Sorting a lists within a dictionary using list data

Question

My problem is I am trying to sort this lists within a dictionary where I am using the dictionary name as the field name and the lists as the data rows. I have a feeling I am using the wrong approach (should this be a tuple or a dataframe or what?)

I have tried using sorted() but can't get past only sorting by the key (e.g. name, score1, score2). I want to maintain the key order but rearrange the values while keeping their relationship across keys.

This is a sample dataset that I want to sort by score1 (or score2):

scores = {'name': ['joe', 'pete', 'betsy', 'susan', 'pat'], 'score1': [99, 90, 84, 65, 100], 'score2': [85, 91, 90, 55, 98]}

After sorting for score1, I would like to see:

pat, joe, pete, betsy, susan

After sorting for score1, I would like to see:

pat, pete, betsy, joe, susan

DeepSpace · Accepted Answer · 2019-04-29 11:59:14Z

2

If you create scores yourself, it will be much easier if it is structured differently, ie [{'name': 'joe', 'score1': 1, 'score2': 2}, ...].

Then the usage of sorted is quite simple:

scores = [{'name': 'joe', 'score1': 99, 'score2': 85},
          {'name': 'pete', 'score1': 90, 'score2': 91},
          {'name': 'betsy', 'score1': 84, 'score2': 90},
          {'name': 'susan', 'score1': 65, 'score2': 55},
          {'name': 'pat', 'score1': 100, 'score2': 98}]

by_score_1 = [d['name'] for d in sorted(scores, key=lambda d: d['score1'], reverse=True)]
print(by_score_1)
by_score_2 = [d['name'] for d in sorted(scores, key=lambda d: d['score2'], reverse=True)]
print(by_score_2)

Outputs:

['pat', 'joe', 'pete', 'betsy', 'susan']
['pat', 'pete', 'betsy', 'joe', 'susan']

answered Apr 29, 2019 at 11:59

DeepSpace

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3 Comments

hiro protagonist Over a year ago

...this is more efficient and clearer than what i suggested!

Mailboxhead Over a year ago

I can restructure scores, so this looks interesting. I then need to add new keys as new tests are created. This seems easily done by: for item in scores: item.update( {'score3': 90}) But what if I then needed to update individually by student? How could I update 'Pat' to a new score 3 of 75?

DeepSpace Over a year ago

@DouglasWood That's a very valid argument, for which the data structure I suggested is a bit less efficient. You will need to iterate over the entire list until you find the correct dictionary: for person_data in scores: if person_data['name'] == 'pat': person_data['score1'] = new_score

Error - Syntactical Remorse · Accepted Answer · 2019-05-09 13:50:43Z

2

The other answer is nice. You could also turn it into a list of tuples then easily sort it:

scores = {
    'name': ['joe', 'pete', 'betsy', 'susan', 'pat'], 
    'score1': [99, 90, 84, 65, 100], 
    'score2': [85, 91, 90, 55, 98]}

t = list(zip(*scores.values()))
print(t)

Output:

[('joe', 99, 85), ('pete', 90, 91), ('betsy', 84, 90), ('susan', 65, 55), ('pat', 100, 98)]

Then you can sort it:

# Sort by score1
print(sorted(t, key=lambda x: (x[1]), reverse=True))

# Sort by score2
print(sorted(t, key=lambda x: (x[2]), reverse=True))

# Sort by both scores:
print(sorted(t, key=lambda x: (x[1], x[2]), reverse=True))

Just a different way to attack the same problem. Doing it this way you can easily print the scores of the individuals as well.

edited May 9, 2019 at 13:50

answered Apr 29, 2019 at 11:58

Error - Syntactical Remorse

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Comments

hiro protagonist · Accepted Answer · 2019-04-29 11:49:46Z

1

this is a possiblity:

print(sorted(scores['name'], reverse=True, 
             key=lambda x: scores['score1'][scores['name'].index(x)]))
# ['pat', 'joe', 'pete', 'betsy', 'susan']
print(sorted(scores['name'], reverse=True, 
             key=lambda x: scores['score2'][scores['name'].index(x)]))
# ['pat', 'pete', 'betsy', 'joe', 'susan']

answered Apr 29, 2019 at 11:49

hiro protagonist

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1 Comment

Mailboxhead Over a year ago

What if I wanted to display the name and the score? Also, what if I wanted to not just print but update the dictionary to be in sorted order by score1?

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