Can you wrap a query function with an if function?
My query can produce 3 results, yes yes, yes no, no no. These results occur in two different cells. I want the if function to then say - yes yes - true and either of the other results false. When doing it simply it just gives an error.
Wrapping of formulas
The IF formula has the syntax IF(logical_expression, value_if_true, value_if_false)
Thereby, the logical expression can be a comparison of values, e.g IF(value="Yes", TRUE, FALSE)
value can be in this case either
a fixed value, e.g. a string
a cell reference, e.g. IF(J2="Yes", TRUE, FALSE)
a nested formula, which can a query
In your case, you are interested in the latter, so:
=IF(query(Sheet2!A1:J7,"select G where A = '"&A2&"' and '"&E2&"' = B",0)="Yes", TRUE, FALSE)
I was able to get an answer with using the IFS function =ifs(query(Sheet2!A3:J9,"select G where A = '"&A4&"' and '"&E4&"' = B",0)="yes","TRUE",query(Sheet2!A3:J9,"select G where A = '"&A4&"' and '"&E4&"' = B",0)="no","FALSE")
That was able to solve the issue that the simple if function could not solve
Related
I have a question on using two functions with an if statement in Google Sheets as one complete function. Both variables have to be true, otherwise it returns false. I need one function to check the date 20 months back from today. If said cell is less than today's date 20 months back it's true, naturally. However, for the complete function to return true it also searches for another text value in another cell and has to be an exact match. Both conditions have to be true (the date and the exact match) for the function to be true. So if the date in the cell is less than today's date 20 months back and the text value in the other cell is an exact match, function is true.
Problem is that it seems like the date function does not seem to apply.
=IF(D2<DATE(YEAR(TODAY()),MONTH(TODAY())-20,DAY(TODAY())),AND(REGEXMATCH(M2,"text")),TRUE,FALSE)
You current formula is not set up correctly (nor logically). Given only what you've shown here, this should work:
=IF(AND( D2<DATE(YEAR(TODAY()),MONTH(TODAY())-20,DAY(TODAY())), REGEXMATCH(M2,"text") ),TRUE,FALSE)
Notice that the AND( ) contains both conditions here, whereas your original formula had it only around the second condition.
However, a shorter version of this would be as follows:
=AND( D2<DATE(YEAR(TODAY()),MONTH(TODAY())-20,DAY(TODAY())), REGEXMATCH(M2,"text") )
... since the result of a properly functioning AND( ) is always TRUE or FALSE anyway.
It looks like you're supplying 4 arguments to the IF statement:
=IF(DATECHECK,AND(TEXTCHECK),TRUE,FALSE)
The IF statement expects 3 arguments instead. 1) the condition, 2) the value if true, and 3) the value if false. You can combine your two conditions using an AND statement like this:
AND(DATECHECK,TEXTCHECK)
The final formula would then be:
=IF(AND(D2<DATE(YEAR(TODAY()),MONTH(TODAY())-20,DAY(TODAY())), REGEXMATCH(M2,"text")),TRUE,FALSE)
I want to check if 6**x1 is bigger than 0 for every positive value of x1. I am using sympy.
I've done the following:
x1 = sm.symbols('x_1',nonnegative=True)
u = 6**x1
def checker(func):
if u > 0:
return True
else:
return False
However, I get the error:
TypeError: cannot determine truth value of Relational
I think this is because the function does not know that x1 is positive. But how do I make sure it knows this? Apparently, its not enough with the sm.symbols definition.
Zebraboard
SymPy only let's you compare things that can be computed to a number with literal > (and similar). If you want to make a query on a symbolic expression use expr.is_positive or (expr.is_extended_positive if you want oo to be considered, too).
>>> u.is_positive
True
This can be None or False, too. None is returned when a definitive determination cannot be made, e.g. symbols('x').is_positive is None -> True.
I'm trying to highlight a row if the number of blank cells between say, C1 and E1 = 3
and then copy this down for every row.
