CODEWITHSUNDEEP
pythonsortingdictionary
Steve
Steve

Reputation: 331

Python: Sorting a nested dict by date

I have a large dict, but similar in concept to this:

data = {'business1': {'1/2':20, '1/4':10, '1/3':30}, 'business2': {'1/2':10, '1/4':20, '1/3':30}}

I want to sort the nested dicts by date. I understand how to do this when no nesting is involved, as shown below. I created a list, and I append the key, value pairs to the list and then sort.

data = {'1/2':20, '1/4':10, '1/3':30}

sorted_data = []
for key,value in data.iteritems():
    temp = (key,value)
    sorted_data.append(temp)
sorted_data.sort(key = lambda item: item[0], reverse=False)
print sorted_data

The problem is how to do this when dicts in dicts are involved, such as what I first mentioned:

data = {'business1': {'1/2':20, '1/4':10, '1/3':30}, 'business2': {'1/2':10, '1/4':20, '1/3':30}}

Upvotes: 1

Views: 929

Answers (3)

Gimo
Gimo

Reputation: 178

from collections import OrderedDict

for k,v in data.items():
    data[k] = OrderedDict(sorted(v.items()))

print data

user input:

 data = {'business1': {'1/2':20, '1/4':10, '1/3':30}, 'business2': {'1/2':10, '1/4':20, '1/3':30}}

output:

 {'business2': OrderedDict([('1/2', 10), ('1/3', 30), ('1/4', 20)]), 'business1': OrderedDict([('1/2', 20), ('1/3', 30), ('1/4', 10)])}

Upvotes: 1

Irshad Bhat
Irshad Bhat

Reputation: 8709

>>> data = {'business1': {'1/2':20, '1/4':10, '1/3':30}, 'business2': {'1/2':10, '1/4':20, '1/3':30}}
>>> {i:sorted(data[i].items(), key=lambda x: x[0]) for i in data}
{'business2': [('1/2', 10), ('1/3', 30), ('1/4', 20)], 'business1': [('1/2', 20), ('1/3', 30), ('1/4', 10)]}

Upvotes: 0

rmunn
rmunn

Reputation: 36668

You haven't said what you want your results to look like, but I would assume you want them to look like this:

result = {'business1': [('1/2',20), ('1/3',30), ('1/4',10)],
          'business2': [('1/2',10), ('1/3',30), ('1/4',20)]}

Here's how I would do it:

result = {}
for business_name, business_data in data.iteritems():
    # The following is basically your single-data loop
    sorted_data = []
    for key,value in business_data.iteritems():
        temp = (key,value)
        sorted_data.append(temp)
    sorted_data.sort(key = lambda item: item[0], reverse=False)
    result[business_name] = sorted_data

Now, you could save a step. The for key,value in business_data.iteritems(): loop is basically repeating what dict.items() does. So you could replace four lines with sorted_data = list(business_data.items()). (The list() call is unnecessary in Python 2 but doesn't hurt anything, and it's needed in Python 3. Since you didn't say which version you're using, I left it in so that my answer will work on either Python 2 or Python 3).

So the final version I would suggest is:

result = {}
for business_name, business_data in data.iteritems():
    sorted_data = list(business_data.items())
    sorted_data.sort(key = lambda item: item[0], reverse=False)
    result[business_name] = sorted_data

Hope this helps.

Upvotes: 0

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