Remove None values or empty values from arrays of dictionaries in python

Question

original

[{'l1Key': 'L1_SHARE', 'l2Key': None}, {'l1Key': 'L1_BROWSE_SOURCE', 'l2Key': None}, {'l1Key': 'L1_BLOCK_SOURCE', 'l2Key': None}, {'l1Key': 'L1_HIDE_NEWS', 'l2Key': None}, {'l1Key': 'NA', 'l2Key': 'HTML_2'}, {'l1Key': 'L1_SHOW_LESS', 'l2Key': 'HTML_2'}, {'l1Key': 'L1_SHOW_MORE', 'l2Key': None}]

expected answer

[{'l1Key': 'L1_SHARE'}, {'l1Key': 'L1_BROWSE_SOURCE'}, {'l1Key': 'L1_BLOCK_SOURCE', }, {'l1Key': 'L1_HIDE_NEWS'}, {'l1Key': 'NA', 'l2Key': 'HTML_2'}, {'l1Key': 'L1_SHOW_LESS', 'l2Key': 'HTML_2'}, {'l1Key': 'L1_SHOW_MORE'}]

i have tried with few codes but i am not getting as expected

[d for d in arr if all(d.values())]
[{'l1Key': 'NA', 'l2Key': 'HTML_2'}, {'l1Key': 'L1_SHOW_LESS', 'l2Key': 'HTML_2'}] //wrong as other key value pairs are all deleted if one of the keys is none or null

Sharan · Accepted Answer · 2020-01-28 11:46:08Z

2

d = [{'l1Key': 'L1_SHARE', 'l2Key': None}, {'l1Key': 'L1_BROWSE_SOURCE', 'l2Key': None}, {'l1Key': 'L1_BLOCK_SOURCE', 'l2Key': None}, {'l1Key': 'L1_HIDE_NEWS', 'l2Key': None}, {'l1Key': 'NA', 'l2Key': 'HTML_2'}, {'l1Key': 'L1_SHOW_LESS', 'l2Key': 'HTML_2'}, {'l1Key': 'L1_SHOW_MORE', 'l2Key': None}]
clean_d = []
for dict in d:
  clean_d.append({k: v for k, v in dict.items() if v is not None})

answered Jan 28, 2020 at 11:46

Sharan

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Jarvis · Accepted Answer · 2020-01-28 11:55:36Z

1

One-liner using list comprehension :

original = [{'l1Key': 'L1_SHARE', 'l2Key': None}, {'l1Key': 'L1_BROWSE_SOURCE', 'l2Key': None}, {'l1Key': 'L1_BLOCK_SOURCE', 'l2Key': None}, {'l1Key': 'L1_HIDE_NEWS', 'l2Key': None}, {'l1Key': 'NA', 'l2Key': 'HTML_2'}, {'l1Key': 'L1_SHOW_LESS', 'l2Key': 'HTML_2'}, {'l1Key': 'L1_SHOW_MORE', 'l2Key': None}]
expected = [{k: v for (k, v) in i.items() if v not in [None, '']} for i in original]

gives

[{'l1Key': 'L1_SHARE'}, {'l1Key': 'L1_BROWSE_SOURCE'}, {'l1Key': 'L1_BLOCK_SOURCE'}, {'l1Key': 'L1_HIDE_NEWS'}, {'l1Key': 'NA', 'l2Key': 'HTML_2'}, {'l1Key': 'L1_SHOW_LESS', 'l2Key': 'HTML_2'}, {'l1Key': 'L1_SHOW_MORE'}]

Checking membership like v not in [...] gives you flexibility to filter with more options.

edited Jan 28, 2020 at 11:55

answered Jan 28, 2020 at 11:50

Jarvis

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1 Comment

Jarvis Over a year ago

@vinaykn Please accept this answer (or any other) if it solved your problem

Shashank V · Accepted Answer · 2020-01-28 11:52:33Z

1

[d for d in arr if all(d.values())]

In this line, you are selecting only those dictionaries which have all values that evaluate to a truthy. So you are missing the whole dictionary even if one value is None in your output.

You can delete the keys in place as follows without creating new dicts.

for curr in d:
    to_delete = [k for k, v in curr.items() if not v]
    for k in  to_delete:
       del curr[k]

answered Jan 28, 2020 at 11:52

Shashank V

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