythonic way to translate this short Ruby code :
>> t=[6,7,8,6,7,9,8,4,3,6,7]
>> i=t.index {|x| x>
>> If you don't know Ruby, the second line means :
>> What is the index, in array t, of the first element x such that x>
>> If can write it in python several ways :
>> t=[6,7,8,6,7,9,8,4,3,6,7]
>> i=0
>> while t[i]>=t[0] : i+=1
>>
>> ... not pythonic I think...
>>
>> Or :
>> t=[6,7,8,6,7,9,8,4,3,6,7]
>> i=[j for j in range(len(t)) if t[j]>
>> ...too cryptic...
>>
> You could go with something like (untested)
> t = [6,7,8,6,7,9,8,4,3,6,7]
> for i, j in enumerate(t):
> if j < t[0]:
> break
> else:
> i = 0
>
> ;-)
>
>
>
>> I'm using Python 3.
>>
>> Thx
>>> t = [6,7,8,6,7,9,8,4,3,6,7]
>>> generator = (element for element in t[1:] if element >= t[0])
>>> print(next(generator))
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Re: Arrays/List, filters, Pytho, Ruby
On Fri, Feb 11, 2011 at 1:24 PM, LL.Snark wrote:
> Hi,
>
> I'm looking for a pythonic way to translate this short Ruby code :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=t.index {|x| x
> If you don't know Ruby, the second line means :
> What is the index, in array t, of the first element x such that x
> If can write it in python several ways :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=0
> while t[i]>=t[0] : i+=1
>
> ... not pythonic I think...
>
> Or :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=[j for j in range(len(t)) if t[j]
> ...too cryptic...
>
> I'm using Python 3.
My version:
t = [6,7,8,6,7,9,8,4,3,6,7]
i = -1
for index, item in enumerate(t):
if item < t[0]:
i = index
break
I'm a big fan of enumerate().
I'm sure an itertools solution is also possible.
Cheers,
Chris
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Re: Arrays/List, filters, Pytho, Ruby
On Friday, February 11, 2011 5:24:15 PM UTC-4, LL.Snark wrote:
> Hi,
>
> I'm looking for a pythonic way to translate this short Ruby code :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=t.index {|x| x
> If you don't know Ruby, the second line means :
> What is the index, in array t, of the first element x such that x
> If can write it in python several ways :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=0
> while t[i]>=t[0] : i+=1
>
> ... not pythonic I think...
>
> Or :
> t=[6,7,8,6,7,9,8,4,3,6,7]
> i=[j for j in range(len(t)) if t[j]
> ...too cryptic...
>
You could go with something like (untested)
t = [6,7,8,6,7,9,8,4,3,6,7]
for i, j in enumerate(t):
if j < t[0]:
break
else:
i = 0
;-)
> I'm using Python 3.
>
> Thx
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Arrays/List, filters, Pytho, Ruby
Hi,
I'm looking for a pythonic way to translate this short Ruby code :
t=[6,7,8,6,7,9,8,4,3,6,7]
i=t.index {|x| x=t[0] : i+=1
... not pythonic I think...
Or :
t=[6,7,8,6,7,9,8,4,3,6,7]
i=[j for j in range(len(t)) if t[j]http://mail.python.org/mailman/listinfo/python-list
