Preferred Python idiom for handling non-existing dictionary keys and why?

[Quentin Crain]

Hello again All!

(First, I would like to mention I did try to google
for the answer here!)

Say I am populating a dictionary with a list and
appending. I have written it thusly:

d={}
for (k,v) in somedata():
try:
d[k].append(v)
except KeyError:
d[k]=[v]

I could have written:

d={}
for (k,v) in somedata():
if (k in d):
d[k].append(v)
else:
d[k]=[v]
Which is perferred and why? Which is "faster"?

Thanks!!

Quentin
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[John J. Lee]

Quentin Crain <cz***@yahoo.com> writes:

(First, I would like to mention I did try to google
for the answer here!)

Say I am populating a dictionary with a list and
appending. I have written it thusly:

[...]

You want the dict.setdefault method.
John

[Terry Reedy]

"Quentin Crain" <cz***@yahoo.com> wrote in message
news:ma********************************@python.org ...

Hello again All!

(First, I would like to mention I did try to google
for the answer here!)
The question you asked is frequent, but hard to isolate, and has
special-case answer (to question you did not ask) even harder to find.
Say I am populating a dictionary with a list and
appending. I have written it thusly:

d={}
for (k,v) in somedata():
try:
d[k].append(v)
except KeyError:
d[k]=[v]

I could have written:

d={}
for (k,v) in somedata():
if (k in d):
d[k].append(v)
else:
d[k]=[v]
Which is perferred and why?
Neither. For me, both are superceded by

d={}
for (k,v) in somedata():
d[k] = d.get(k, []).append(v)

Read Library Reference 2.2.7 Mapping Types to learn current dict
methods.
Which is "faster"?

Tradeoff is small extra overhead every loop (the condition) versus
'occasional' big overhead (exception catching). Choice depends on
frequency of exceptions. As I remember, one data-based rule of thumb
from years ago is to use conditional if frequency more that 10%. You
could try new timeit() on all three versions.

Terry J. Reedy

[Skip Montanaro]

Say I am populating a dictionary with a list and appending. I have
written it thusly:

John> You want the dict.setdefault method.

d.setdefault() never made any sense to me (IOW, to use it I always had to
look it up). The semantics of what it does just never stick in my brain.
Consequently, even though it's less efficient I generally write such loops
like this:

d = {}
for (key, val) in some_items:
lst = d.get(key) or []
lst.append(val)
d[key] = lst

Note that the first statement of the loop is correct (though perhaps not
obvious at first glance), since once initialized, d[key] never tests as
False. FYI, timeit tells the performace tale:

% timeit.py -s 'd={}' 'x = d.setdefault("x", [])'
1000000 loops, best of 3: 1.82 usec per loop
% timeit.py -s 'd={}' 'x = d.get("x") or [] ; d["x"] = x'
100000 loops, best of 3: 2.34 usec per loop

But my way isn't bad enough for me to change. ;-)

Skip

[Gerrit Holl]

Skip Montanaro wrote:

>> Say I am populating a dictionary with a list and appending. I have
>> written it thusly:

John> You want the dict.setdefault method.

d.setdefault() never made any sense to me (IOW, to use it I always had to
look it up).

I have had a wrong idea about setdefault for a very long time. To me, it
sounds like: "set this value the default value of dict, so after this
call, let each non-existing key result in this value".

Gerrit.

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[Dave Benjamin]

In article <jO********************@comcast.com>, Terry Reedy wrote:

"Quentin Crain" <cz***@yahoo.com> wrote in message
news:ma********************************@python.org ...

Which is perferred and why?

Neither. For me, both are superceded by

d={}
for (k,v) in somedata():
d[k] = d.get(k, []).append(v)

Not quite... append returns None, so you'll need to write that as two
separate statements, ie.:

d[k] = d.get(k, [])
d[k].append(v)

Or, just:

d.setdefault(k, []).append(v)

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: d r i n k i n g l i f e o u t o f t h e c o n t a i n e r :

[Peter Otten]

Terry Reedy wrote:

Neither. For me, both are superceded by

d={}
for (k,v) in somedata():
d[k] = d.get(k, []).append(v)

This would have to look up the key twice, so it has no advantage over

if k in d:
d[k].append(v)
else:
d[k] = [v]

Anyway, list.append() returns always None, so it does not work.
I think you mean

d = {}
for (k, v) in somedata:
d.setdefault(k, []).append(v)

There is a small overhead for throwing away a new list object if the key is
already in the dictionary, but I don't really care.
(Or is the compiler smart enough to reuse the empty list?)

Peter

[Alex Martelli]

Skip Montanaro wrote:
...

% timeit.py -s 'd={}' 'x = d.setdefault("x", [])'
1000000 loops, best of 3: 1.82 usec per loop
% timeit.py -s 'd={}' 'x = d.get("x") or [] ; d["x"] = x'
100000 loops, best of 3: 2.34 usec per loop

But my way isn't bad enough for me to change. ;-)

Actually, you can still do a bit better w/o using setdefault:

[alex@lancelot pop]$ timeit.py -s'd={}' 'x=d.setdefault("x",[])'
1000000 loops, best of 3: 0.925 usec per loop
[alex@lancelot pop]$ timeit.py -s'd={}' 'x=d.get("x") or []; d["x"]=x'
1000000 loops, best of 3: 1.21 usec per loop
[alex@lancelot pop]$ timeit.py -s'd={}' 'x=d.get("x",[]); d["x"]=x'
1000000 loops, best of 3: 1.13 usec per loop

as d.get takes a second optional argument, you can still save the 'or'.
Alex

[Terry Reedy]

"Terry Reedy" <tj*****@udel.edu> wrote in message
news:jO********************@comcast.com...

Read Library Reference 2.2.7 Mapping Types to learn current dict
methods.

Seeing the other responses, I see I need to do the same and read about
setdefault() ;-)