dict1=({"EMP$$1":1,"EMP$$2":2,"EMP$$3":3})
How to check if EMP exists in the dictionary using python
dict1.get("EMP##") ??
dict1=({"EMP$$1":1,"EMP$$2":2,"EMP$$3":3})
How to check if EMP exists in the dictionary using python
dict1.get("EMP##") ??
There's no way to match dictionary keys like this. I suggest you rethink your data structure for this problem. If this has to be extra quick you could use something like a suffix tree.
It's not entirely clear what you want to do.
You can loop through the keys in the dict
selecting keys using the startswith()
method:
>>> for key in dict1:
... if key.startswith("EMP$$"):
... print "Found",key
...
Found EMP$$1
Found EMP$$2
Found EMP$$3
You can use a list comprehension to get all the values that match:
>>> [value for key,value in dict1.items() if key.startswith("EMP$$")]
[1, 2, 3]
If you just want to know if a key matches you could use the any()
function:
>>> any(key.startswith("EMP$$") for key in dict1)
True
This approach strikes me as contrary to the intent of a dictionary.
A dictionary is made up of hash keys which have had values associated with them. The benefit of this structure is that it provides very fast lookups (on the order of O(1)). By searching through the keys, you're negating that benefit.
I would suggest reorganizing your dictionary.
dict1 = {"EMP$$": {"1": 1}, {"2", 2}, {"3": 3} }
Then, finding "EMP$$" is as simple as
if "EMP$$" in dict1:
#etc...
You need to be a lot more specific with what you want to do. However, assuming the dictionary you gave:
dict1={"EMP$$1":1, "EMP$$2":2, "EMP$$3":3}
If you wanted to know if a specific key was present before trying to request it you could:
dict1.has_key('EMP$$1')
True
Returns True
as dict1
has the a key EMP$$1
.
You could also forget about checking for keys and rely on the default return value of dict1.get()
:
dict1.get('EMP$$5',0)
0
Returns 0
as default given dict1
doesn't have a key EMP$$5
.
In a similar way you could also use a `try/except/ structure to catch and handle missed keys:
try:
dict1['EMP$$5']
except KeyError, e:
# Code to deal w key error
print 'Trapped key error in dict1 looking for %s' % e
The other answers to this question are also great, but we need more info to be more precise.