Print and join elements of a Json with Python

Question

I have a json and I need to create a list of list like:

lista = [['334507', 'XXX', '334507', 36.07, 3985499.0],
['271018','YYY', '271007', 23.11, 1335150.0]]

all_sites = {
    u'displayValue': {
        u'siteId': {u'334507': u'XXX', u'271018': u'YYY'}
    },
    u'rows': [[u'334507', 36.07, 3985499.0],
              [u'271018', 23.11, 1335150.0]],
    u'alert': None,
    u'columns': [u'siteId', u'revenue', u'paidImpressions'],
    u'currency': u'USD'

}

I tried with something like this:

sites = all_sites['displayValue']['siteId'].items()
sites_data = all_sites['rows']

data = []

for item in sites:
    data.append(list(item))

for item in sites_data:
    data.append(item)

But how can I merge the list with the first item of the list??

Well for starters all_sites['displayValue']['siteId'] is of type dictionary and to access both key and value you need to use for key,value in all_sites['displayValue']['siteId'].items(). But your data is stored in another variable: all_sites['displayValue']['rows']. Your attempt to fetch all data is incomplete. — Anton vBR
– Anton vBR, Commented Oct 18, 2018 at 17:21
I guess, your json_ variable should've been called all_sites. — Eli Korvigo
– Eli Korvigo, Commented Oct 18, 2018 at 17:23
It'll be inefficient to find the specific index in the array calls rows where your siteId exists. I would recommend making rows a dictionary instead where the key would be your siteId. Otherwise, you'd have to iterate though all indices until you find the index where the element (sub-array) has your siteId as the first index. — LeKhan9
– LeKhan9, Commented Oct 18, 2018 at 17:30

Thiago Melo · Accepted Answer · 2018-10-18 17:44:04Z

2

Just giving what you asked for :)

[[k,v]+all_sites['rows'][i] for i, (k,v) in enumerate(all_sites['displayValue']['siteId'].items())]

First, iterate over the siteId:

>>> [t for t in all_sites['displayValue']['siteId'].items()]
[('334507', 'XXX'), ('271018', 'YYY')]

Now, you know that it will give you a tuple in t, which you might replace as:

>>> [[k,v] for k,v in all_sites['displayValue']['siteId'].items()]
[['334507', 'XXX'], ['271018', 'YYY']]

Now, the other items are in the "rows" key, but you need to know the row's index. One way of knowing the index of each (k,v) tuple is:

>>> [[k,v]+[i] for i, (k,v) in enumerate(all_sites['displayValue']['siteId'].items())]
[['334507', 'XXX', 0], ['271018', 'YYY', 1]]

See that it merges the list [k,v] with the list [i]. Now, all you have to do is take the row you are interested for each index:

>>> [[k,v]+all_sites['rows'][i] for i, (k,v) in enumerate(all_sites['displayValue']['siteId'].items())]
[['334507', 'XXX', '334507', 36.07, 3985499.0],
 ['271018', 'YYY', '271018', 23.11, 1335150.0]]

I hope that helps.

edited Oct 18, 2018 at 17:44

answered Oct 18, 2018 at 17:33

Thiago Melo

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3 Comments

Martin Bouhier Over a year ago

Yes that is what I need. Can you explain me how can you do the function??

Thiago Melo Over a year ago

sure, I'll improve it for clarity, glad it helped

Thiago Melo Over a year ago

I think Oluwafemi's answer bellow is a way smarter tough :)

Oluwafemi Sule · Accepted Answer · 2018-10-18 17:51:03Z

1

You can enumerate the siteId field items.
Then zip this with the list in the rows field.

Iterate over the zip and extend each item (note that item should be converted to a list before extending it) with its matching row.

siteid_items = json_['displayValue']['siteId'].items()
rows = json_["rows"]

result = list(
    list(item) + row
    for item, row in zip(siteid_items, rows)
)

print(result)

''' 
stdout:

[['334507', 'XXX', '334507', 36.07, 3985499.0],
['271018', 'YYY', '271018', 23.11, 1335150.0]]
'''

edited Oct 18, 2018 at 17:51

answered Oct 18, 2018 at 17:37

Oluwafemi Sule

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Comments

Chandella07 · Accepted Answer · 2018-10-18 17:52:47Z

0

Below example may help you.

json_ = {u'displayValue': {u'siteId': {u'334507': u'XXX', u'271018': u'YYY'}},
     u'rows': [[u'334507', 36.07, 3985499.0], [u'271007', 23.11, 1335150.0]],
     u'alert': None,
     u'columns': [u'siteId', u'revenue', u'paidImpressions'],
     u'currency': u'USD'}

result = []
count = len(json_["displayValue"]["siteId"])

for element in range(count):
    item = list(json_["displayValue"]["siteId"].items()[element])+json_["rows"][element]
    result.append(item)

print result

Output:

[[u'334507', u'XXX', u'334507', 36.07, 3985499.0], [u'271018', u'YYY', u'271007', 23.11, 1335150.0]]

edited Oct 18, 2018 at 17:52

answered Oct 18, 2018 at 17:42

Chandella07

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