maintaining list order

[perfringo]

Now my clumsy brain has processed it and as a result I refactored your original code a little bit:

comparelist = [0, 3000, 6000] 
elements = [[3], [4, 400], [3,4222]] 
result = []

for element in elements: 
    result.append([])                            # add sublist for every list in elements
        for num in element:                    
            for value in reversed(comparelist):  
                if value < num: 
                    result[-1].append(value)    # append first match and break
                    break 

I assume that 'comparelist' is always sorted and starting from end is required to return closest match (in last case 3000 instead of 0)

However, there is some ambiguity: what should be returned when condition fully or partially not met? (for example comparelist = [4500, 5000, 6000])