I've tried using:
=IF(COUNTBLANK($C1:$E1)=3)
But it's not working, can anybody help?
Under conditional formatting, your formula should be the following based on what you've given. The reason is conditional format is trying to see the result as TRUE or False. The IF statement is trying to tell the computer what to do when it's TRUE or FALSE.
COUNTBLANK($C1:$E1)=3
if you want to use IF you will need to do it like this:
=IF(COUNTBLANK($C1:$E1)=3, 1)
I want to evaluate if a constraint is respected or not in Pyomo when the values of the variables contained in constraint expression are known.
Use case: We know that one particular constraint sometimes makes the problem infeasible, depending on the value of the variable. Instead of sending the problem to the solver to test if the problem is feasible, converting the constraint expression to a boolean type would be enough to determine if the constraint is the culprit.
For the sake of providing a feasible example, here would be the code:
from pyomo.environ import ConcreteModel, Var, Constraint
model = ConcreteModel()
model.X = Var()
def constraint_rule(model):
return model.X <= 1
model.a_constraint = Constraint(rule=constraint_rule)
Now, let's try to work with the expression to evaluate:
# Let's define the expression in this way:
expression = constraint_rule(model)
# Let's show that the expression is what we expected:
print(str(expression))
The previous statement should print X <= 1.0.
Now, the tricky part is how to evaluate the expression.
if expression == True:
print("feasible")
else:
print("infeasible")
creates an TypeError Exception (TypeError: Cannot create an EqualityExpression where one of the sub-expressions is a relational expression: X <= 1.0).
The last example doesn't work because constraint_rule doesn't return a boolean but a Pyomo expression.
Finally, I know that something like
def evaluate_constraint_a_expression(model):
return value(model.X) <= 1
would work, but I can't assume that I will always know the content of my constraint expression, so I need a robust way of evaluating it.
Is there a clever way of achieving this? Like, evaluating the expression as a boolean and evaluating the left hand side and right hand side of the expression at the same time?
The solution is to use value function. Even if it says that it evaluates an expression to a numeric value, it also converts the expression to a boolean value if it is an equality/inequality expression, like the rule of a constraint.
Let's suppose that the model is defined the way it is in the question, then the rest of the code should be:
from pyomo.environ import value
if value(expression) == True:
print("feasible")
else:
print("infeasible")
where expression is defined as written in the question.
However, be advised that numeric precision in Python using this method can be different than the one provided by the solver. Therefore, it is possible that this method will show that a constraint is infeasible while it is just a matter of numeric imprecision of under 1e-10. So, while it is useful in finding if most constraints are feasible, it also generates some false positives.
disclaimer: My title may not be accurate as far as what I would like to accomplish, but I can update if someone can correct my terminology
I have 2 functions, each with a separate purpose and usable on its own, but occasionally I would like to combine the two to perform both actions at once and return a single result, and to do this I would like to assign to a variable name
I know I can create a 3rd function that does basically what I want as it is really simple.. though it's become a bit of a challenge to myself to find a way of doing this
def str2bool(string):
return string.lower() in ("yes", "true", "t", "1")
def get_setting(string):
if string == 'cat':
return 'yes'
else:
return 'no'
VALID_BOOL = str2bool(get_setting)
print VALID_BOOL('cat')
So basically I would like to assign the combination of the 2 functions to a variable that I can call and pass in the string parameter to evaluate
In my real world code, get_setting() would retrieve a user setting and return the value, I would then like to test that value and return it as a boolean
Again I know I can just create a 3rd function that would get the value and do the quick test.. but this is more for learning to see if it can be done as I'm trying to do.. and so far my different variations of assigning and calling aren't working, is it even possible or would it turn too complex?
Using lambda is easy, but i don't know if it is exactly what you are looking for.
Example:
f = lambda astring : str2bool(get_setting(astring))
Outputs:
>>> f('cat')
True