Re: dict comprehension question.
On Tuesday, January 1, 2013 11:10:48 AM UTC+8, Steven D'Aprano wrote:
On Sat, 29 Dec 2012 18:56:57 -0500, Terry Reedy wrote:
On 12/29/2012 2:48 PM, Quint Rankid wrote:
Given a list like:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
I would like to be able to do the following as a dict comprehension.
a = {}
for x in w:
a[x] = a.get(x,0) + 1
results in a having the value:
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
Let me paraphrase this: I have nice, clear, straightforward,
*comprehensible* code that I want to turn into an incomprehensible mess
with a 'comprehension. That is the ironic allure of comprehensions.
But... but... one liner! ONE LINNR Won't somebody think
of the lines I'll save
*wink*
In case it's not obvious, I'm 100% agreeing with Terry here. List comps
and dict comps are wonderful things, but they can't do everything, and
very often even if they can do something they shouldn't because it makes
the code inefficient or unreadable.
There's nothing wrong with a two or three liner.
--
Steven
This is useful for not being choked in sorting a list
by the notorious quick-sort.
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Re: dict comprehension question.
On Sat, 29 Dec 2012 18:56:57 -0500, Terry Reedy wrote:
On 12/29/2012 2:48 PM, Quint Rankid wrote:
Given a list like:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
I would like to be able to do the following as a dict comprehension.
a = {}
for x in w:
a[x] = a.get(x,0) + 1
results in a having the value:
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
Let me paraphrase this: I have nice, clear, straightforward,
*comprehensible* code that I want to turn into an incomprehensible mess
with a 'comprehension. That is the ironic allure of comprehensions.
But... but... one liner! ONE LINNR Won't somebody think
of the lines I'll save
*wink*
In case it's not obvious, I'm 100% agreeing with Terry here. List comps
and dict comps are wonderful things, but they can't do everything, and
very often even if they can do something they shouldn't because it makes
the code inefficient or unreadable.
There's nothing wrong with a two or three liner.
--
Steven
--
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dict comprehension question.
Newbie question. I've googled a little and haven't found the answer.
Given a list like:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
I would like to be able to do the following as a dict comprehension.
a = {}
for x in w:
a[x] = a.get(x,0) + 1
results in a having the value:
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
I've tried a few things
eg
a1 = {x:self.get(x,0)+1 for x in w}
results in error messages.
And
a2 = {x:a2.get(x,0)+1 for x in w}
also results in error messages.
Trying to set a variable to a dict before doing the comprehension
a3 = {}
a3 = {x:a3.get(x,0)+1 for x in w}
gets this result, which isn't what I wanted.
{1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1}
I'm not sure that it's possible to do this, and if not, perhaps the
most obvious question is what instance does the get method bind to?
TIA
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Re: dict comprehension question.
In article
724d4fea-606a-4503-b538-87442f6bc...@ci3g2000vbb.googlegroups.com,
Quint Rankid qbr...@gmail.com wrote:
Newbie question. I've googled a little and haven't found the answer.
Given a list like:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
I would like to be able to do the following as a dict comprehension.
a = {}
for x in w:
a[x] = a.get(x,0) + 1
Why are you trying to do this mind-blowing thing? Other than as an
entry in an obfuscated code contest, what is this for?
Anyway, I don't think this is possible with a dict comprehension.
Entries in the dict depend on entries previously put into the dict. I
don't see any way a dict comprehension can deal with this.
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Re: dict comprehension question.
On 12/29/2012 02:48 PM, Quint Rankid wrote:
Newbie question. I've googled a little and haven't found the answer.
Given a list like:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
I would like to be able to do the following as a dict comprehension.
a = {}
for x in w:
a[x] = a.get(x,0) + 1
results in a having the value:
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
I've tried a few things
eg
a1 = {x:self.get(x,0)+1 for x in w}
results in error messages.
And
a2 = {x:a2.get(x,0)+1 for x in w}
also results in error messages.
Trying to set a variable to a dict before doing the comprehension
a3 = {}
a3 = {x:a3.get(x,0)+1 for x in w}
gets this result, which isn't what I wanted.
{1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1}
I'm not sure that it's possible to do this, and if not, perhaps the
most obvious question is what instance does the get method bind to?
TIA
Will this do?:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
{x: w.count(x) for x in w}
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
- mitya
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Re: dict comprehension question.
On 12/29/2012 03:01 PM, Mitya Sirenef wrote:
On 12/29/2012 02:48 PM, Quint Rankid wrote:
Newbie question. I've googled a little and haven't found the answer.
Given a list like:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
I would like to be able to do the following as a dict comprehension.
a = {}
for x in w:
a[x] = a.get(x,0) + 1
results in a having the value:
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
I've tried a few things
eg
a1 = {x:self.get(x,0)+1 for x in w}
results in error messages.
And
a2 = {x:a2.get(x,0)+1 for x in w}
also results in error messages.
Trying to set a variable to a dict before doing the comprehension
a3 = {}
a3 = {x:a3.get(x,0)+1 for x in w}
gets this result, which isn't what I wanted.
