Hello,
Could any UiPath Veteran help me solve this issue:
I have dictionary (key, value) of double, double. Im looking for a simplest way to count how many keys have value 0 connected with them.
Anyone ?
Kind regards
Krzysztof
Hello,
Could any UiPath Veteran help me solve this issue:
I have dictionary (key, value) of double, double. Im looking for a simplest way to count how many keys have value 0 connected with them.
Anyone ?
Kind regards
Krzysztof
Maybe we need some more inputs. Give a try at
Assign Acitvity
intCount | DataType: int32 =
yourDictVar.Count(Function (x) x.Value.equals(0))
yourDictVar.AsEnumerable().Count(Function (x) x.Value.Equals(0))
Update1 - Statement Correction
As yourDictVar.Count refers to the Count Property, so the LINQ Operator “Count” is not called. With the “AsEnumerable()” part, the usage of the LINQ Operator can be invoked.
Verification:

both solutions give same error: compiler errors encountered processing expression “dict.Count(Function(x) x.Value.equals(0))”.(2) :error BC32016: ‘Public overloads ReadOnly Property Count As Integer’ has no parameters and its return type cannot be indexed.
any ideas
It should be
yourDictVar.Values.Count(Function (x) x.Value = 0)
Stil same error:
compiler errors encountered processing expression “dict.Values.Count(Function(x) x.Value=0)”.(2) :error BC32016: ‘Public overloads ReadOnly Property Count As Integer’ has no parameters and its return type cannot be indexed.
im not sure i writen it correctly: i need to count how many KEYS are paired to value=0
please dont give up on me ![]()
Use this expression
myDictionary.Values.Where(Function(value) value = 0).Count()
Dictionary:
Logic:
Output:
dict.Where(Function(x) cint(x.Value)=0).Count
see if this helps.
regards
SG.
Welcome to the community
you only need to add as enumerable to make it proper
Dict.Values.AsEnumerable().Count(Function(v) v = 0)
cheers
Thank you do much
Mydict.where… worked
Have a great dat !
This topic was automatically closed 3 days after the last reply. New replies are no longer allowed.