How to select multiple JSON Objects using python

Question

I have a Json data as following. The Json has many such objects with same NameId's:

[{
    "NameId": "name1",
    "exp": {
        "exp1": "test1"
    }
}, {
    "NameId": "name1",
    "exp": {
        "exp2": "test2"
    }
}

]

Now, what I am after is to create a new Json Object that has a merged exp and create a file something like below, so that I do not have multiple NameId:

[{ "NameId": "name1", "exp": { "exp1": "test1", "exp2": "test2" } } ]

Is there a possibility I can achive it using Python?

Wyrzutek · Accepted Answer · 2022-10-27 12:36:01Z

1

You can do the manual work, merging the entries while rebuilding the structure. You can keep a dictionary with the exp to merge them.

import json

jsonData = [{
    "NameId": "name1",
    "exp": {
        "exp1": "test1"
    }
}, {
    "NameId": "name1",
    "exp": {
        "exp2": "test2"
    }
}, {
    "NameId": "name2",
    "exp": {
        "exp3": "test3"
    }
}]

result = []
expsDict = {}
for entry in jsonData:
    nameId = entry["NameId"]
    exp = entry["exp"]
    if nameId in expsDict:
        # Merge exp into resultExp.
        # Note that resultExp belongs to both result and expsDict,
        # changes made will be reflected in both containers!
        resultExp = expsDict[nameId]
        for (expName, expValue) in exp.items():
            resultExp[expName] = expValue
    else:
        # Copy copy copy, otherwise merging would modify jsonData too!
        exp = exp.copy()
        entry = entry.copy()
        entry["exp"] = exp
        # Add a new item to the result
        result.append(entry)
        # Store exp to later merge other entries with the same name.
        expsDict[nameId] = exp

print(result)

edited Oct 27, 2022 at 12:36

answered Oct 27, 2022 at 10:09

Wyrzutek

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7 Comments

Geebee Over a year ago

Thank you so much, you made it look like it was very easy :) and I have wasted 2 days on it already ....

Wyrzutek Over a year ago

@Geebee Happy to help. Please note my edit, resultExp was not shared only between the result and the dictionary, but also the original data! You can print(jsonData) at the end to see the difference

Wyrzutek Over a year ago

If I understand the question, the magic is that dictionaries are stored by reference. When you do list.append(x) or dict[key]=x and then modify x (e.g. x[a]=b), also the element of the list and the dict are modified, because they are different references of the same object. Also dict[key][a]=b modifies x. So the modifications done to resultExp are reflected to expsDict[nameId], the exp that was assigned to it in the else part, and the entry["exp"] that has been added to the result, because they are all the same object!

Wyrzutek Over a year ago

There is a difference between assigning a variable and modifying it. When you assign, you modify only that variable, because you are changing what it references; for instance a={}, b=a, b=3, print(a) will display {}, because at the end b references to 3 and a has nothing to do with it. Instead, when the variable is modified, all references get modified; for instance, a={}, b=a, b[0]=3, print(a) will display {0:3}, because a and b are references of the same dictionary. It is difficult indeed!

Wyrzutek Over a year ago

Note that the operator x+=y behaves like an assignment, not a modification, because it is only a shortcut for x=x+y. For instance, a=0, b=a, b+=3, print(a) will display 0, just like if there was b=b+3 instead of b+=3. b is assigned to something different and hence loses the reference to a.

|

Deepak Tripathi · Accepted Answer · 2022-10-27 09:37:07Z

0

You can use itertools.groupby and functools.reduce

d = [{
    "NameId": "name1",
    "exp": {
        "exp1": "test1"
    }
}, {
    "NameId": "name1",
    "exp": {
        "exp2": "test2"
    }
}]
from itertools import groupby
[ {'NameId': k, 'exp': reduce(lambda x,y : {**x["exp"], **y["exp"]} , v) } for k,v in groupby(sorted(d, key=lambda x: x["NameId"]), lambda x: x["NameId"]) ]

#output
[{'NameId': 'name1', 'exp': {'exp1': 'test1', 'exp2': 'test2'}}]

answered Oct 27, 2022 at 9:37

Deepak Tripathi

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2 Comments

Geebee Over a year ago

Thanks @Deepak, I did not get a chance to go through your solution. I will go through it and will let you know the outcome.

Deepak Tripathi Over a year ago

Sure @Geebee Take your time

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