翻译强力驱动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?
If you manipulate data, often it helps that each column be an observed variable (name, age), and each row be an observation (e.g. 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']}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']}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']}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']}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.