Converting and writing a dictionary to YAML format-CodePudding

I'm retrieving the names of all the columns in each table from a database and create a dictionary where table name is the key and value is a list of column names in that table.

my_dict = {
    'table_1': ['col_1', 'col_2', 'col_3', 'col_4']
}

I'm trying to convert and write the above dictionary into YAML format in a .yml file

I want my output in YAML file to be as:

tables:
  - name: table_1
    columns:
      - name: col_1
      - name: col_2
      - name: col_3
      - name: col_4

My try:

import yaml
import ruamel.yaml as yml

def to_yaml():
    my_dict = {
    'table_1' : ['col_1', 'col_2', 'col_3', 'col_4']
    }

    my_file_path = "/Users/Desktop/python/tables_file"

    with open(f"{my_file_path}/structure.yml", "w") as file:
    yaml.dump(yaml.load(my_dict), default_flow_style=False)

if __name__ == '__main__'
    to_yaml()

The above code didn't give the expected output. I also tried different approaches based on some YAML related Stack Overflow posts but couldn't get it work.

CodePudding user response：

Use:

import yaml
import ruamel.yaml

def to_yaml():
    my_dict = {
    'table_1' : ['col_1', 'col_2', 'col_3', 'col_4']
    }

    my_file_path = "/Users/Desktop/python/tables_file"

    with open(f"{my_file_path}/structure.yml", "w") as file:
        # yaml.dump need a dict and a file handler as parameter
        yaml = ruamel.yaml.YAML()
        yaml.indent(sequence=4, offset=2)
        yaml.dump(my_dict, file)

if __name__ == '__main__':
    to_yaml()

Content of structure.yml:

table_1:
  - col_1
  - col_2
  - col_3
  - col_4

If you want your expected output, you need to change the structure of your dict. There is nothing magical here.