遍历有序字典，就好像它是一个列表一样

Question

I have this ordered dictionary:

numerals = collections.OrderedDict([('I', 1),('V', 5),('X', 10),('L', 50),
                                    ('C', 100),('D', 500),('M', 1000)])

And I want to write a function which creates the rule for numerals, that 'I' can only be subtracted from 'V' and 'X', 'X' can only be subtracted from 'L' and 'C', and 'C' can only be subtracted from 'D' and 'M'. If it were a list, I could make a simple rule that elements with an even-numbered index can be subtracted from the two consecutive elements of the list. However, as dictionaries aren't indexed I'm not sure how I would go about this. Is there an equivalent way of making that rule using a list, but using a dictionary instead?

Answer 1

You could call enumerate on the OrdereDdict to generate indices for each key-value pair in the dictionary:

>>> for index, key in enumerate(numerals):
...    print(index, key, numerals[key])
...
0 I 1
1 V 5
2 X 10
3 L 50
4 C 100
5 D 500
6 M 1000 

You would then use the generated index values to differentiate key-value pairs at even indices from those at odd valued indices.

Answer 2

It's somewhat inefficient, but you could do create a mapping function as shown to lookup random indices on-the-fly. If you're doing to do it a lot, it could be optimized to not recreate the rev_map everytime it was called.

import collections

numerals = collections.OrderedDict([('I', 1),('V', 5),('X', 10),('L', 50),
                                    ('C', 100),('D', 500),('M', 1000)])

def index_to_key(od, index):
    rev_map = {i:k for i,k in enumerate(od.keys())}
    return rev_map[index]

for i in range(len(numerals)):
    key = index_to_key(numerals, i)
    value = numerals[key]
    print('numerals[{}]: key: {!r}, value: {}'.format(i, key, value))

Output:

numerals[0]: key: 'I', value: 1
numerals[1]: key: 'V', value: 5
numerals[2]: key: 'X', value: 10
numerals[3]: key: 'L', value: 50
numerals[4]: key: 'C', value: 100
numerals[5]: key: 'D', value: 500
numerals[6]: key: 'M', value: 1000