how to sort a dictionary with lists of multiple numeric values?

Question

I have a dictionary as the following data structure:

d = {'TRANSFERRED': [2281, 1031, 1775, 867, 1242],
     'CLOSED':      [239, 269, 645, 540, 388], 
     'DEFERRED':    [89, 5, 68, 48, 37],
     'OPEN':        [3, 0, 2, 1, 0],
     'IN PROGRESS': [0, 2, 4, 0, 5],
     'QUEUED':      [0, 0, 0, 0, 0]}

The dictionary contains lists with numeric values and I would like to order them from the lowest to the highest value, something like this:

d = {'TRANSFERRED': [867, 1031, 1242, 1775, 2281],
     'CLOSED':      [239, 269, 388, 540, 645], 
     'DEFERRED':    [5, 37, 48, 68, 89],
     'OPEN':        [0, 0, 1, 2, 3],
     'IN PROGRESS': [0, 0, 2, 4, 5],
     'QUEUED':      [0, 0, 0, 0, 0]}

I am concerned that dictionaries cannot be sorted because they are inherently orderless but other types such as lists and tuples are not orderless, moreover, I have been using the following trick to order dictionaries with lists that contain single items such as:

d2 = {'TRANSFERRED': [-2281],
      'CLOSED':      [239], 
      'DEFERRED':    [489],
      'OPEN':        [34],
      'IN PROGRESS': [0],
      'QUEUED':      [-10]}
sorted(d2.items(), key=lambda x: x[1], reverse=True)

The result gives the following sorted data structure:

[('DEFERRED', [489]), 
 ('CLOSED', [239]), 
 ('OPEN', [34]), 
 ('IN PROGRESS', [0]), 
 ('QUEUED', [-10]), 
 ('TRANSFERRED', [-2281])]

I want to replicate this same result but with a dictionary with lists of multiple numeric values. How can I achieve this? Please, feel free to use the following link repl.it - sort dictionary with lists of multiple items to test your results. Feedback or comments to improve this question are welcome.

Ajax1234 · Accepted Answer · 2017-12-11 19:53:42Z

2

You can use dict.items():

d = {'TRANSFERRED': [2281, 1031, 1775, 867, 1242],
 'CLOSED':      [239, 269, 645, 540, 388], 
 'DEFERRED':    [89, 5, 68, 48, 37],
 'OPEN':        [3, 0, 2, 1, 0],
 'IN PROGRESS': [0, 2, 4, 0, 5],
 'QUEUED':      [0, 0, 0, 0, 0]}
new_d = {a:sorted(b) for a, b in d.items()}

Output:

{'IN PROGRESS': [0, 0, 2, 4, 5], 'TRANSFERRED': [867, 1031, 1242, 1775, 2281], 'DEFERRED': [5, 37, 48, 68, 89], 'CLOSED': [239, 269, 388, 540, 645], 'OPEN': [0, 0, 1, 2, 3], 'QUEUED': [0, 0, 0, 0, 0]}

answered Dec 11, 2017 at 19:53

Ajax1234

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1 Comment

Joe Iddon Over a year ago

Always so fast, just typing out the same solution!

Fuji Komalan · Accepted Answer · 2017-12-12 06:24:22Z

d = {
     'TRANSFERRED':2281,
     'CLOSED':239, 
     'DEFERRED':89,
     'OPEN':3,
     'IN PROGRESS':2,
     'QUEUED':0
}

# Some usefull method of dictionary
# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#
#
# To get dict keys:
# ----------------
#
# d.keys() = > dict_keys(['TRANSFERRED', 'CLOSED', 'DEFERRED', 'OPEN', 'IN PROGRESS', 'QUEUED'])
# The output is an iterable.
#
# To get dict values
#-------------------
#
# d.values() => dict_values([2281, 239, 89, 3, 2, 0])
# The out is an iterable.
#
# To get both key and values
# --------------------------
#
# d.items() => dict_items([('TRANSFERRED', 2281), ('CLOSED', 239), ('DEFERRED', 89), ('OPEN', 3),
# ('IN PROGRESS', 2), ('QUEUED', 0)])
# The out is iterable of tuples. Each tuple contains key as the first tuple item and value as
# second item. (k,v)

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