How to test if a key in a dictionary exists?

[Frank]

Hi,

does anyone know how one can test if, e.g., a dictionary 'name' has a
key called 'name_key'?

This would be possible:

keys_of_names = names.keys()
L = len(keys_of_names)

for i in range(L):
if keys_of_names[i] == name_key:
print 'found'
But certainly not efficient. I would expect there is something like:

name.is_key(name_key)

I appreciate your help!

Frank

[Irmen de Jong]

Frank wrote:

Hi,

does anyone know how one can test if, e.g., a dictionary 'name' has a
key called 'name_key'?

name_key in name

e.g.

>>name={"john": 42}
"john" in name

True

>>"julie" in name

False

--Irmen

[Rune Strand]

Yes, you have name.has_key(name_key) and perhaps better, the in
operator:

if name_key in name:
do something

[Jeff McNeil]

Sure, you can use "if key in dict" to test for membership:
Python 2.3.5 (#1, Jan 13 2006, 20:13:11)
[GCC 4.0.1 (Apple Computer, Inc. build 5250)] on darwin
Type "help", "copyright", "credits" or "license" for more information.

>>exampledict = {"a" : 1, "b" : 2}
"a" in exampledict

True

>>"q" in exampledict

False

>>>

Alternatively, "exampledict.has_key("a")" is also valid.

See:

http://docs.python.org/tut/node7.htm...00000000000000

-Jeff

On 10 Mar 2007 13:17:11 -0800, Frank <su******@gmail.comwrote:

Hi,

does anyone know how one can test if, e.g., a dictionary 'name' has a
key called 'name_key'?

This would be possible:

keys_of_names = names.keys()
L = len(keys_of_names)

for i in range(L):
if keys_of_names[i] == name_key:
print 'found'
But certainly not efficient. I would expect there is something like:

name.is_key(name_key)

I appreciate your help!

Frank

--
http://mail.python.org/mailman/listinfo/python-list

[Bjoern Schliessmann]

Frank wrote:

does anyone know how one can test if, e.g., a dictionary 'name'
has a key called 'name_key'?

Yes. It's already posted; next time have a look at the concise
library reference:

http://docs.python.org/lib/typesmapping.html

This would be possible:

keys_of_names = names.keys()
L = len(keys_of_names)

for i in range(L):
if keys_of_names[i] == name_key:
print 'found'

Coming in from Java? 8)

Iterating through something using len(X) is quite rare in Python.

If we had no simple "key in dict" test, I'd write the above as:

for key in names.keys():
if key == name_key:
print "found"
break
else:
print "not found"

(It would even suffice to write "for key in names".)

Regards,
Björn

--
BOFH excuse #331:

those damn raccoons!

[mensanator]

On Mar 10, 3:17?pm, "Frank" <[e-mail]:

Hi,

does anyone know how one can test if, e.g., a dictionary 'name' has a
key called 'name_key'?

This would be possible:

keys_of_names = names.keys()
L = len(keys_of_names)

for i in range(L):
if keys_of_names[i] == name_key:
print 'found'

But certainly not efficient. I would expect there is something like:

name.is_key(name_key)

I appreciate your help!

I often build histograms using dictionaries.

if hist.has_key(outcome):
hist[outcome] += 1 # if key already exists, increment
else:
hist[outcome] = 1 # else create a new key

>
Frank

[Paul Rubin]

if hist.has_key(outcome):

hist[outcome] += 1 # if key already exists, increment
else:
hist[outcome] = 1 # else create a new key

You could write that:

hist[outcome] = 1 + hist.get(outcome, 0)

Or with Python 2.5 you could use

hist = defaultdict(int)
...
hist[outcome] += 1 # automatically initializes to 0 if outcome not found

[Paul McGuire]

On Mar 10, 9:12 pm, Paul Rubin <http://[e-mail]:

if hist.has_key(outcome):
hist[outcome] += 1 # if key already exists, increment
else:
hist[outcome] = 1 # else create a new key

You could write that:

hist[outcome] = 1 + hist.get(outcome, 0)

Or with Python 2.5 you could use

hist = defaultdict(int)
...
hist[outcome] += 1 # automatically initializes to 0 if outcome not found

Often when building a histogram, one knows in advance what the keys
will be. For instance, when working with data from 0 to 100 and
looking for frequencies by decade, you can initialize the histogram-
dict with:

for i in range(10):
histodict[i] = 0

and then just update the appropriate bucket without having to do any
testing at all:

fibSeqUpTo100 = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
for val in fibSeqUpTo100:
key = val/10
histodict[ key ] += 1

This has the added benefit of including entries for the empty buckets
as well.