{1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1}
I'm not sure that it's possible to do this, and if not, perhaps the
most obvious question is what instance does the get method bind to?
TIA
Will this do?:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
{x: w.count(x) for x in w}
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
- mitya
I should probably add that this might be inefficient for large lists as
it repeats count for each item. If you need it for large lists, profile
against the 'for loop' version and decide if performance is good enough
for you, for small lists it's a nice and compact solution.
In a more general case, you can't refer to the list/dict/etc
comprehension as it's being constructed, that's just not a design goal
of comprehensions.
-m
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Re: dict comprehension question.
On Sat, Dec 29, 2012 at 3:09 PM, Mitya Sirenef msire...@lightbird.netwrote:
On 12/29/2012 03:01 PM, Mitya Sirenef wrote:
On 12/29/2012 02:48 PM, Quint Rankid wrote:
Newbie question. I've googled a little and haven't found the answer.
Given a list like:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
I would like to be able to do the following as a dict comprehension.
a = {}
for x in w:
a[x] = a.get(x,0) + 1
results in a having the value:
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
I've tried a few things
eg
a1 = {x:self.get(x,0)+1 for x in w}
results in error messages.
And
a2 = {x:a2.get(x,0)+1 for x in w}
also results in error messages.
Trying to set a variable to a dict before doing the comprehension
a3 = {}
a3 = {x:a3.get(x,0)+1 for x in w}
gets this result, which isn't what I wanted.
{1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1}
I'm not sure that it's possible to do this, and if not, perhaps the
most obvious question is what instance does the get method bind to?
TIA
Will this do?:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
{x: w.count(x) for x in w}
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
- mitya
I should probably add that this might be inefficient for large lists as
it repeats count for each item. If you need it for large lists, profile
against the 'for loop' version and decide if performance is good enough
for you, for small lists it's a nice and compact solution.
In a more general case, you can't refer to the list/dict/etc
comprehension as it's being constructed, that's just not a design goal
of comprehensions.
Would this help:
w = [1,2,3,1,2,4,4,5,6,1]
s = set(w)
s
set([1, 2, 3, 4, 5, 6])
{x:w.count(x) for x in s}
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
-m
--
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Re: dict comprehension question.
Quint Rankid wrote:
Newbie question. I've googled a little and haven't found the answer.
Given a list like:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
I would like to be able to do the following as a dict comprehension.
a = {}
for x in w:
a[x] = a.get(x,0) + 1
results in a having the value:
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
I've tried a few things
eg
a1 = {x:self.get(x,0)+1 for x in w}
results in error messages.
And
a2 = {x:a2.get(x,0)+1 for x in w}
also results in error messages.
Trying to set a variable to a dict before doing the comprehension
a3 = {}
a3 = {x:a3.get(x,0)+1 for x in w}
gets this result, which isn't what I wanted.
{1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1}
I'm not sure that it's possible to do this, and if not, perhaps the
most obvious question is what instance does the get method bind to?
The name a3 will not be rebound until after the right side is evaluate. To
spell it with a loop:
a3 = {}
_internal = {} # You have no access to this var.
# Even if you can beat a particular
# Python implementation -- you shouldn't
for x in w:
_internal[x] = a3.get(x, 0) + 1
a3 = _internal
That should make it clear that x will be looked up in the old empty a3
dict.
The closest you can get to a self-updating dict is probably
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
a1 = {}
a1.update((x, a1.get(x, 0)+1) for x in w)
a1
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
but that it works doesn't mean it is a good idea.
If your Python version supports it the obvious choice is
from collections import Counter
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
Counter(w)
Counter({1: 3, 2: 2, 4: 2, 3: 1, 5: 1, 6: 1})
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Re: dict comprehension question.
On 2012-12-29 19:48, Quint Rankid wrote:
Newbie question. I've googled a little and haven't found the answer.
Given a list like:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
I would like to be able to do the following as a dict comprehension.
a = {}
for x in w:
a[x] = a.get(x,0) + 1
results in a having the value:
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
I've tried a few things
eg
a1 = {x:self.get(x,0)+1 for x in w}
results in error messages.
And
a2 = {x:a2.get(x,0)+1 for x in w}
also results in error messages.
Trying to set a variable to a dict before doing the comprehension
a3 = {}
a3 = {x:a3.get(x,0)+1 for x in w}
gets this result, which isn't what I wanted.
{1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1}
I'm not sure that it's possible to do this, and if not, perhaps the
most obvious question is what instance does the get method bind to?
You can't do it with a comprehension.
The best way is probably with the 'Counter' class:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
from collections import Counter
Counter(w)
Counter({1: 3, 2: 2, 4: 2, 3: 1, 5: 1, 6: 1})
If you want the result be a dict, then just the result to 'dict':
dict(Counter(w))
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
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Re: dict comprehension question.
