How to create different objects depending on input?

Question

What I have so far:

import app.models as models

if table_name ==  "Appliances":
    model = models.Appliances
elif table_name == "ArtsCraftsAndSewing":
    model = models.ArtsCraftsAndSewing
...

for index, row in df.iterrows():
    model = model(param1=row[column1], param2=row[column2], ...)
    model.save()

TypeError: 'Appliances' object is not callable

Basically I'm trying to call model = models.Appliances(param1=row[column1], param2=row[column2], ...) except the specific object changes depending on input.

Edit: Appliances and ArtsCraftsAndSewing have the same parameters.

Will the table_name will be same as model class name always? — JPG
– JPG, Commented Jul 19, 2020 at 16:20
@ArakkalAbu Yes, but I didn't want to parse the input string as python code because it'll introduce security issues. — NoName
– NoName, Commented Jul 19, 2020 at 18:08

willeM_ Van Onsem · Accepted Answer · 2020-07-19 16:20:22Z

4

By writing:

model = model(…)

in the second iteration, model no longer refers to the model, but now to an object of that model. You should change the name of the object, for example:

for index, row in df.iterrows():
    model_object = model(param1=row[column1], param2=row[column2], …)
    # …

You might want to use a dictionary to do the mapping:

models = [models.Appliances, models.ArtsCraftsAndSewing]
model_dict = {
    m.__name__: m for m in models
}
model = model_dict[table_name]

answered Jul 19, 2020 at 16:20

willeM_ Van Onsem

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1 Comment

NoName Over a year ago

Oooooh okay, so I actually just replaced the selected model class. It works for the first iteration, but not the rest. I was trying to condense the number of variables, but that was the wrong condensation.

JPG · Accepted Answer · 2020-07-19 16:24:27Z

3

You can use getattr() here,

from app_name import models

table_name = 'Appliances' # or something else
ModelKlass = getattr(models, table_name)

for index, row in df.iterrows():
    model_instance = ModelKlass(param1=row[column1], param2=row[column2], ...)

Note: I assume table_name will be the Django model name.

answered Jul 19, 2020 at 16:24

JPG

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Comments

saran3h · Accepted Answer · 2020-07-19 16:20:25Z

1

Just create a dictionary of all the different model types and pass your inputs as keys:

model_types = dict('appliances' = Appliances, 'arts_crafts_and_sewing' = ArtsCraftsAndSewing)

your_object = model_types[type](param1=arg1, param2=arg2, ...) // here `type` is your input which you can enter via some form or whatever.

answered Jul 19, 2020 at 16:20

saran3h

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