How do I sort a key:list dictionary by values in list?

Question

I have a dictionary

mydict = {'name':['peter', 'janice', 'andy'], 'age':[10, 30, 15]}

How do I sort this dictionary based on key=="name" list?

End result should be:

mydict = {'name':['andy', 'janice', 'peter'], 'age':[15, 30, 10]}

Or is dictionary the wrong approach for such data?

Answer 1

If you manipulate data, often it helps that each column be an observed variable (name, age), and each row be an observation (eg a sampled person). More on tidy data in this PDF link

Bad programmers worry about the code. Good programmers worry about data structures and their relationships - Linus Torvalds

A list of dictionaries lends itself better to operations like this. Below I present a beginner-friendly snippet to tidy your data. Once you have a good data structure, sorting by any variable is trivial even for a beginner. No one-liner Python kung-fu :)

>>> mydict = {'name':['peter', 'janice', 'andy'], 'age':[10, 30, 15]}

Let's work on a better data structure first

>>> persons = []
>>> for i, name in enumerate(mydict['name']):
...     persons.append({'name': name, 'age': mydict['age'][i]})
... 
>>> persons
[{'age': 10, 'name': 'peter'}, {'age': 30, 'name': 'janice'}, {'age': 15, 'name': 'andy'}]

Now it's easier to work on this data structure which is similar to "data frames" in data analysis environments. Let's sort it by person.name

>>> persons = sorted(persons, key=lambda person: person['name'])

Now bring it back to your format if you want to

>>> {'name': [p['name'] for p in persons], 'age': [p['age'] for p in persons]}
{'age': [15, 30, 10], 'name': ['andy', 'janice', 'peter']}

Answer 2

zip is used to make tuples (name, age)

dict = {'name':[], 'age':[]}
for name, age in sorted(zip(mydict['name'], mydict['age'])):
    dict['name'].append(name)
    dict['age'].append(age)

output:

{'age': [15, 30, 10], 'name': ['andy', 'janice', 'peter']}

Answer 3

A two-line version:

>>> s = sorted(zip(mydict['name'], mydict['age']))
>>> dict([('name', [x[0] for x in s]), ('age', [x[1] for x in s])])
{'age': [15, 30, 10], 'name': ['andy', 'janice', 'peter']}

Answer 4

You can use defaultdict ( for faster performance ) and zip - firstly zip person to corresponding age then sort and after all generate defaultdict and dictionary comprehension.

from collections import defaultdict
dd= defaultdict(list)
mydict = {'name':['peter', 'janice', 'andy'], 'age':[10, 30, 15]}
d=zip(mydict['name'],mydict['age'])

for i in sorted(d,key=lambda x: x[0]):
    dd['names'].append(i[0])
    dd['age'].append(i[1])
print {k:v for k,v in dd.items()}

Output-

{'age': [15, 30, 10], 'names': ['andy', 'janice', 'peter']}

Answer 5

Yes maybe your approach with the dictionnary isn't the best.Here you're just storing list in a dictionnary. But you may also use a list to store those list. That would be almost the same What i would suggest first is dict of dict.

myPeople = {}
myPeople["peter"] = {"age":20, "country" : "USA"}
myPeople["john"] = {"age":30, "country" : "newzealand"}
myPeople["fred"] = {"age":32, "country" : "France"}
# or if you have only one descript per name to store, 
# you may use only a dict
# myPeopleName["peter"] = 20
# myPeopleName["john"] = 30

#then you iterate the dictionnary as before and the result will be print
# out sorted by key

for k in sorted(myPeople.keys()): 
    print myPeople[k]

Edit : My code was totaly wrong. And collection do not sort value by key but preserve the order of the key which you add in the dictionary.