How to sort a dictionary in python by value when the value is a list and I want to sort it by the first index of that list

Question

is it possible to sort a python dictionary by value if the value is a list, and I want it to be sorted by the first value of that list. E.G:

data = {
"Joe" : [1, "Joe", "password", "Joe@Email.com"], 
"Toby" : [2, "Toby", "password", "Toby@Email.com"], 
"John" : [4, "John", "password", "John@Email.com"], 
"Julie" : [3, "Julie", "password", "Julie@Email.com"]
}

I would like it to be like: Where 'i' is the key

for i in range(len(data)):
    print("UserID: ", str(data[i][0]), ". Username: ", data[i][1])

>> UserID: 1. Username: Joe
>> UserID: 2. Username: Toby
>> UserID: 3. Username: Julie
>> UserID: 4. Username: John

Many thanks.

Answer 1

You cannot sort a dict in place, because Python dicts are unordered. You have at least 2 alternatives :

Create a sorted list of tuples

You can use sorted with a key= argument. In this case, it would be the first element of the dict value :

sorted(data.items(), key= lambda x: x[1][0])
# [('Joe', [1, 'Joe', 'password', 'Joe@Email.com']), ('Toby', [2, 'Toby', 'password', 'Toby@Email.com']), ('Julie', [3, 'Julie', 'password', 'Julie@Email.com']), ('John', [4, 'John', 'password', 'John@Email.com'])]

It returns a sorted list of tuples, which you can use to iterate and print the result :

data = {
    "Joe": [1, "Joe", "password", "Joe@Email.com"],
    "Toby": [2, "Toby", "password", "Toby@Email.com"],
    "John": [4, "John", "password", "John@Email.com"],
    "Julie": [3, "Julie", "password", "Julie@Email.com"]
}

for name, lst in sorted(data.items(), key=lambda x: x[1][0]):
    print("UserID : %d. Username : %s" % (lst[0], name))

# UserID : 1. Username : Joe
# UserID : 2. Username : Toby
# UserID : 3. Username : Julie
# UserID : 4. Username : John

Create an OrderedDict

If you want to sort data and keep the functionality of a dict, you can create an OrderedDict :

from collections import OrderedDict

data = {
    "Joe": [1, "Joe", "password", "Joe@Email.com"],
    "Toby": [2, "Toby", "password", "Toby@Email.com"],
    "John": [4, "John", "password", "John@Email.com"],
    "Julie": [3, "Julie", "password", "Julie@Email.com"]
}

data = OrderedDict(sorted(data.items(), key=lambda x: x[1][0]))
# OrderedDict([('Joe', [1, 'Joe', 'password', 'Joe@Email.com']), ('Toby', [2, 'Toby', 'password', 'Toby@Email.com']), ('Julie', [3, 'Julie', 'password', 'Julie@Email.com']), ('John', [4, 'John', 'password', 'John@Email.com'])])

Note : For both examples, key=lambda x: x[1] would also be enough.

Answer 2

You can use itemgetter, this should sort the entries by the list (you can sort lists of lists and it will by default sort by the first element). This will give back a list of tuples, with the first element being the key, which you can loop through and print.

import operator

data = {
"Joe" : [1, "Joe", "password", "Joe@Email.com"], 
"Toby" : [2, "Toby", "password", "Toby@Email.com"], 
"John" : [4, "John", "password", "John@Email.com"], 
"Julie" : [3, "Julie", "password", "Julie@Email.com"]
}


sorted_data = sorted(data.items(), key=operator.itemgetter(1))

for entry in (sorted_data):
    print("UserID: " + str(entry[1][0]) + ". Username: " + entry[0])

Output:

UserID: 1. Username: Joe
UserID: 2. Username: Toby
UserID: 3. Username: Julie
UserID: 4. Username: John

Note: The reason sorted_data is a list of tuples instead of a dictionary is that python dictionaries are inherently unordered. See answers to this for an explanation: Why is python ordering my dictionary like so?