<SYSTEM_TASK:>
Solve the following problem using Python, implementing the functions described below, one line at a time
<END_TASK>
<USER_TASK:>
Description:
 def apply_conditions(self, value):
 """Applies any boundary conditions to the given value. The value is manipulated according based on the following conditions: * If `self.cyclic` is True then `value` is wrapped around to the minimum (maximum) bound if `value` is `>= self.max` (`< self.min`) bound. For example, if the minimum and maximum bounds are `0, 2*pi` and `value = 5*pi`, then the returned value will be `pi`. * If `self.min` is a reflected boundary then `value` will be reflected to the right if it is `< self.min`. For example, if `self.min = 10` and `value = 3`, then the returned value will be 17. * If `self.max` is a reflected boundary then `value` will be reflected to the left if it is `> self.max`. For example, if `self.max = 20` and `value = 27`, then the returned value will be 13. * If `self.min` and `self.max` are both reflected boundaries, then `value` will be reflected between the two boundaries until it falls within the bounds. The first reflection occurs off of the maximum boundary. For example, if `self.min = 10`, `self.max = 20`, and `value = 42`, the returned value will be 18 ( the first reflection yields -2, the second 22, and the last 18). * If neither bounds are reflected and cyclic is False, then the value is just returned as-is. Parameters value : float The value to apply the conditions to. Returns ------- float The value after the conditions are applied; see above for details. """

        retval = value
        if self._cyclic:
            retval = apply_cyclic(value, self)
        retval = self._reflect(retval)
        if isinstance(retval, numpy.ndarray) and retval.size == 1:
            try:
                retval = retval[0]
            except IndexError:
                retval = float(retval)
        return retval