Why does one work, but not the other?

[j_mckitrick]

I've done this before:

data = [self.cong.tm[k] for k in self.cong.tm.li]
#li is list, tm is dict

instead of:

for k in self.cong.tm.li :
data.append(sel f.cong.tm[k])

but when I try:

self.liststore = [[item] for item in data]


instead of:

for item in data:
self.liststore. append([item])

I get an empty list! What gives??

jonathon

[Erik Max Francis]

Re: Why does one work, but not the other?

j_mckitrick wrote:
[color=blue]
> but when I try:
>
> self.liststore = [[item] for item in data]
>
> instead of:
>
> for item in data:
> self.liststore. append([item])
>
> I get an empty list! What gives??[/color]

You're probably doing something else wrong; this fragment works fine:
[color=blue][color=green][color=darkred]
>>> data = [1, 2, 3, 4]
>>> [[item] for item in data][/color][/color][/color]
[[1], [2], [3], [4]]

--
__ Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
/ \ San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
\__/ Ferocious, aren't I?
-- Lt. Vincent Hanna

[Adonis]

Re: Why does one work, but not the other?

----- Original Message -----
From: "j_mckitric k" <j_mckitrick@bi gfoot.com>
Newsgroups: comp.lang.pytho n
Sent: Thursday, June 17, 2004 9:54 PM
Subject: Why does one work, but not the other?

[color=blue]
> I've done this before:
>
> data = [self.cong.tm[k] for k in self.cong.tm.li]
> #li is list, tm is dict
>
> instead of:
>
> for k in self.cong.tm.li :
> data.append(sel f.cong.tm[k])
>
> but when I try:
>
> self.liststore = [[item] for item in data]
>
>
> instead of:
>
> for item in data:
> self.liststore. append([item])
>
> I get an empty list! What gives??
>
> jonathon[/color]
[color=blue][color=green][color=darkred]
>>> tm = {'a':1, 'b':2, 'c':3}
>>> li = tm.keys()
>>> data = [tm[k] for k in li]
>>> data[/color][/color][/color]
[1, 3, 2][color=blue][color=green][color=darkred]
>>> # also you can do
>>> data2 = [tm[k] for k in tm]
>>> data2[/color][/color][/color]
[1, 3, 2]

Check the dictionary, make sure it has entries?
When supplying the dictionary with elements given by the list, make sure the
keys exist.

or you can do:[color=blue][color=green][color=darkred]
>>> data = [tm.get(k) for k in li][/color][/color][/color]

Will return None if a key does not exist to further debug.

Hope this helps.

Adonis

[j_mckitrick]

Re: Why does one work, but not the other?

Thanks for the help, guys. I found the problem. The liststore cannot
be redefined once set up for the TreeView. I can only clear/append to
it.

But I'm still on my mission to replace 'for' with list comprehensions
where possible, according to the article on optimization on the python
site.


That being said, is there a way to write this as a comprehension? I
can't figure out how to do so and get k into the key correctly. I'm
just trying to save a dictionary via anydbm.

for k, v in self.options.it ems():
db[k] = str(v)

jonathon

[Peter Otten]

Re: Why does one work, but not the other?

j_mckitrick wrote:
[color=blue]
> But I'm still on my mission to replace 'for' with list comprehensions
> where possible, according to the article on optimization on the python
> site.[/color]

I don't know the article, but I assume it doesn't tell list comprehensions
are always faster/better.
[color=blue]
> That being said, is there a way to write this as a comprehension? I
> can't figure out how to do so and get k into the key correctly. I'm
> just trying to save a dictionary via anydbm.
>
> for k, v in self.options.it ems():
> db[k] = str(v)[/color]

Yes,
[color=blue][color=green][color=darkred]
>>> dk = {1:2, 3:4}
>>> options = {1:4, 2:6, 3:8}
>>> dk.update(dict([(k, str(v)) for (k, v) in options.iterite ms()]))
>>> dk[/color][/color][/color]
{1: '4', 2: '6', 3: '8'}[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

but why would you trade a muddy comprehension for a clean loop? The for loop
is clearer (and faster, I suppose) here. Remember that list comprehensions
are a means rather than an end.

With 2.4 that may be a different story, as the above will reduce (I think)
to

dk.update((k, str(v)) for (k, v) in options.iterite ms())

However, some overhead (generating throwaway tuples) is likely to remain.

Peter

[j_mckitrick]

Re: Why does one work, but not the other?

Peter Otten <__peter__@web. de> wrote in message news:<cauk4d$b4 n$00$1@news.t-online.com>...[color=blue]
> j_mckitrick wrote:
>[color=green]
> > But I'm still on my mission to replace 'for' with list comprehensions
> > where possible, according to the article on optimization on the python
> > site.[/color]
>
> I don't know the article, but I assume it doesn't tell list comprehensions
> are always faster/better.[/color]

from http://www.python.org/doc/essays/list2str.html:

Try to use map(), filter() or reduce() to replace an explicit for
loop, but only if you can use a built-in function: map with a built-in
function beats for loop, but a for loop with in-line code beats map
with a lambda function!


I remember another, but can't find it right now.

[Terry Reedy]

Re: Why does one work, but not the other?


"j_mckitric k" <j_mckitrick@bi gfoot.com> wrote in message
news:ec6dce8b.0 406171754.5d5d8 4db@posting.goo gle.com...[color=blue]
> I've done this before:
>
> data = [self.cong.tm[k] for k in self.cong.tm.li]
> #li is list, tm is dict
>
> instead of:
>
> for k in self.cong.tm.li :
> data.append(sel f.cong.tm[k])
>
> but when I try:
>
> self.liststore = [[item] for item in data]
>
>
> instead of:
>
> for item in data:
> self.liststore. append([item])
>
> I get an empty list! What gives??[/color]

For questions like this, about supposedly anomalous behavior, you usually
need to give actual input and output, reduced to the minimum code needed to
show the purported behavior. Otherwise, the easiest guess is that you did
not use the same value of data in the two snippets;-)

Terry J. Reedy