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I have an python object list. for example "all_names" I can retrieve values from that list in this way:

for a in all_names:
    print a.name,a.age

My target is to quickly find out if a certain name exist in the list all_names.

so I'm doing it in this way.

all = []
for a in all_names:
    all.append(a.name)
if "any_name" in all:
    print "Name exit"
else:
    print "Not found"

But if size of "all_names" is very huge (ex. if len(all_names) > 100,000), then I think It will not be an efficient way to do the same.

So wanted to know, if is there any more efficient way than this??

Thanks in Adv.

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3
  • Is storing all of the names in a global list as you're creating the objects out of the picture? Commented Aug 18, 2013 at 17:47
  • all_names is a really bad variable name for a list of things which are not names Commented Aug 18, 2013 at 18:05
  • 1
    although all is an even worse variable name, since it shadows the builtin all Commented Aug 18, 2013 at 18:09

4 Answers 4

Reset to default
1

Try:

if any(i.name == "any_name" for i in all_names):
    print "Name exists"
else:
    print "Not found"

If you need to find a name only once, this is the way. If you need to find more names - use set:

names = set(i.name for i in all_names)
if "any_name" in names:
    print "Name exists"
else:
    print "Not found"
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5 Comments

@AnttiHaapala, yes set is faster, but take into account time and memory for its construction.
@AnttiHaapala: I beleive the original was actually O(N^2), due to the dynamic reallocation of the list. This allocates constant memory.
No, the dynamic reallocation of the list is O(N). Anyway I removed my comment since you made a set version.
@AnttiHaapala: Then why is string concatenation (instead of ''.join) often described as having quadratic complexity?
Yes, string concatenation has quadratic complexity, and so does tuple concatenation, and list concatenation (a + b), but list append + extend has linear complexity.
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create all_name as
all_name={a.name for a in all_names}
rest everything is fine in your code.. .


Note:don't use reserved built-ins(in your case all) names

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3 Comments

Please don't use the builtin all as a variable name
thanks Eric for pointing out that. . i have just forgotten that user has used that keyword. .
all is not a keyword. I hadn't noticed that the OP also made this mistake
0

Use a set.

all_set = set(a.name for a in all_names)

if "any_name" in all_set:
    print "Name exists"
else:
    print "Not found"

The set works like a dictionary, with O(1) access time. Of course to appreciate the speed of the set, you need to reuse it (ask for several names on the same set).

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Comments

0

What about this:

for i in all_names:
    if i.name == "any_name":
        print("Name found!")
        break
else:
    # The loop wasn't broken out of
    print("Name not found")

Alternatively:

if any(i.name == "any_name" for i in all_names):
    print("Name found!")
else:
    print("Name not found")
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2 Comments

this is only 50 % faster than his original approach
Using a set would require much more memory, although I'd agree it would be a better approach if it was reused.

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