current community

your communities

more stack exchange communities

Stack Exchange
tour help
stack overflow careers
Take the 2-minute tour ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free, no registration required.
up vote 431 down vote favorite
85

I wanted to test if a key exists in a dictionary before updating the value for the key. I wrote the following code:

if 'key1' in dict.keys():
  print "blah"
else:
  print "boo"

I think this is not the best way to accomplish this task. Is there a better way to test for a key in the dictionary?

share|improve this question
3  
Calling dict.keys() creates a list of keys, according to the documentation docs.python.org/2/library/stdtypes.html#dict.keys but I'd be surprised if this pattern wasn't optimised for, in a serious implementation, to translate to if 'key1' in dict:. –  Evgeni Sergeev Aug 12 '13 at 8:51
    
So I finally found out why many of my Python scripts were so slow :) :(. That's because I've been using x in dict.keys() to check for keys. And that happened because the usual way to iterate over keys in Java is for (Type k : dict.keySet()), this habit causing for k in dict.keys() to feel more natural than for k in dict (which should still be fine in terms of performance?), but then checking keys becomes if k in dict.keys() too, which is a problem... –  Evgeni Sergeev Aug 12 '13 at 8:58
    
@EvgeniSergeev if k in dict_: tests for presence of k in the KEYS of dict_, so you still don't need dict_.keys(). (This has bit me, as it reads to me like its testing for a value in dict. But it isn't.) –  ToolmakerSteve Dec 16 '13 at 23:34
    
@ToolmakerSteve That's right, but not only do you not need it, it's not a good practice. –  Evgeni Sergeev Dec 17 '13 at 1:51
1  
Try "key in dict" –  marcelosalloum Jun 26 at 17:18

8 Answers 8

up vote 473 down vote accepted

in is the intended way to test for the existence of a key in a dict.

d = dict()

for i in xrange(100):
    key = i % 10
    if key in d:
        d[key] += 1
    else:
        d[key] = 1

If you wanted a default, you can always use dict.get():

d = dict()

for i in xrange(100):
    key = i % 10
    d[key] = d.get(key, 0) + 1

... and if you wanted to always ensure a default value for any key you can use defaultdict from the collections module, like so:

from collections import defaultdict

d = defaultdict(lambda: 0)

for i in xrange(100):
    d[i % 10] += 1

... but in general, the in keyword is the best way to do it.

share|improve this answer
12  
I usually just use get if I'm going to be pulling the item out of the dictionary anyway. No sense in using in and pulling the item out of the dictionary. –  Jason Baker Oct 21 '09 at 19:12
5  
I fully agree. But if you only need to know if a key exists, or you need to distinguish between a case where the key is defined and a case where you are using a default, in is the best way of doing it. –  Chris B. Oct 21 '09 at 19:16
1  
Thanks guys... great feedback. I am understanding the subtleties of the language better with each of your replies. –  Mohan Gulati Oct 21 '09 at 19:19
up vote 269 down vote

You don't have to call keys:

if 'key1' in dict:
  print "blah"
else:
  print "boo"

That will be much faster as it uses the dictionary's hashing as opposed to doing a linear search, which calling keys would do.

share|improve this answer
2  
That is great. I was under the impression that it would internally still traverse the list of keys, but I see this works more like testing membership in a set. –  Mohan Gulati Oct 21 '09 at 19:20
9  
@Mohan Gulati: You understand that a dictionary is a hashtable of keys mapped to values, right? A hashing algorithm converts the key to an integer and the integer is used to find a location in the hash table that matches. en.wikipedia.org/wiki/Hash_table –  hughdbrown Oct 22 '09 at 2:31
up vote 68 down vote

There are two different ways to do it:

  1. key in dict
  2. dict.has_key(key)

You might also want to check out the Python Quick Reference.

EDIT:

As noted, "has_key" is deprecated in Python 3.0+.

share|improve this answer
20  
dict.has_key(key) has been deprecated in favor of key in dict –  David Locke Oct 21 '09 at 19:28
    
@David, thanks... I haven't really looked at 3.0, yet. –  Michael Aaron Safyan Oct 21 '09 at 20:04
1  
Technically, has_key is deprecated for Python 2.x+ (not merely for 3.0+). That is, new code is recommended to not use it, even when writing in Python 2.x. (Because it is a feature known to be going away in future versions, and there is a perfectly good substitute to use instead.) What happens in 3.0 is that it is removed completely. –  ToolmakerSteve Dec 16 '13 at 23:19
up vote 20 down vote

I would recommend using the setdefault method instead. It sounds like it will do everything you want.

