[x for x in <> while <>]?

[urikaluzhny]

It seems that I rather frequently need a list or iterator of the form
[x for x in <> while <>]
And there is no one like this.
May be there is another short way to write it (not as a loop). Is
there?
Thanks

 

[Terry Reedy]

| It seems that I rather frequently need a list or iterator of the form
| [x for x in <> while <>]

I can think of two ways to interpret that.

| And there is no one like this.
| May be there is another short way to write it (not as a loop). Is
| there?

Using loops to write an example of what you mean would make the above
clearer.

tjr

 

[Bruno Desthuilliers]

urikaluzhny a écrit :

It seems that I rather frequently need a list or iterator of the form
[x for x in <> while <>]
And there is no one like this.
May be there is another short way to write it (not as a loop). Is
there?

The answer is very probably in the itertools module.

 

[Paul Rubin]

It seems that I rather frequently need a list or iterator of the form
[x for x in <> while <>]
And there is no one like this.
May be there is another short way to write it (not as a loop). Is there?

itertools.takewhile(condition, seq)

 

[urikaluzhny]

| It seems that I rather frequently need a list or iterator of the form
| [x for x in <> while <>]

I can think of two ways to interpret that.

I mean like [x for x in <A> if <B>], only that it breaks the loop when
the expression <B> is false.

 

[Geoffrey Clements]

| It seems that I rather frequently need a list or iterator of the form
| [x for x in <> while <>]

I can think of two ways to interpret that.

I mean like [x for x in <A> if <B>], only that it breaks the loop when
the expression <B> is false.

def gen(a):
for x in a:
if B: break
yield x

a_gen = gen(A)

# now iterate over a_gen

 

[afrobeard]

The following proposed solution is not intended to be a solution, it
goes completely against the zen of python. [Type import this into the
python command interpreter]

I brought it down to two lines:-

l = range(6)
[1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]

itertools would still be a better approach in my opinion.

Just because I'm curious to know, can anyone bring it shorter[even if
its cryptic] than this without invoking any Python Library.

P.S. Once again I would not recommend using this as Explicit is better
than Implicit
P.P.S. It is strongly undesirable for us humans to use anything
starting with __

"urikaluzhny" <[email protected]> wrote in message

| It seems that I rather frequently need a list or iterator of the form
| [x for x in <> while <>]

I can think of two ways to interpret that.

I mean like [x for x in <A> if <B>], only that it breaks the loop when
the expression <B> is false.

def gen(a):
    for x in a:
        if B: break
        yield x

a_gen = gen(A)

# now iterate over a_gen

 

[castironpi]

The following proposed solution is not intended to be a solution, it
goes completely against the zen of python. [Type import this into the
python command interpreter]

I brought it down to two lines:-

l = range(6)
[1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]

itertools would still be a better approach in my opinion.

Just because I'm curious to know, can anyone bring it shorter[even if
its cryptic] than this without invoking any Python Library.

P.S. Once again I would not recommend using this as Explicit is better
than Implicit
P.P.S. It is strongly undesirable for us humans to use anything
starting with __

"urikaluzhny" <[email protected]> wrote in message

| It seems that I rather frequently need a list or iterator of the form
| [x for x in <> while <>]
I can think of two ways to interpret that.
I mean like [x for x in <A> if <B>], only that it breaks the loop when
the expression <B> is false.

def gen(a):
    for x in a:
        if B: break
        yield x

a_gen = gen(A)

# now iterate over a_gen

- Show quoted text -

In your original, you have:

l = range(6)
[1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]

You may be hyperextending the use of '..if..else..', which is one of
my fears regarding 'with x as y'. "l.__delslice__(0,len(l))" is not
an expression.

 

[afrobeard]

According to http://en.wikipedia.org/wiki/Expression_(programming)

"An expression in a programming language is a combination of values,
variables, operators, and functions that are interpreted (evaluated)
according to the particular rules of precedence and of association for
a particular programming language, which computes and then produces
(returns, in a stateful environment) another value."

l.__delslice__(0,len(l)) is an expression because it returns None
[which also happens to be a value] in this case.

The following proposed solution is not intended to be a solution, it
goes completely against the zen of python. [Type import this into the
python command interpreter]

I brought it down to two lines:-

l = range(6)
[1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]

itertools would still be a better approach in my opinion.

Just because I'm curious to know, can anyone bring it shorter[even if
its cryptic] than this without invoking any Python Library.

P.S. Once again I would not recommend using this as Explicit is better
than Implicit
P.P.S. It is strongly undesirable for us humans to use anything
starting with __

| It seems that I rather frequently need a list or iterator of the form
| [x for x in <> while <>]
I can think of two ways to interpret that.
I mean like [x for x in <A> if <B>], only that it breaks the loop when
the expression <B> is false.
def gen(a):
    for x in a:
        if B: break
        yield x
a_gen = gen(A)
# now iterate over a_gen

- Show quoted text -

In your original, you have:

l = range(6)
[1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]

You may be hyperextending the use of '..if..else..', which is one of
my fears regarding 'with x as y'.  "l.__delslice__(0,len(l))" is not
an expression.

