CODEWITHSUNDEEP
pythonlistpython-2.7dictionary
Tony
Tony

Reputation: 1057

How to do two-level nested map

This is my code:

html_tags = [{'tag': 'a',
              'attribs': [('class', 'anchor'),
                          ('aria-hidden', 'true')]}]

I just can do it by one-level for-loop and one-level map as follow:

for index, tag in enumerate(html_tags):
    html_tags[index]['attribs'] = map(lambda x: '@{}="{}"'.format(*x), tag['attribs'])
print html_tags

However, this is my output (result):

[{'attribs': ['@class="anchor"', '@aria-hidden="true"'], 'tag': 'a'}]

How to do two-level nested map and output the same result.

Upvotes: 1

Views: 480

Answers (1)

Kasravnd
Kasravnd

Reputation: 107297

I suggest a dictionary comprehension :

>>> html_tags = [{i:map(lambda x: '@{}="{}"'.format(*x), j) if i=='attribs' else j for i,j in html_tags[0].items()}]
>>> html_tags
[{'attribs': ['@class="anchor"', '@aria-hidden="true"'], 'tag': 'a'}]
>>>

Also instead of using map with lambda as a more efficient way you can use a list comprehension :

>>> html_tags = [{i:['@{}="{}"'.format(*x) for x in j] if i=='attribs' else j for i,j in html_tags[0].items()}]
>>> html_tags
[{'attribs': ['@class="anchor"', '@aria-hidden="true"'], 'tag': 'a'}]
>>>

Upvotes: 2

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