Python: Append Multiple Values for One Key in Nested Dictionary

Question

I have the below list of tuples:

p = [("01","Master"),("02","Node"),("03","Node"),("04","Server")]

I want my output to look like:

y = {
     "Master":{"number":["01"]},
     "Node":{"number":["02", "03"]},
     "Server":{"number":["04"]}
     }

I have tried the below code:

y = {}
for line in p:
     if line[1] in y:
          y[line[1]] = {}
          y[line[1]]["number"].append(line[0])
     else:
          y[line[1]] = {}
          y[line[1]]["number"] = [line[0]]

And I get the below error:

 Traceback (most recent call last):
         File "<stdin>", line 4, in <module>
 KeyError: 'number'

How do I solve this?

is it always in ascending order by the first ID??

jamylak
– jamylak

2014-12-26 13:33:55 +00:00
Commented Dec 26, 2014 at 13:33 — jamylak
– jamylak, Commented Dec 26, 2014 at 13:33
@jamylak No not necessarily

user1452759
– user1452759

2014-12-29 07:13:12 +00:00
Commented Dec 29, 2014 at 7:13 — user1452759
– user1452759, Commented Dec 29, 2014 at 7:13

Padraic Cunningham · Accepted Answer · 2014-12-26 13:46:45Z

2

from collections import defaultdict

d = defaultdict(lambda: defaultdict(list))
for v, k in p:
    d[k]["number"].append(v)


print(d)

  defaultdict(<function <lambda> at 0x7f8005097578>, {'Node': defaultdict(<type 'list'>, {'number': ['02', '03']}), 'Master': defaultdict(<type 'list'>, {'number': ['01']}), 'Server': defaultdict(<type 'list'>, {'number': ['04']})})

without defaultdict:

d = {}
from pprint import pprint as pp
for v, k in p:
    d.setdefault(k,{"number":[]})
    d[k]["number"].append(v)


pp(d)

{'Master': {'number': ['01']},
 'Node': {'number': ['02', '03']},
 'Server': {'number': ['04']}}

edited Dec 26, 2014 at 13:46

answered Dec 26, 2014 at 13:25

Padraic Cunningham

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2 Comments

Ashwini Chaudhary Over a year ago

If you want to use defaultdict, then take full advantage of it: d = defaultdict(lambda: defaultdict(list)). Now all you need to do is: d[k]["number"].append(v).

Padraic Cunningham Over a year ago

@AshwiniChaudhary true, changed

Jivan · Accepted Answer · 2014-12-26 13:29:18Z

1

It's because you don't initialize your dictionary when needed, and you reset it when not needed.

Try this:

p = [("01","Master"),("02","Node"),("03","Node"),("04","Server")]

y = {}
for (number, category) in p:
    if not y.get(category, False):
        # initializes your sub-dictionary
        y[category] = {"number": []}

    # adds the correct number to the sub-dictionary
    y[category]["number"].append(number)

Note that using a tuple unpacking for (number, category) in p allows your code to be more readable inside your loop.

answered Dec 26, 2014 at 13:29

Jivan

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1 Comment

user1452759 Over a year ago

This is much more sleek! Thank you!!

UltraInstinct · Accepted Answer · 2014-12-26 13:30:44Z

1

You are resetting the dictionary!

for line in p:
     if line[1] in y:
          #y[line[1]] = {}   -- RESET! ["number"] will now disappear.
                               #.. which leads to error in the next line.
          y[line[1]]["number"].append(line[0])
     else:
          y[line[1]] = {}
          y[line[1]]["number"] = [line[0]]

A more pythonic way of achieving the same thing would be by using a defaultdict as demonstrated in other answers.

answered Dec 26, 2014 at 13:30

UltraInstinct

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1 Comment

user1452759 Over a year ago

Thank you! This works!! I knew I was making a stupid mistake somewhere! Thanks a lot!

Vivek Sable · Accepted Answer · 2014-12-26 13:34:50Z

1

Do not assign {} to key when key is already present in y.

y = {}
for line in p:
     try:
          y[line[1]]["number"].append(line[0])
     except:
          y[line[1]] = {}
          y[line[1]]["number"] = [line[0]]

OR Use defaultdict use:-

>>> from collections import defaultdict
>>> p = [("01","Master"),("02","Node"),("03","Node"),("04","Server")]
>>> d = defaultdict(list)
>>> for k, v in p:
...    d[v].append(k)
... 
>>> d
defaultdict(<type 'list'>, {'Node': ['02', '03'], 'Master': ['01'], 'Server': ['04']})

edited Dec 26, 2014 at 13:34

answered Dec 26, 2014 at 13:29

Vivek Sable

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