Convert dictionary (with a tuple as the values) to a list?

Question

I made a dictionary:

sort = {
    str(4213) : ("STEM Center",0),
    str(4201) : ("Foundations Lab",1),
    str(4204) : ("CS Lab",2),
    str(4218) : ("Workshop Room",3),
    str(4205) : ("Tiled Room",4),
    "Out" : ("Outside",5)
}

How do I convert it into a list where each item in the list is a tuple? (using a loop)

ex:

[('4213', 'STEM Center', 0),...]

Why `str(4213)` instead of `'4213'`? – RoadRunner Jul 15 '18 at 06:06 — RoadRunner, Jul 15 '18 at 06:06

score 4 · Answer 1 · answered Jul 15 '18 at 05:56

Easiest way to achieve that is using list comprehension:

sort = {
    str(4213) : ("STEM Center",0),
    str(4201) : ("Foundations Lab",1),
    str(4204) : ("CS Lab",2),
    str(4218) : ("Workshop Room",3),
    str(4205) : ("Tiled Room",4),
    "Out" : ("Outside",5)
}

print([(k, name, count) for k, (name, count) in sort.items()])

Prints:

[('4213', 'STEM Center', 0), ('4201', 'Foundations Lab', 1), ('4204', 'CS Lab', 2), ('4218', 'Workshop Room', 3), ('4205', 'Tiled Room', 4), ('Out', 'Outside', 5)]

You can extend this to work for any number of elements in the value by using `(k, *v) for k, v in sort.items()` — user3483203, Jul 15 '18 at 06:03

PydPiper · Answer 2 · 2018-07-15T06:15:34.673

2

Use dictionary comprehension to unpack a dict into it's keys/values (k/v) and since your example the values are tuples with a content of index-0 and index-1, and your ouput format desired is to pack it all together, you can accomplish that by calling each index of the tuple-value (v[0], v[1])

[(k,v[0], v[1]) for k, v in sort.items()]

edited Jul 15 '18 at 06:15

answered Jul 15 '18 at 05:56

PydPiper

411
5
18

score 2 · Answer 3 · answered Jul 15 '18 at 05:59

You can iterate over the items in the dictionary and join them as tuples:

sort = { str(4213) : ("STEM Center",0), str(4201) : ("Foundations Lab",1), str(4204) : ("CS Lab",2), str(4218) : ("Workshop Room",3), str(4205) : ("Tiled Room",4), "Out" : ("Outside",5) }

[(k,) + v for k, v in sort.items()]

Output:

[('4213', 'STEM Center', 0),
 ('4204', 'CS Lab', 2),
 ('4205', 'Tiled Room', 4),
 ('Out', 'Outside', 5),
 ('4201', 'Foundations Lab', 1),
 ('4218', 'Workshop Room', 3)]

^ will work regardless of length of dictionary value tuples – cosmic_inquiry Jul 15 '18 at 06:00 — cosmic_inquiry, Jul 15 '18 at 06:00

score 1 · Answer 4 · answered Jul 15 '18 at 06:03

Since the question specifically mentions that it should be in a loop (perhaps there is further processing that needs to be done that was taken out), here's what that would look like (modifying @Andrej Kesely's strong approach)

sort_ = { str(4213) : ("STEM Center",0), str(4201) : ("Foundations Lab",1), str(4204) : ("CS Lab",2), str(4218) : ("Workshop Room",3), str(4205) : ("Tiled Room",4), "Out" : ("Outside",5) }

result = []
for key, (name, count) in sort_.items():
    result.append((key, name, count))

I renamed sort to sort_. Even though it isn't in this namespace (sort is a method on lists), I don't like using key words.

[('4213', 'STEM Center', 0),
 ('4201', 'Foundations Lab', 1),
 ('4204', 'CS Lab', 2),
 ('4218', 'Workshop Room', 3),
 ('4205', 'Tiled Room', 4),
 ('Out', 'Outside', 5)]

Subhi Alqudsi · Answer 5 · 2018-07-15T06:15:18.243

-1

new_list = [(k,sort[k][0],sort[k][1]) for k in sort]
print(new_list)

=> [('4213', 'STEM Center', 0), ('4201', 'Foundations Lab', 1)]

of course you can do that too

l=[] for k in sort: l.append((k,sort[k][0],sort[k][1]))

[('4213', 'STEM Center', 0), ('4201', 'Foundations Lab', 1)]

edited Jul 15 '18 at 06:15

answered Jul 15 '18 at 06:09

Subhi Alqudsi

1
1

score -1 · Answer 6 · answered Jul 15 '18 at 06:25

-1

sorted_list = list(sort.items())

This will give you [('Out', ('Outside', 5)), ('4201', ('Foundations Lab', 1)), ('4204', ('CS Lab', 2)), ('4218', ('Workshop Room', 3)), ('4213', ('STEM Center', 0)), ('4205', ('Tiled Room', 4))]

answered Jul 15 '18 at 06:25

Rohan Shenoy

805
10
23

Convert dictionary (with a tuple as the values) to a list?

6 Answers6