Json count elements depending on value of another using Python

Question

I have a JSON object that looks like this:

[{'key1': 'yes', 'key2': 'yes', 'key3': 'yes'},
 {'key1': 'yes', 'key2': 'no', 'key3': 'yes'},
 {'key1': 'yes', 'key2': 'null', 'key3': 'yes'}]

I am trying to count the occurrences of Key1 when the value of Key2 is not yes and I can use that to rank Key1 by the appearance of the value yes in Key2.

Hi @alecxe I edited the question slightly. I must be really low on Caffeine. What I intend to do is find us my initial count of Key1 which I have as Yes : 409, No: 334, N/A:40 etc and rank its values depending on the appearance of yes in Key2. So my output i something like Yes:60%, No:30%, N/A: 30% — petergt
– petergt, Commented Jan 30, 2016 at 2:04

alecxe · Accepted Answer · 2016-01-30 01:37:50Z

2

You can use sum():

>>> l = [
    {"key1": "yes", "key2": "yes", "key3": "yes"}, 
    {"key1": "yes", "key2": "no", "key3": "yes"}, 
    {"key1": "yes", "key2": "null", "key3": "yes"}
]
>>> sum(item["key2"] != "yes" for item in l)
2

answered Jan 30, 2016 at 1:37

alecxe

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Tony · Accepted Answer · 2016-01-30 02:23:00Z

0

I have another answer, it's implemented by 'filter' and 'len':

l = [
    {"key1": "yes", "key2": "yes", "key3": "yes"}, 
    {"key1": "yes", "key2": "no", "key3": "yes"}, 
    {"key1": "yes", "key2": "null", "key3": "yes"}
]

f = filter(lambda x: x['key2']!='yes', l)
print f
print len(f)

The print f outputs the filter result. The print len(f) is to further output the number of the filter result.

[{'key3': 'yes', 'key2': 'no', 'key1': 'yes'}, {'key3': 'yes', 'key2': 'null', 'key1': 'yes'}]
2

answered Jan 30, 2016 at 2:23

Tony

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