CODEWITHSUNDEEP
pythonyaml
JrRockeTer
JrRockeTer

Reputation: 35

Python creating partial YAML file from a complexYAML file

I don't find an easy way to do this..

I have a large complex yaml file, I want to load it and write only a couple of the keys into a NEW yaml file.

So for example:

Existing:

a:
  bsect: 
    thing: one
ab:
  bbsect:
    morething: more 
thisone: dbname
ac:
  ccc:
    cccthing: here
bd:
  bbb:
    bbbthing: again

So I load the yaml file using import yaml and I load using: cfg = yaml.load(file) within a with open.

I want to create a NEW yaml file just with with cfg[ab] and cfg[thisone]

But when I create the yaml file using yaml.dump I obviously end up with the content of cfg[ab] and cfg[thisone] without the keys and it looks like this.

bbsect:
   morething: more
dbname.

It makes sense but what is the best way to end up with a new yaml file that contains:

ab:
  bbsect:
    morething: more 
thisone: dbname

I hope this makes sense.

Thanks in advance.

Upvotes: 2

Views: 232

Answers (1)

Equinox
Equinox

Reputation: 6758

You can create a new blank config and write data to it.

x = '''
a:
  bsect: 
    thing: one
ab:
  bbsect:
    morething: more 
thisone: dbname
ac:
  ccc:
    cccthing: here
bd:
  bbb:
    bbbthing: again
'''
import yaml
cfg = yaml.load(x)
new_cfg = {}
new_cfg['ab'] = cfg['ab']
new_cfg['thisone'] = cfg['thisone']
yaml.dump(new_cfg, default_flow_style=False)

Output:

ab:
  bbsect:
    morething: more
thisone: dbname

Upvotes: 1

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