 

[afrobeard]

l.__delslice__(0,len(l)) is an expression as it returns None which is
a value

The following proposed solution is not intended to be a solution, it
goes completely against the zen of python. [Type import this into the
python command interpreter]

I brought it down to two lines:-

l = range(6)
[1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]

itertools would still be a better approach in my opinion.

Just because I'm curious to know, can anyone bring it shorter[even if
its cryptic] than this without invoking any Python Library.

P.S. Once again I would not recommend using this as Explicit is better
than Implicit
P.P.S. It is strongly undesirable for us humans to use anything
starting with __

| It seems that I rather frequently need a list or iterator of the form
| [x for x in <> while <>]
I can think of two ways to interpret that.
I mean like [x for x in <A> if <B>], only that it breaks the loop when
the expression <B> is false.
def gen(a):
    for x in a:
        if B: break
        yield x
a_gen = gen(A)
# now iterate over a_gen

- Show quoted text -

In your original, you have:

l = range(6)
[1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]

You may be hyperextending the use of '..if..else..', which is one of
my fears regarding 'with x as y'.  "l.__delslice__(0,len(l))" is not
an expression.

 

[castironpi]

l.__delslice__(0,len(l)) is an expression as it returns None which is
a value

The following proposed solution is not intended to be a solution, it
goes completely against the zen of python. [Type import this into the
python command interpreter]
I brought it down to two lines:-
l = range(6)
[1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]
itertools would still be a better approach in my opinion.
Just because I'm curious to know, can anyone bring it shorter[even if
its cryptic] than this without invoking any Python Library.
P.S. Once again I would not recommend using this as Explicit is better
than Implicit
P.P.S. It is strongly undesirable for us humans to use anything
starting with __
On May 15, 5:10 pm, "Geoffrey Clements"
| It seems that I rather frequently need a list or iterator of the form
| [x for x in <> while <>]
I can think of two ways to interpret that.
I mean like [x for x in <A> if <B>], only that it breaks the loop when
the expression <B> is false.
def gen(a):
    for x in a:
        if B: break
        yield x
a_gen = gen(A)
# now iterate over a_gen

In your original, you have:

l = range(6)
[1 if b!=4 else l.__delslice__(0,len(l)) for b in l][:-1]

You may be hyperextending the use of '..if..else..', which is one of
my fears regarding 'with x as y'.  "l.__delslice__(0,len(l))" is not
an expression.- Hide quoted text -

- Show quoted text -

Functional programming is really important to a former professor of
me. I like to say that None returns as a value. I think you can call
functional "evaluational".

 

[Alexandr N Zamaraev]

| It seems that I rather frequently need a list or iterator of the form
| [x for x in <> while <>]
I can think of two ways to interpret that.

I mean like [x for x in <A> if <B>], only that it breaks the loop when
the expression <B> is false.

How do you plan to modify B during iteration?
May be
[x for x in itertools.takewhile(<B>, <A>)]
when <B> function accept element <A> and return True or False

 

[Ruediger]

It seems that I rather frequently need a list or iterator of the form
[x for x in <> while <>]
And there is no one like this.
May be there is another short way to write it (not as a loop). Is
there?
Thanks

I usually have the same problem and i came up with an solution like that:

from operator import ne
def test(iterable, value, op=ne):
_n = iter(iterable).next
while True:
_x = _n()
if op(_x, value):
yield _x
else:
raise StopIteration

l = range(6)
print [x for x in test(l, 4)]

r@linux:~/tmp> python test18.py
[0, 1, 2, 3]

 

[Paul Hankin]

It seems that I rather frequently need a list or iterator of the form
[x for x in <> while <>]
And there is no one like this.
May be there is another short way to write it (not as a loop). Is
there?
Thanks

I usually have the same problem and i came up with an solution like that:

from operator import ne
def test(iterable, value, op=ne):
    _n = iter(iterable).next
    while True:
        _x = _n()
        if op(_x, value):
            yield _x
        else:
            raise StopIteration

This is better written using takewhile...
itertools.takewhile(lambda x: x != value, iterable)

But if you really need to reinvent the wheel, perhaps this is simpler?

def test(iterable, value, op=operator.ne):
for x in iterable:
if not op(x, value):
return
yield x

 

[Ruediger]

Paul Hankin wrote:

This is better written using takewhile...
itertools.takewhile(lambda x: x != value, iterable)

But if you really need to reinvent the wheel, perhaps this is simpler?

def test(iterable, value, op=operator.ne):
for x in iterable:
if not op(x, value):
return
yield x

yes you are right it is. However as i mentioned in my post i came up with an
solution 'like' that. In fact my original code was to complex to post.
While simplifying it, i've overseen the obvious solution.

For special cases where you need to do more complex tests, the best solution
is IMHO to hide it in an generator function like above.