CODEWITHSUNDEEP
pythondictionaryimportyaml
78282219
78282219

Reputation: 159

Save a yaml file to a generator object to a dictionary python

I am having a lot of issues saving a YAML file to a python dictionary. I will display two routes I take below, both being suboptimal.

Yaml file:

modules:
    module_1: True
    module_2: True
    module_3: False
scenarios:
    constant:
        start_date: "2018-09-30"
        end_date: "2019-09-30"
    adverse:
        start_date: "2019-09-30"
        end_date: "2022-09-30"

Route 1: Saving YAML file directly to a dictionary without specifying a loader which is now depreciated

import yaml
filepath = "C:\\user\\path\\file.yaml"
_dict = yaml.load(open(filepath))
print(type(_dict))
>>> <class 'dict'>

error message: Calling yaml.load() without loader=... is depreciated, as it is unsafe

Route 2: Loading in as a generator (which is not subscriptable)

import yaml
filepath = "C:\\user\\path\\file.yaml"
document = open(filepath, "r")
dictionary = yaml.safe_load_all(document)

print(type(dictionary)
>>> <generator object>

print(dictionary["modules"]["module_1"]
>>> generator object is not subscriptable

Is there a way I can import my yaml file into a dictionary safely? I wish to use the dictionary in my python project instead of creating global variables, etc.

Example:

if _dict["modules"]["module_1"]:
    # Do something

Upvotes: 3

Views: 2109

Answers (1)

Adeom
Adeom

Reputation: 248

Only calling without loader was depracted. You can always pass SafeLoader to the load function.

import yaml

with open(filepath, 'r') as stream:
    dictionary = yaml.load(stream, Loader=yaml.SafeLoader)

This should return your dictionary.

edit:

And as for yaml.safe_load_all, you only need to call generator.__next__() to obtain the dictionary.

import yaml
filepath = "C:\\user\\path\\file.yaml"
document = open(filepath, "r")
generator = yaml.safe_load_all(document)
dictionary = generator.__next__()

I would recomend the first option for your use.

Upvotes: 4

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