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Lets say I had a column of numbers (5 digits)
71053
71187
12378
72678
My required function was
f(7,1,’x’,”x”,”x”) =710537 and 71187
f(“x”,2,”x”,”x”,”x”)=12378 and 72678
f(“x”,x,”x”,”x”,”8”)=12378 and 72678
f(“x”,x,”6”,”x”,”8”)= 72678
etc
What would be the best approach ?
Solved! Go to Solution.
Step 1 Separate data for each digit.
Step 2 Check the condition.
Here is different approach.
The filter mask is provided as one single string, so there is no limit on number of digits in the data numbers or size of filter mask. To keep it all self-contained I found it better to provide the data vector as a second argument to the function.
The function assumes leading zeros for the numbers and treats any non-numeric character in the mask string as a wildcard (meaning "any digit").
P6 worksheet attached