> But in the general case can you do the same job as
> reduce in a list comprehension?
>>> def f(x, y):
... return x + y
...
>>> tmp = [0]
>>> [f(x, y) for x in arr for y in [tmp[-1]] if tmp.append(f(x, y)) or
>>> True][-1]
45
Let's try some more...
>>> def f(x, y):
... return x*y
...
>>> tmp
On 25/11/05, Nick Lunt <[EMAIL PROTECTED]> wrote:
> > -Original Message-
> > >>> def f(x, y):
> > ... return x*y
> > ...
> > >>> arr = range(1, 10)# Don't want to include 0!
> > >>> reduce(f, arr) # Our target
> > 362880
> > >>> tmp = [1]
> > >>> [f(x, y) for x in arr for y i
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> >>> def f(x, y):
> ... return x + y
> ...
> >>> arr = range(10)
> >>> sum(arr) # Our target
> 45
> >>> tmp = [0]
> >>> [f(x, y) for x in arr for y in [tmp[-1]] if tmp.append(f(x,
> y)) or True][-1]
>
On 22/11/05, Alan Gauld <[EMAIL PROTECTED]> wrote:
> But in the general case can you do the same job as
> reduce in a list comprehension? ie apply an operation
> to the first two elements of a sequence and replace
> them with the result, then repeat until the list becomes
> a single value?
Well ..
Hi John,
> Everything is possible with list comprehensions!
>>> [x for z in a for y in z for x in y]
> [1, 2, 3, 31, 32, 4, 5, 6, 7, 71, 72, 8, 9]
impressive!
But in the general case can you do the same job as
reduce in a list comprehension? ie apply an operation
to the first two elements of
Hi John,
thanks it.
It is great. I looked for it, but I couldn't made it.
I have tried it with wrong order:
# I have tried it so
[x for x in y for y in a]
[[8], [8], [8], [9], [9], [9]] # that is wrong,
# Instead of that you wrote
[x for y in a for x in y]
[[1], [2], [3, 31, 32], [4], [5], [6],
On 21/11/05, János Juhász <[EMAIL PROTECTED]> wrote:
> I can't imagine how this could be made with list comprehension.
>
> >>> import operator
> >>> a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9]))
> >>> reduce(operator.add, a) # it makes a long list now
> ([1], [2], [3, 31, 32], [4]
János Juhász wrote:
> Hi,
>
> I can't imagine how this could be made with list comprehension.
>
import operator
a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9]))
reduce(operator.add, a) # it makes a long list now
>
> Is it possible to substitute reduce with comprehensio
>>> a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9]))
>>> reduce(operator.add, a) # it makes a long list now
([1], [2], [3, 31, 32], [4], [5], [6], [7, 71, 72], [8], [9])
When I make list comprehension, the list hierarchy is allways the same or
>>> [item for item in a] # the deepnes
János Juhász said unto the world upon 2005-11-21 01:20:
> Hi,
>
> I can't imagine how this could be made with list comprehension.
>
>
import operator
a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9]))
reduce(operator.add, a) # it makes a long list now
>
> ([1], [2], [3,
Hi,
I can't imagine how this could be made with list comprehension.
>>> import operator
>>> a = (([1],[2],[3,31,32],[4]), ([5],[6],[7, 71, 72]), ([8],[9]))
>>> reduce(operator.add, a) # it makes a long list now
([1], [2], [3, 31, 32], [4], [5], [6], [7, 71, 72], [8], [9])
When I make list compre
11 matches
Mail list logo