-- Paul

[Alex Martelli]

Paul McGuire <pt***@austin.rr.comwrote:

will be. For instance, when working with data from 0 to 100 and
looking for frequencies by decade, you can initialize the histogram-
dict with:

for i in range(10):
histodict[i] = 0

A better way, of course (also saving the histodict = {} which you didn't
mention but surely intended) would be:

histodict = dict.fromkeys(range(10), 0)

In Python, in general, the more conceptually compact, direct, simple and
higher-abstraction approach tends to be faster as well, BTW.
Alex

[John Machin]

On Mar 12, 3:19 pm, [e-mail] (Alex Martelli) wrote:

Paul McGuire <[e-mail]:

will be. For instance, when working with data from 0 to 100 and
looking for frequencies by decade, you can initialize the histogram-
dict with:

for i in range(10):
histodict[i] = 0

A better way, of course (also saving the histodict = {} which you didn't
mention but surely intended) would be:

histodict = dict.fromkeys(range(10), 0)

In Python, in general, the more conceptually compact, direct, simple and
higher-abstraction approach tends to be faster as well, BTW.

That's so true, especially if it's matched to the problem space, isn't
too new, and doesn't need looking up in the docs :-)

Python 2.1.3 (#35, Apr 8 2002, 17:47:50) [MSC 32 bit (Intel)] on
win32
Type "copyright", "credits" or "license" for more information.

>>histodict = dict.fromkeys(range(10), 0)

Traceback (most recent call last):
File "<stdin>", line 1, in ?
NameError: name 'dict' is not defined

>>histolist = [0] * 10
histolist[3] += 1
histolist

[0, 0, 0, 1, 0, 0, 0, 0, 0, 0]

>>>

Cheers,
John

[Larry Bates]

Frank wrote:

Hi,

does anyone know how one can test if, e.g., a dictionary 'name' has a
key called 'name_key'?

This would be possible:

keys_of_names = names.keys()
L = len(keys_of_names)

for i in range(L):
if keys_of_names[i] == name_key:
print 'found'
But certainly not efficient. I would expect there is something like:

name.is_key(name_key)

I appreciate your help!

Frank

Others have posted the direct answer but often you are wanting to
do something with the key if it exists and the best method is:

try: names['name_key']+=1
except KeyError:
#
# Key doesn't exist, create it, display error, or whatever you want
# to do if it doesn't exist.
#

-Larry

[mensanator]

On Mar 11, 11:49 pm, "John Machin" <[e-mail]:

On Mar 12, 3:19 pm, [e-mail] (Alex Martelli) wrote:

Paul McGuire <[e-mail]:

will be. For instance, when working with data from 0 to 100 and
looking for frequencies by decade, you can initialize the histogram-
dict with:

for i in range(10):
histodict[i] = 0

A better way, of course (also saving the histodict = {} which you didn't
mention but surely intended) would be:

histodict = dict.fromkeys(range(10), 0)

In Python, in general, the more conceptually compact, direct, simple and
higher-abstraction approach tends to be faster as well, BTW.

That's so true, especially if it's matched to the problem space, isn't
too new, and doesn't need looking up in the docs :-)

Python 2.1.3 (#35, Apr 8 2002, 17:47:50) [MSC 32 bit (Intel)] on
win32
Type "copyright", "credits" or "license" for more information.>>histodict = dict.fromkeys(range(10), 0)

Traceback (most recent call last):
File "<stdin>", line 1, in ?
NameError: name 'dict' is not defined>>histolist = [0] * 10

>histolist[3] += 1
histolist

[0, 0, 0, 1, 0, 0, 0, 0, 0, 0]

In defense of my original if:else: idiom, I'll point
out that there may be more to be done when the key is
absent than simply creating a new one and initializing it.

Also, a dictionary doesn't need an integer index, so I can
create bins such as '2.01','2.02',etc.

And in some cases, it's not known what the range of the
keys will be making it difficult to pre-allocate the list.

And yes, the empty buckets will need to be created for
graphing purposes, but that can be done after the test
has completed so that the only empty buckets that need
be created are those between the outliers.

>
Cheers,
John