Adding a value to an existing key in python

Question

I have a list:

Marks =[
    [12345,75,'English'],
    [23452,83,'Physics'],
    [23560,81,'Statistics'],
    [23415,61,'Computer'],
    [23459,90,'Physics'],
    [12345,75,'Computer'],
    [23452,100,'Statistics']
]

and I want to make a dictionary where the class name is the key and the two numbers are the values. I can make the initial dictionary like this:

for i in list1:
    dict1[i[2]]=i[0],i[1]

But when I print the dictionary, only one of each of the pairs of numbers associated is printed. I'm looking for a result like:

{ 
  'English' : [[12345,75]],
  'Physics' : [[23452,83], [23459,90]],
  'Statistics' : [[23560,81], [23452,100]],
  'Computer' : [[23415,61], [12345,75]]
}

Is there any way I can extend the values associated with a key?

I tried:

if i[2] in dict1:
    dict1.extend

The issue that I'm having is that there would be two sets of tuples for the same course. It should be 'Statistics: [23560, 81], [23452, 100]. And I can add the first set of numbers, but not hte second — user10613120
– user10613120, Commented Apr 9, 2019 at 12:17
@user10613120 It's a good idea to include an example of the output you're looking for. I've edited in an example. — glibdud
– glibdud, Commented Apr 9, 2019 at 12:22
{'Computer': [12345, 75], [23145, 65] , 'English': [12345, 75], 'Physics': [23459, 90], [12345,67], 'Statistics': [23452, 100], [98765,30]} — user10613120
– user10613120, Commented Apr 9, 2019 at 12:25

Rakesh · Accepted Answer · 2019-04-09 12:25:56Z

2

Using collections.defaultdict

Ex:

from collections import defaultdict

Marks = [[12345,75,'English'], [23452,83,'Physics'], [23560,81,'Statistics'], [23415,61,'Computer'], [23459,90,'Physics'], [12345,75,'Computer'], [23452,100,'Statistics']]

result = defaultdict(list)
for i in Marks:
    result[i[-1]].append(i[:2])
print(result)

Output:

defaultdict(<type 'list'>, {'Statistics': [[23560, 81], [23452, 100]], 'Physics': [[23452, 83], [23459, 90]], 'Computer': [[23415, 61], [12345, 75]], 'English': [[12345, 75]]})

edited Apr 9, 2019 at 12:25

answered Apr 9, 2019 at 12:10

Rakesh

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Comments

holidayfun · Accepted Answer · 2019-04-09 12:09:18Z

1

You can only have a single value for a key in your dictionary. But you could save the two values as either a list or a tuple like so:

for i in list1:
    dict1[i[2]]=i[0:1]

or

for i in list1:
    dict1[i[2]]=(i[0], i[1])

answered Apr 9, 2019 at 12:09

holidayfun

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1 Comment

user10613120 Over a year ago

but there are two sets of two numbers associated with each course. My poor explanation, but that's the issue that I'm having. It should be 'Statistics: [23560, 81], [23452, 100]

accdias · Accepted Answer · 2019-04-09 12:31:58Z

1

You can use comprehensions to achieve that easily:

>>> Marks = [[12345,75,'English'], [23452,83,'Physics'], [23560,81,'Statistics'], [23415,61,'Computer'], [23459,90,'Physics'], [12345,75,'Computer'], [23452,100,'Statistics']]
>>> d = { m[2]: [m[0], m[1]] for m in Marks }
>>> d
{'English': [12345, 75], 'Physics': [23459, 90], 'Statistics': [23452, 100], 'Computer': [12345, 75]}
>>>

EDIT: This will create the dictionary taking into account those repeated keys.

d = {}
for mark in Marks:
     if not mark[2] in d:
             d[mark[2]] = [[mark[0], mark[1]]]
     else:
             d[mark[2]] += [[mark[0], mark[1]]]

The dictionary will end up with this shape:

>>> d
{'English': [[12345, 75]], 'Physics': [[23452, 83], [23459, 90]], 'Statistics': [[23560, 81], [23452, 100]], 'Computer': [[23415, 61], [12345, 75]]}
>>>

edited Apr 9, 2019 at 12:31

answered Apr 9, 2019 at 12:11

accdias

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4 Comments

user10613120 Over a year ago

but there are two sets of two numbers associated with each course. My poor explanation, but that's the issue that I'm having. It should be 'Statistics: [23560, 81], [23452, 100]

accdias Over a year ago

Sorry. I completely missed that detail. I will try to update my example.

user10613120 Over a year ago

Perfect! This is what I was trying to do with either extending or adding to the value. Thanks!

accdias Over a year ago

Take a look on Rakesh's answer. It is a better and more Pythonic way of achieving the same result.

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