Iterating JSON structure of list object and finding next keys

Question

I have a Json list and I want to print all the keys from a given key till the end of dictionary. But the code I wrote is very complex. How to do it with less complexity ? I'm using Python 3

dictionary = [{"a": "1"}, {"b": "2"}, {"c": "3"}, {"d": "4"}]

try:
    for token in dictionary:
            if "b" in list(token.keys())[0]:
                new_dict = dictionary[len(list(token.keys())[0]):]
                for i in new_dict:
                    print(new_dict[len(list(i.keys())[0]):])
                break
            else:
                print("Inception")
except Exception as error:
    print(str(error))

DESIRED
Input: b
Output: c, d

My Output:

Inception
[{'c': '3'}, {'d': '4'}]
[{'c': '3'}, {'d': '4'}]
[{'c': '3'}, {'d': '4'}]

What's your desired output for the given data ? The question is unclear. — llllllllll, Mar 23 '18 at 12:07
Sounds like you want [`itertools.dropwhile()`](https://docs.python.org/3/library/itertools.html#itertools.dropwhile).. — Martijn Pieters, Mar 23 '18 at 12:09
`if 'b' in list(token.keys())[0]` should just be `if 'b' in token`. Dictionaries have no order, your test would be subject to the whim of insertion order and the current random hash seed if you have more than one key. — Martijn Pieters, Mar 23 '18 at 12:10

Martijn Pieters · Answer 1 · 2018-03-23T12:28:05.733

Use itertools.dropwhile() to skip all dictionaries that don't have a 'b' key:

from itertools import dropwhile

filtered = dropwhile(lambda t: 'b' not in t, dictionary)
next(filtered)  # skip the dictionary with `b` in it.

for token in filtered:
    print(token)

This prints all dictionaries after the first. If you only need to print their keys, do so explicitly:

filtered = dropwhile(lambda t: 'b' not in t, dictionary)
next(filtered)  # skip the dictionary with `b` in it.

for token in filtered:
    print(*token, sep='\n')

This prints the keys on separate lines; if there is just one key, then that's all that'll be printed for each token dictionary.)

As a side note: you really do not want to use list(dict.keys())[0]. Before Python 3.6, dictionaries have no set order (instead being subject to insertion and deletion history and the current random hash seed), so if you have more than one key what value you'll get is a gamble. And all you want to do is see if a key is present, so use a key in dictobject membership test.

To get the first key out of each dictionary, I'd use next(iter(dictobject)), avoiding creating a list:

first_key = next(iter(dictobject))

I'd only use this if I had single-key dictionaries. I'd also avoid using dictionaries for such a scenario; perhaps you really wanted to use an ordered dictionary instead (in Python < 3.6, use collections.OrderedDict(), otherwise use the regular dict type).

Thanks Martijin :) This helps. – LITDataScience Mar 23 '18 at 12:20 — LITDataScience, Mar 23 '18 at 12:20

jpp · Answer 2 · 2018-03-23T12:15:25.950

2

This is one way. The idea is to find the index of the dictionary with desired key. Then filter your list accordingly.

keys = [next(iter(x)) for x in dictionary]
res = keys[keys.index('b')+1:]
# ['c', 'd']

edited Mar 23 '18 at 12:15

answered Mar 23 '18 at 12:09

jpp

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Thanks for the update jpp :) But I just need c, d (i.e, only the keys). How can I find the keys? – LITDataScience Mar 23 '18 at 12:13
Thanks jpp :) This helps. – LITDataScience Mar 23 '18 at 12:19

score 0 · Answer 3 · answered Mar 23 '18 at 12:22

0

You can re-write your code as follows for your desired result.

found = False
for data in dictionary:
    if found:
        print(list(data.keys())[0])
    if 'b' in data.keys():
        found = True

answered Mar 23 '18 at 12:22

akhilsp

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Iterating JSON structure of list object and finding next keys

3 Answers3