The easiest and purest method without relying on C headers is PyYaml (documentation), which can be installed via pip install pyyaml:
#!/usr/bin/env python
import yaml
with open("example.yaml", 'r') as stream:
try:
print(yaml.safe_load(stream))
except yaml.YAMLError as exc:
print(exc)
And that's it. A plain yaml.load() function also exists, but yaml.safe_load() should always be preferred unless you explicitly need the arbitrary object serialization/deserialization provided in order to avoid introducing the possibility for arbitrary code execution.
Note the PyYaml project supports versions up through the YAML 1.1 specification. If YAML 1.2 specification support is needed, see ruamel.yaml as noted in this answer.
To access any element of a list in a YAML file like this:
global:
registry:
url: dtr-:5000/
repoPath:
dbConnectionString: jdbc:oracle:thin:@x.x.x.x:1521:abcd
You can use following python script:
import yaml
with open("/some/path/to/yaml.file", 'r') as f:
valuesYaml = yaml.load(f, Loader=yaml.FullLoader)
print(valuesYaml['global']['dbConnectionString'])
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.
Example:
defaults.yaml
url: https://www.google.com
environment.py
from ruamel import yaml
data = yaml.safe_load(open('defaults.yaml'))
data['url']
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)
# -*- 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.safe_load(stream)
print(data == data_loaded)
a list:
- 1
- 42
- 3.141
- 1337
- help
- €
a string: bla
another dict:
foo: bar
key: value
the answer: 42
.yml and .yaml
For your application, the following might be important:
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
First install pyyaml using pip3.
Then import yaml module and load the file into a dictionary called 'my_dict':
import yaml
with open('filename.yaml') as f:
my_dict = yaml.safe_load(f)
That's all you need. Now the entire yaml file is in 'my_dict' dictionary.
#!/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:]))
Source: Stackoverflow.com