Sorting a list containing Dictionary in Python

Question

I have list that contains dictionary that need to be sorted based on the alphabetic order

[
    {
        'index': False,
        'definition': {
            'id': 1111111L,
            'value': u'Large Content'
        },
        'id': 1234567L,
        'name': {
            'id': 9999999999L,
            'value': u'INTRODUCTION'
        }
    },
    {
        'index': False,
        'definition': {
            'id': 22222222L,
            'value': u'Large Content'
        },
        'id': 2L,
        'name': {
            'id': 3333333333333l,
            'value': u'Abstract'
        }
    },
    {
        'index': False,
        'definition': {
            'id': 8888888888L,
            'value': u'Large Content'
        },
        'id': 1L,
        'name': {
            'id': 343434343434L,
            'value': u'Bulletin'
        }
    }
    {
        'index': False,
        'definition': {
            'id': 1113311L,
            'value': u'Large Content'
        },
        'id': 333434L,
        'name': {
            'id': 9999999999L,
            'value': u'<b>END</b>'
        }
    },
]

I need to sort based on ['name']['value'] to result

Abstract
Bulletin
INTRODUCTION
END

But when i do it i get the capital letters first

bg = []
for n in a:
   bg = sorted(a, key=lambda n: n["name"]["value"])

INTRODUCTION
END
Abstract
Bulletin

I can't reproduce your results. For the data you have given, I get [u'<b>END</b>', u'Abstract', u'Bulletin', u'INTRODUCTION']. — LaC
– LaC, Commented May 22, 2012 at 7:07
I remove the html tags using the tip from love-python.blogspot.com/2008/07/… — Rakesh
– Rakesh, Commented May 22, 2012 at 7:26
I just meant that your example doesn't show the problem you were having. In fact, for those four words, case-sensitivity makes no difference. — LaC
– LaC, Commented May 22, 2012 at 8:03

lvc · Accepted Answer · 2012-05-22 07:03:34Z

6

To make the sort case insensitive, put everything in lower case in your sort key:

bg = sorted(a, key=lambda n: n["name"]["value"].lower())

answered May 22, 2012 at 7:03

lvc

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hamstergene · Accepted Answer · 2012-05-22 07:04:13Z

2

Because capitals are lexicographically smaller. Drop letter case before sorting:

key=lambda n: n["name"]["value"].lower()

answered May 22, 2012 at 7:04

hamstergene

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