如何在Python 3中创建这个复杂的字典？

Question

I am having a heck of a time trying to create a dictionary for a script I am trying to write that handles different IP addresses.

My question is this:

Given d1 and d2 , and assuming d1 and d2 will always have an equal number of key:pair items, how can I create d3 ?

d1 = {1:a, 2:b, 3:b, 4:c, 5:c}

d2 = {one:a, two:b, three:b, four:c, five:c}

d3 = {a:[{1:one}], b:[{2:two},{3:three}], c:[{4:four},{5:five}]}

You can see d3 contains keys that are the same as the values from d1 and d2 , and for each key in d3 , it has a value of a list, in which there are more dictionaries with key:pair values that correspond to the original keys from d1 and d2 .

I have been trying to create d3 for a while now but I just can't seem to reason out how to do it. Your help is greatly appreciated!

Answer 1

First tip before giving out the answer :

reverse_d1 = {}
for k, v in d1.iteritems():
  reverse_d1.set_default(v, []).append(k)

Answer 2

All, thank you for the input. I really appreciate it. Unfortunately I realized that the way I currently posed the question, it is completely invalid. The keys in d2 correspond to subnet masks, which are not all unique, so the way I had it structured is not even a valid dictionary.

After a lot of fussing, I came up with a working solution:

p = 0
sysNames = ['R3','R2','R1']
d1 = OrderedDict([('200.200.200.2','R3'),('200.200.200.1','R2'),('172.172.172.1','R2'),('172.172.172.100','R1'),('192.168.1.151','R1')])

oid1 = '1.3.6.1.2.1.4.20.1.3.200.200.200.2'
oid2 = '1.3.6.1.2.1.4.20.1.3.200.200.200.1'
oid3 = '1.3.6.1.2.1.4.20.1.3.172.172.172.1'
oid4 = '1.3.6.1.2.1.4.20.1.3.172.172.172.100'
oid5 = '1.3.6.1.2.1.4.20.1.3.192.168.1.151'

d2 = {oid1:'255.255.255.0',oid2:'255.255.255.0',oid3:'255.255.255.0',oid4:'255.255.255.0',oid5:'255.255.255.0'}

d1keys = list(d1.keys())
d1values = list(d1.values())
maskOid = '1.3.6.1.2.1.4.20.1.3.'

d3 = {}

for i in range(len(sysNames)):
    ipMaskList = list()
    numInterfaces = d1values.count(sysNames[i])

    for dummy in range(numInterfaces):
        ipMaskList.append({d1keys[p]:d2.get(maskOid+d1keys[p])})
        p += 1

    d3[sysNames[i]] = ipMaskList

As typical, I'm sure this is a horribly inefficient and convoluted way to accomplish my goal. I am absolutely not a great programmer, and am just happy when things run correctly.

Again thank you guys you your help, and if anybody would like to post a more efficient solution, please feel free :)