Python 2.7 : List to Dict

Question

I have a list like this :

['Wi-Fi', '10.0.10.22', '255.255.0.0', '34-F3-9A-4C-D0-C4', 'VMware Network Adapter VMnet1', '192.168.10.111', '255.255.255.0', '00-50-56-C0-00-01', 'Loopback Pseudo-Interface 1', '127.0.0.1', '255.0.0.0', '00-00-00-00-00-00-00-E0']

How can i convert it to dictionary like below :

{'1' : ['Wi-Fi', '10.0.10.22', '255.255.0.0', '34-F3-9A-4C-D0-C4'], '2' : ['VMware Network Adapter VMnet1', '192.168.10.111', '255.255.255.0', '00-50-56-C0-00-01'], '3' : ['Loopback Pseudo-Interface 1', '127.0.0.1', '255.0.0.0', '00-00-00-00-00-00-00-E0']}

Many thanks, Quang

@AntonvBR I guess every four items get assigned to a new key — Ma0, Apr 18 '18 at 11:30
Guessing is not knowing! @Quang: what have you tried so far? Also, you don't have a string, but a list.... — hellow, Apr 18 '18 at 11:32

score 1 · Answer 1 · answered Apr 18 '18 at 11:32

If you want to split by 4 use this function for instance: How do you split a list into evenly sized chunks?

l = ['Wi-Fi', '10.0.10.22', '255.255.0.0', '34-F3-9A-4C-D0-C4', 'VMware Network Adapter VMnet1', '192.168.10.111', '255.255.255.0', '00-50-56-C0-00-01', 'Loopback Pseudo-Interface 1', '127.0.0.1', '255.0.0.0', '00-00-00-00-00-00-00-E0']

def chunks(l, n):
    """Yield successive n-sized chunks from l."""
    for i in range(0, len(l), n):
        yield l[i:i + n]

d = {ind:i for ind, i in enumerate(chunks(l, 4), 1)}

d returns:

{1: ['Wi-Fi', '10.0.10.22', '255.255.0.0', '34-F3-9A-4C-D0-C4'],
 2: ['VMware Network Adapter VMnet1',
  '192.168.10.111',
  '255.255.255.0',
  '00-50-56-C0-00-01'],
 3: ['Loopback Pseudo-Interface 1',
  '127.0.0.1',
  '255.0.0.0',
  '00-00-00-00-00-00-00-E0']}

score 1 · Answer 2 · answered Apr 18 '18 at 11:32

1

Assuming that your rule is to split the list every four items, you can do it like so:

res = {i+1: n[i*4:4*(i+1)] for i in range(len(n)//4)}

which gives:

{1: ['Wi-Fi', '10.0.10.22', '255.255.0.0', '34-F3-9A-4C-D0-C4'], 
 2: ['VMware Network Adapter VMnet1', '192.168.10.111', '255.255.255.0', '00-50-56-C0-00-01'], 
 3: ['Loopback Pseudo-Interface 1', '127.0.0.1', '255.0.0.0', '00-00-00-00-00-00-00-E0']}

Or if you are allergic to slicing:

n_i = iter(n)
res = {i+1: [next(n_i) for _ in range(4)] for i in range(len(n)//4)}

answered Apr 18 '18 at 11:32

Ma0

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Your first solution is the most pythonic one. I don't think the second one improves anything – sciroccorics Apr 18 '18 at 11:37
@sciroccorics Well, it doesn't.. The only excuse is *slicing-allergy*. – Ma0 Apr 18 '18 at 11:38
If someone suffers from slicing-allergy, maybe should he switch from Python to C ;-) – sciroccorics Apr 18 '18 at 11:47

score 1 · Accepted Answer · answered Apr 18 '18 at 11:55

Inspired by the grouper recipe in itertools documentation:

l = ['Wi-Fi', '10.0.10.22', '255.255.0.0', '34-F3-9A-4C-D0-C4', 
     'VMware Network Adapter VMnet1', '192.168.10.111', '255.255.255.0', '00-50-56-C0-00-01', 
     'Loopback Pseudo-Interface 1', '127.0.0.1', '255.0.0.0', '00-00-00-00-00-00-00-E0']

d = {i: sublist for i, sublist in enumerate(zip(*([iter(l)]*4)))}

print(d)

#{0: ('Wi-Fi', '10.0.10.22', '255.255.0.0', '34-F3-9A-4C-D0-C4'), 
# 1: ('VMware Network Adapter VMnet1', '192.168.10.111', '255.255.255.0', '00-50-56-C0-00-01'), 
# 2: ('Loopback Pseudo-Interface 1', '127.0.0.1', '255.0.0.0', '00-00-00-00-00-00-00-E0')}

score 1 · Answer 4 · answered Apr 18 '18 at 13:56

input = ['Wi-Fi', '10.0.10.22', '255.255.0.0', '34-F3-9A-4C-D0-C4', 'VMware Network Adapter VMnet1', '192.168.10.111', '255.255.255.0', '00-50-56-C0-00-01', 'Loopback Pseudo-Interface 1', '127.0.0.1', '255.0.0.0', '00-00-00-00-00-00-00-E0']

d = {}
for i in range(1,len(input)/3):
    d[i] = []

item = iter(input)
for i in range(len(d)):
    d[i+1] = [next(item) for x in range(len(d)+1)]
print d

The result as d would be {1: ['Wi-Fi', '10.0.10.22', '255.255.0.0', '34-F3-9A-4C-D0-C4'], 2: ['VMware Network Adapter VMnet1', '192.168.10.111', '255.255.255.0', '00-50-56-C0-00-01'], 3: ['Loopback Pseudo-Interface 1', '127.0.0.1', '255.0.0.0', '00-00-00-00-00-00-00-E0']}

Thanks so much, you help me alot – Park Yo Jin Apr 19 '18 at 08:12 — Park Yo Jin, Apr 19 '18 at 08:12

Python 2.7 : List to Dict

4 Answers4