On 12/29/2012 03:15 PM, Joel Goldstick wrote:
On Sat, Dec 29, 2012 at 3:09 PM, Mitya Sirenef msire...@lightbird.net
mailto:msire...@lightbird.net wrote:
On 12/29/2012 03:01 PM, Mitya Sirenef wrote:
On 12/29/2012 02:48 PM, Quint Rankid wrote:
Newbie question. I've googled a little and haven't found the
answer.
Given a list like:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
I would like to be able to do the following as a dict
comprehension.
a = {}
for x in w:
a[x] = a.get(x,0) + 1
results in a having the value:
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
I've tried a few things
eg
a1 = {x:self.get(x,0)+1 for x in w}
results in error messages.
And
a2 = {x:a2.get(x,0)+1 for x in w}
also results in error messages.
Trying to set a variable to a dict before doing the comprehension
a3 = {}
a3 = {x:a3.get(x,0)+1 for x in w}
gets this result, which isn't what I wanted.
{1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1}
I'm not sure that it's possible to do this, and if not, perhaps the
most obvious question is what instance does the get method bind to?
TIA
Will this do?:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
{x: w.count(x) for x in w}
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
- mitya
I should probably add that this might be inefficient for large
lists as
it repeats count for each item. If you need it for large lists,
profile
against the 'for loop' version and decide if performance is good
enough
for you, for small lists it's a nice and compact solution.
In a more general case, you can't refer to the list/dict/etc
comprehension as it's being constructed, that's just not a design goal
of comprehensions.
Would this help:
w = [1,2,3,1,2,4,4,5,6,1]
s = set(w)
s
set([1, 2, 3, 4, 5, 6])
{x:w.count(x) for x in s}
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
Indeed, this is much better -- I didn't think of it..
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Re: dict comprehension question.
On 12/29/2012 4:40 PM, Mitya Sirenef wrote:
On 12/29/2012 03:15 PM, Joel Goldstick wrote:
Would this help:
w = [1,2,3,1,2,4,4,5,6,1]
s = set(w)
s
set([1, 2, 3, 4, 5, 6])
{x:w.count(x) for x in s}
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
Indeed, this is much better -- I didn't think of it..
It still turns an O(n) problem into an O(k*n) problem, where k is the
number of distinct items.
--
Terry Jan Reedy
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Re: dict comprehension question.
On 12/29/2012 2:48 PM, Quint Rankid wrote:
Given a list like:
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
I would like to be able to do the following as a dict comprehension.
a = {}
for x in w:
a[x] = a.get(x,0) + 1
results in a having the value:
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
Let me paraphrase this: I have nice, clear, straightforward,
*comprehensible* code that I want to turn into an incomprehensible mess
with a 'comprehension. That is the ironic allure of comprehensions.
Comprehensions do not allow for interactions between the source items.
Mitya and Joel worked around this with solutions that do redundant
calculation and multiply the time order.
Reductions do allow for interactions. Doing everything as a reduction
was the fad before comprehensions came along ;-)
from functools import reduce
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
def update(dic, n):
Mutate and return dic (contrary to usual Python policy)
dic[n] = dic.get(n, 0) + 1
return dic
counts = reduce(update, w, {})
print(counts == {1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1})
# prints True
The above is how to rewrite your code in a functional language that does
not have statements and explicit iteration. In Python, I would only
bother to wrap the body of the loop in a function if I needed the same
body in multiple places.
Comprehensions are filtered mappings and that both filter and map can be
written as reduction, so reduction included comprehension. It is more
powerful because it can also do sequential interaction. Indeed, I would
say that it should only be used when there is sequential interaction.
--
Terry Jan Reedy
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Re: dict comprehension question.
On 12/29/12 15:40, Mitya Sirenef wrote:
w = [1,2,3,1,2,4,4,5,6,1]
s = set(w)
s
set([1, 2, 3, 4, 5, 6])
{x:w.count(x) for x in s}
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
Indeed, this is much better -- I didn't think of it..
Except that you're still overwhelmed by iterating over every element
in w for every distinct element. So you've gone from O(N**2) to
O(k*N).
The cleanest way to write it (IMHO) is MRAB's
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
from collections import Counter
results = dict(Counter(w))
which should gather all the statistics in one single pass across w
making it O(N), and it's Pythonically readable.
-tkc
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Re: dict comprehension question.
On Sat, Dec 29, 2012 at 7:26 PM, Tim Chase python.l...@tim.thechases.comwrote:
On 12/29/12 15:40, Mitya Sirenef wrote:
w = [1,2,3,1,2,4,4,5,6,1]
s = set(w)
s
set([1, 2, 3, 4, 5, 6])
{x:w.count(x) for x in s}
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
Indeed, this is much better -- I didn't think of it..
Except that you're still overwhelmed by iterating over every element in
w for every distinct element. So you've gone from O(N**2) to O(k*N).
The cleanest way to write it (IMHO) is MRAB's
w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
from collections import Counter
results = dict(Counter(w))
which should gather all the statistics in one single pass across w
making it O(N), and it's Pythonically readable.
-tkc
I like this too. I haven't learned about collections module yet. Thanks
for the pointer
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