Creating a complex nested dictionary from multiple lists in Python

Question

I am struggling to create a nested dictionary with the following data:

Team,       Group,  ID,  Score,  Difficulty
OneTeam,    A,      0,   0.25,   4
TwoTeam,    A,      1,   1,      10
ThreeTeam,  A,      2,   0.64,   5
FourTeam,   A,      3,   0.93,   6
FiveTeam,   B,      4,   0.5,    7
SixTeam,    B,      5,   0.3,    8
SevenTeam,  B,      6,   0.23,   9
EightTeam,  B,      7,   1.2,    4

Once imported as a Pandas Dataframe, I turn each feature into these lists: teams, group, id, score, diff.

Using this stack overflow answer Create a complex dictionary using multiple lists I can create the following dictionary:

{'EightTeam': {'diff': 4, 'id': 7, 'score': 1.2},
 'FiveTeam': {'diff': 7, 'id': 4, 'score': 0.5},
 'FourTeam': {'diff': 6, 'id': 3, 'score': 0.93},
 'OneTeam': {'diff': 4, 'id': 0, 'score': 0.25},
 'SevenTeam': {'diff': 9, 'id': 6, 'score': 0.23},
 'SixTeam': {'diff': 8, 'id': 5, 'score': 0.3},
 'ThreeTeam': {'diff': 5, 'id': 2, 'score': 0.64},
 'TwoTeam': {'diff': 10, 'id': 1, 'score': 1.0}}

using the code:

{team: {'id': i, 'score': s, 'diff': d} for team, i, s, d in zip(teams, id, score, diff)}

But what I'm after is having 'Group' as the main key, then team, and then id, score and difficulty within the team (as above).

I have tried:

{g: {team: {'id': i, 'score': s, 'diff': d}} for g, team, i, s, d in zip(group, teams, id, score, diff)}

but this doesn't work and results in only one team per group within the dictionary:

{'A': {'FourTeam': {'diff': 6, 'id': 3, 'score': 0.93}},
 'B': {'EightTeam': {'diff': 4, 'id': 7, 'score': 1.2}}}

Below is how the dictionary should look, but I'm not sure how to get there - any help would be much appreciated!

{'A:': {'EightTeam': {'diff': 4, 'id': 7, 'score': 1.2},
  'FiveTeam': {'diff': 7, 'id': 4, 'score': 0.5},
  'FourTeam': {'diff': 6, 'id': 3, 'score': 0.93},
  'OneTeam': {'diff': 4, 'id': 0, 'score': 0.25}},
 'B': {'SevenTeam': {'diff': 9, 'id': 6, 'score': 0.23},
  'SixTeam': {'diff': 8, 'id': 5, 'score': 0.3},
  'ThreeTeam': {'diff': 5, 'id': 2, 'score': 0.64},
  'TwoTeam': {'diff': 10, 'id': 1, 'score': 1.0}}}

Can you post your original data in text not image?

MyNameIsCaleb
– MyNameIsCaleb

2019-05-24 14:22:30 +00:00
Commented May 24, 2019 at 14:22 — MyNameIsCaleb
– MyNameIsCaleb, Commented May 24, 2019 at 14:22

JohnO · Accepted Answer · 2019-05-24 15:42:58Z

3

A dict comprehension may not be the best way of solving this if your data is stored in a table like this.

Try something like

from collections import defaultdict
groups = defaultdict(dict)
for g, team, i, s, d in zip(group, teams, id, score, diff):
    groups[g][team] = {'id': i, 'score': s, 'diff': d }

By using defaultdict, if groups[g] already exists, the new team is added as a key, if it doesn't, an empty dict is automatically created that the new team is then inserted into.

Edit: you edited your answer to say that your data is in a pandas dataframe. You can definitely skip the steps of turning the columns into list. Instead you could then for example do:

from collections import defaultdict
groups = defaultdict(dict)
for row in df.itertuples():
    groups[row.Group][row.Team] = {'id': row.ID, 'score': row.Score, 'diff': row.Difficulty}

edited May 24, 2019 at 15:42

answered May 24, 2019 at 14:30

JohnO

7876 silver badges10 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

bobman Over a year ago

i convert the features into lists (which i should have mentioned earlier - now edited) - would that mean dict comprehension could be used?

JohnO Over a year ago

As bracco23 pointed out, you can use a dict comprehension. The question is would you want to and the answer is probably no because "readability counts". I'll add to my answer how you might do it directly from the dataframe.

bracco23 · Accepted Answer · 2019-05-24 15:25:45Z

If you absolutely want to use comprehension, then this should work:

z = zip(teams, group, id, score, diff)
s = set(group)
d = { #outer dict, one entry for each different group
    group: ({ #inner dict, one entry for team, filtered for group
        team: {'id': i, 'score': s, 'diff': d} 
        for team, g, i, s, d in z
        if g == group
        }) 
    for group in s 
    }

I added linebreaks for clarity

EDIT:

After the comment, to better clarify my intention and out of curiosity, I run a comparison:

# your code goes here

from collections import defaultdict
import timeit

teams = ['OneTeam', 'TwoTeam', 'ThreeTeam', 'FourTeam', 'FiveTeam', 'SixTeam', 'SevenTeam', 'EightTeam']
group = ['A', 'A', 'A', 'A', 'B', 'B', 'B', 'B']
id = [0, 1, 2, 3, 4, 5, 6, 7]
score = [0.25, 1, 0.64, 0.93, 0.5, 0.3, 0.23, 1.2] 
diff = [4, 10, 5, 6, 7, 8, 9, 4]

def no_comprehension():
    global group, teams, id, score, diff
    groups = defaultdict(dict)
    for g, team, i, s, d in zip(group, teams, id, score, diff):
        groups[g][team] = {'id': i, 'score': s, 'diff': d }

def comprehension():
    global group, teams, id, score, diff
    z = zip(teams, group, id, score, diff)
    s = set(group)
    d = {group: ({team: {'id': i, 'score': s, 'diff': d} for team, g, i, s, d in z if g == group}) for group in s}

print("no comprehension:")
print(timeit.timeit(lambda : no_comprehension(), number=10000))
print("comprehension:")
print(timeit.timeit(lambda : comprehension(), number=10000))

executable version

Output:

no comprehension:
0.027287796139717102
comprehension:
0.028979241847991943

They do look the same, in terms of performance. With my sentence above, I was just highlighting this as an alternative solution to the one already posted by @JohnO.

Do you think this would be the fastest approach? I get the impression from you saying 'if you absolutely want to use' that it may not be the best method. @bracco23

Collectives™ on Stack Overflow

Creating a complex nested dictionary from multiple lists in Python

2 Answers 2

2 Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related