functional programming - python equivalent of quote in lisp -

in python equivalent of quote operator? finding need delay evaluation. example, suppose in following lisp psuedocode have:

a = '(func, 'g) g = (eval a) 

what doing deferring evaluation of g till later time. necessary because want define g later. equivalent idea of psuedocode in python?

a = lambda: func(g) g = a() 

this isn't quite literal translation - literal translation use string , eval - it's best fit. quoting isn't wanted in lisp anyway; wanted delay or create lambda. note func , g closure variables in lambda function, rather symbols, if call a environment different bindings func or g, it'll still use variables a's environment of definition.