>>> d = {'foo':'bar'}
>>> q = d.setdefault('foo','baz') #Do not override the existing key
>>> print q #The value takes what was originally in the dictionary
bar
>>> print d
{'foo': 'bar'}
>>> r = d.setdefault('baz',18) #baz was never in the dictionary
>>> print r #Now r has the value supplied above
18
>>> print d #The dictionary's been updated
{'foo': 'bar', 'baz': 18}
share|improve this answer
3  
What does setdefault have to do with the OP's question? –  hughdbrown Oct 21 '09 at 19:11
8  
@hughdbrown "I wanted to test if a key exists in a dictionary before updating the value for the key." Sometimes posts include code that generate a flurry of responses to something that's not quite the original goal. To accomplish the goal stated in the first sentence, setdefault is the most effective method, even though it's not a drop-in replacement for the sample code posted. –  David Berger Oct 21 '09 at 19:14
2  
This is the superior answer because it satisfies OP's goal instead of just giving the technically correct answer. See: nedbatchelder.com/blog/201207/… –  Niels Bom Jul 24 '12 at 13:07
    
+1 for an informative answer, that taught me something. However, whether it is the best solution depends on what the coder has in mind; e.g. the meaning of "before updating the value of the key". Maybe he's going to throw an exception if it is not present (== no permission to add new keys). Maybe its a dictionary of counts, and he's going to add 1 to the existing count, in which case `d[key] = d.get(key, 0) + 1' is the cleanest solution (as Chris shows, after your answer was written). (I only bother mentioning this, in case future readers come here, with different tasks in mind.) –  ToolmakerSteve Dec 16 '13 at 23:42
    
@NielsBom ... IMHO setdefault is only the superior solution when an existing entry should not be overwritten. (An important case, but not the only reason to be testing existence of a key.) –  ToolmakerSteve Dec 16 '13 at 23:46
up vote 19 down vote

You can shorten this:

if 'key1' in dict:
    ...

However, this is at best a cosmetic improvement. Why do you believe this is not the best way?

share|improve this answer
35  
This is much more than a cosmetic improvement. The time to find a key using this method is O(1) whereas calling keys would generate a list and be O(n). –  Jason Baker Oct 21 '09 at 19:08
2  
Good point. +1 to your answer. –  Greg Hewgill Oct 21 '09 at 19:09
up vote 4 down vote

Just an FYI adding to Chris. B (best answer):

d = defaultdict(int)

Works as well; the reason is that calling int() returns 0 which is what defaultdict does behind the scenes (when constructing a dictionary), hence the name "Factory Function" in the documentation.

share|improve this answer
    
+1: Just what I needed to know, for creating a dictionary of counts. –  ToolmakerSteve Dec 16 '13 at 23:56
    
(I gave you the +1, because Chris' defaultdict(lambda: 0) seemed obscure to me. saying "this is a dictionary of int's, so they start with the default value of an int, e.g. int() e.g. 0" I like.) –  ToolmakerSteve Dec 17 '13 at 0:03
    
If you're creating a dictionary of counts, you should be using Counter (assuming Python 2.7). And I used defaultdict(lambda: 0) instead of defaultdict(int) because I think it's clearer what's going on; the reader doesn't need to know you get 0 if you call int() without arguments. YMMV. –  Chris B. Feb 3 at 19:35
up vote 3 down vote

What about using EAFP (easier to ask forgiveness than permission):

try:
   blah = dict["mykey"]
   # key exists in dict
except:
   # key doesn't exist in dict

See other SO posts:

Using try vs if in python or

Checking for member existence in Python

share|improve this answer
1  
Try/except may be more expensive if it's likely that the key often doesn't exist. From the post you referenced: "[I]f you expect that 99 % of the time result will actually contain something iterable, I'd use the try/except approach. It will be faster if exceptions really are exceptional. If result is None more than 50 % of the time, then using if is probably better.[...][A]n if statement always costs you, it's nearly free to set up a try/except block. But when an Exception actually occurs, the cost is much higher." stackoverflow.com/a/1835844/1094092 –  billrichards Aug 19 at 20:07
up vote 1 down vote

you can use the has_key() method:

if dict.has_key('xyz')==1:
    #update the value for the key
else:
    pass
share|improve this answer
3  
.has_key() has been deprecated; you should use in as shown in other answers. –  Brad Koch May 11 '13 at 16:50
2  
BTW, I recommend reading ALL existing answers to an OLD question, before answering it. This answer added nothing, since the suggestion already existed in Michael's answer, from '09. (I don't mean to discourage an attempt to add something useful to a discussion. Keep trying.) –  ToolmakerSteve Dec 16 '13 at 23:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.