How can I use a variable as a comparator in python?

Question

I have some KPIs and for some, beyond the target value is good while for others it's bad.

Can I do something like

comparator = '<'
value = 100
target = 200

and then go on to say

value comparator target

So that python sees this as 100 < 200 and returns True?

For context, I have a table of KPIs which follow the format:

KPI1: < 100 On Target, > 110 Action Required

KPI2: > 50 On Target, <

and I plan to loop through them and through their associated data to apply RAG ratings.

The `operator` module provides function versions of each comparison operator. — chepner, Jun 27 '19 at 16:33

Michael Bianconi · Answer 1 · 2019-06-27T16:35:58.647

0

You can use first-class methods. This allows you to skip importing the operator module, and is safer than using eval():


def lt(a, b): return a < b
def gt(a, b): return a > b
def eq(a, b): return a == b

comparator = lt
print(comparator(4, 5))  # >>> True

edited Jun 27 '19 at 16:35

answered Jun 27 '19 at 16:34

Michael Bianconi

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These functions are already predefined in the `operator` module. – chepner Jun 27 '19 at 16:35
If he wanted to use custom comparators, this would be the way to go. – Michael Bianconi Jun 27 '19 at 16:36

score -2 · Answer 2 · answered Jun 27 '19 at 16:34

-2

This can work (using eval)

comparator_1  = '>'
x = 7
y = 12
print(eval('{} {} {}'.format(x,comparator_1,y)))
comparator_2  = '<'
print(eval('{} {} {}'.format(x,comparator_2,y)))

output

False
True

answered Jun 27 '19 at 16:34

balderman

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There's no need for `eval` here. – chepner Jun 27 '19 at 16:35
@chepner It is an option (But I agree it is not the best one..) – balderman Jun 27 '19 at 16:37
Just because something is *an* option doesn't mean it's a *good* one or should be recommended. – chepner Jun 27 '19 at 16:38

How can I use a variable as a comparator in python?

2 Answers2