Parsing elements from named list in Python

Question

I am getting a named list as an output from a Python program.

[{'path': u'/home/mycomp/Documents/folder1/906350_16379886.JPG', 'score': 7.937254, 'dist': 0.0, 'id': u'AWW3yZOubIicFyUL0_lv', 'metadata': None}, {'path': u'/home/mycomp/Documents/folder1/907675_16389607.JPG', 'score': 0.5119519, 'dist': 0.2922064602034243, 'id': u'AWW3bhtybIicFyUL0uVN', 'metadata': None}]

The length of the output in this case is 2 and I do not have control on the length of the output. I want to get only the filename from the path printed. I understand i can use

os.path.basename(path)

to get the filename. However, when I tried getting the path from the list using the following code:

for (i, imagePath) in enumerate(imagePaths):
    a=ses.search_image(imagePath)
    for k,v in enumerate(a):
        print(v)

imagePath = provides the path to the image

a=ses.search_image(imagePath) = searches for similar image and stores all the matching image parameters for the image in the path to a. That output of a is what I have given above.

for k,v in enumerate(a):
     print(v)

actually prints each list as a separate line like this:

{'path': u'/home/mycomp/Documents/folder1/906350_16379886.JPG', 'score': 7.937254, 'dist': 0.0, 'id': u'AWW3yZOubIicFyUL0_lv', 'metadata': None}
{'path': u'/home/mycompDocuments/folder1/907675_16389607.JPG', 'score': 0.5119519, 'dist': 0.2922064602034243, 'id': u'AWW3bhtybIicFyUL0uVN', 'metadata': None}

The desired output is just this:

906350_16379886.JPG
907675_16389607.JPG

Sorry to bother you, but there is no such thing as a "named list" in python. The item in your first example is a list with two elements. Also, you say "prints each list...". Those are not lists, they are dictionaries. Dictionaries and lists are different types of objects in python. Following the naming conventions will surely make solving problems easier in the future. — Chillie
– Chillie, Commented Sep 10, 2018 at 8:04
@Chillie Thank you for the clarification...I am on learning path of programming...got lost with the nomenclature. will be careful next time. — Apricot
– Apricot, Commented Sep 10, 2018 at 14:39
Good luck! Make sure you read some of the official documentation it has plenty of useful information with examples. — Chillie
– Chillie, Commented Sep 10, 2018 at 15:03

salparadise · Accepted Answer · 2018-09-10 04:45:37Z

2

Use os.basename to extract the filename, and grab the path key and not the entire dictionary. (Also, don't see a need to use enumerate in here)

import os # goes at the top of your script
for v in a:
    print(os.path.basename(v['path']))

answered Sep 10, 2018 at 4:45

salparadise

5,8751 gold badge28 silver badges32 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Austin · Accepted Answer · 2018-09-10 04:50:18Z

0

Without importing os:

lst = [{'path': u'/home/mycomp/Documents/folder1/906350_16379886.JPG', 'score': 7.937254, 'dist': 0.0, 'id': u'AWW3yZOubIicFyUL0_lv', 'metadata': None}, {'path': u'/home/mycomp/Documents/folder1/907675_16389607.JPG', 'score': 0.5119519, 'dist': 0.2922064602034243, 'id': u'AWW3bhtybIicFyUL0uVN', 'metadata': None}]

print([x['path'].rsplit('/', 1)[1] for x in lst])
# ['906350_16379886.JPG', '907675_16389607.JPG']

answered Sep 10, 2018 at 4:50

Austin

26.1k4 gold badges28 silver badges52 bronze badges

Collectives™ on Stack Overflow

Parsing elements from named list in Python

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related