How can I parse a YAML file in Python

Question

How can I parse a YAML file in Python?

Answer 1

Example:

defaults.yaml

url: https://www.google.com

environment.py

from ruamel import yaml

data = yaml.safe_load(open('defaults.yaml'))
data['url']

Answer 2

Import yaml module and load the file into a dictionary called 'my_dict':

import yaml
my_dict = yaml.load(open('filename'))

That's all you need. Now the entire yaml file is in 'my_dict' dictionary.

Answer 3

I use ruamel.yaml. Details & debate here.

from ruamel import yaml

with open(filename, 'r') as fp:
    read_data = yaml.load(fp)

Usage of ruamel.yaml is compatible (with some simple solvable problems) with old usages of PyYAML and as it is stated in link I provided, use

from ruamel import yaml

instead of

import yaml

and it will fix most of your problems.

EDIT: PyYAML is not dead as it turns out, it's just maintained in a different place.

Answer 4

If you have YAML that conforms to the YAML 1.2 specification (released 2009) then you should use ruamel.yaml (disclaimer: I am the author of that package). It is essentially a superset of PyYAML, which supports most of YAML 1.1 (from 2005).

If you want to be able to preserve your comments when round-tripping, you certainly should use ruamel.yaml.

Upgrading @Jon's example is easy:

import ruamel.yaml as yaml

with open("example.yaml") as stream:
    try:
        print(yaml.safe_load(stream))
    except yaml.YAMLError as exc:
        print(exc)

Use safe_load() unless you really have full control over the input, need it (seldom the case) and know what you are doing.

If you are using pathlib Path for manipulating files, you are better of using the new API ruamel.yaml provides:

from ruamel.yaml import YAML
from pathlib import Path

path = Path('example.yaml')
yaml = YAML(typ='safe')
data = yaml.load(path)

Answer 5

Read & Write YAML files with Python 2+3 (and unicode)

# -*- coding: utf-8 -*-
import yaml
import io

# Define data
data = {'a list': [1, 42, 3.141, 1337, 'help', u'€'],
        'a string': 'bla',
        'another dict': {'foo': 'bar',
                         'key': 'value',
                         'the answer': 42}}

# Write YAML file
with io.open('data.yaml', 'w', encoding='utf8') as outfile:
    yaml.dump(data, outfile, default_flow_style=False, allow_unicode=True)

# Read YAML file
with open("data.yaml", 'r') as stream:
    data_loaded = yaml.load(stream)

print(data == data_loaded)

Created YAML file

a list:
- 1
- 42
- 3.141
- 1337
- help
- €
a string: bla
another dict:
  foo: bar
  key: value
  the answer: 42

Common file endings

.yml and .yaml

Alternatives

• CSV: Super simple format (read & write)
• JSON: Nice for writing human-readable data; VERY commonly used (read & write)
• YAML: YAML is a superset of JSON, but easier to read (read & write, comparison of JSON and YAML)
• pickle: A Python serialization format (read & write)
• MessagePack (Python package): More compact representation (read & write)
• HDF5 (Python package): Nice for matrices (read & write)
• XML: exists too *sigh* (read & write)

For your application, the following might be important:

• Support by other programming languages
• Reading / writing performance
• Compactness (file size)

See also: Comparison of data serialization formats

In case you are rather looking for a way to make configuration files, you might want to read my short article Configuration files in Python

Answer 6

#!/usr/bin/env python

import sys
import yaml

def main(argv):

    with open(argv[0]) as stream:
        try:
            #print(yaml.load(stream))
            return 0
        except yaml.YAMLError as exc:
            print(exc)
            return 1

if __name__ == "__main__":
    sys.exit(main(sys.argv[1:]))