Possible solution for AttributeError: 'Index' object has no attribute 'replace'

Question

Let's assume I have this code below which uses the flatten function to flatten a JSON object and convert it into a Pandas data frame.

from flatten_json import flatten
import pandas as pd

data = [{ 
    "_id" : 1, 
    "IdList" : [
        6422
    ], 
    "levels" : [
        {
            "active" : "true", 
            "level" : 3, 
            "actions" : [
                {
                    "isActive" : "true"
                }]
        }]
}
,
{ 
    "_id" : 2, 
    "IdList" : [
        6442
    ], 
    "levels" : [
        {
            "active" : "true", 
            "level" : 1, 
            "actions" : [
                {
                    "isActive" : "true"
                }]
        }]
}

]

dic_flattened = [flatten(i) for i in data]
result = pd.DataFrame(dic_flattened)
result.columns = result.columns.str.replace("_0_", ".")
print(result)

Current Output:

   _id  IdList_0 levels.active  levels.level levels.actions.isActive
0    1      6422          true             3                    true
1    2      6442          true             1                    true

The function uses a _ as a separator, however, I do not need the underscore along with the integers in the column name so I have added a replace which gets rid of them. My only issue is when I try to replace the _0 from the IdList_0 column I get the following error:

AttributeError: 'Index' object has no attribute 'replace'

Is there any way for me to get rid of that _0 from the column name so the desired output can be like the following:

Desired Output:

   _id    IdList levels.active  levels.level levels.actions.isActive
0    1      6422          true             3                    true
1    2      6442          true             1                    true

Laurent · Accepted Answer · 2021-03-23 17:53:26Z

4

Try:

result.columns = ["_id", "IdList", "levels.active", "levels.level", "levels.actions.isActive"]

Or a "dynamic" version as per your second question:

result.columns = [col.replace("_0", "") for col in result.columns]

edited Mar 23, 2021 at 17:53

answered Mar 23, 2021 at 16:28

Laurent

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4 Comments

jcoke Over a year ago

it still gives me the same error when I try that solution

Laurent Over a year ago

Following your comment, I updated my answer with an alternate method.

jcoke Over a year ago

it now returns NaN in the IdList column, also is there a way to do this dynamically so without explicitly specifying the column names

Laurent Over a year ago

This time, it should be good (works for me), see updated answer.

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