将 YAML 文件转换为 python dict

Question

I am having the following problem of mapping documents within a YAML file to a dict and properly mapping them.

I have the following YAML file, which represents a server ( db.yml ):

instanceId: i-aaaaaaaa
     environment:us-east
     serverId:someServer
     awsHostname:ip-someip
     serverName:somewebsite.com
     ipAddr:192.168.0.1
     roles:[webserver,php]

I load this YAML file, which I can do without any problems, I think I understand that.

instanceId = getInstanceId()
stream = file('db.yml', 'r')
dict = yaml.load_all(stream)

for key in dict:
    if key in dict == "instanceId":
        print key, dict[key]

I'd like the logic to work like the following:

• load yaml, map to dict
• look in every dict in the document, if the instanceId matches that which was set by getInstanceId() , then print out all of the keys and values for that document.

If I look at the map data structure from the command line, I get:

{'instanceId': 'i-aaaaaaaa environment:us-east serverId:someServer awsHostname:ip-someip serverName:someserver ipAddr:192.168.0.1 roles:[webserver,php]'}

I think I might be creating the data structure for the YAML file improperly, and on matching the contents on the dict , I am a bit lost.

Side note: I cannot load all of the documents in this file using yaml.load() , I tried yaml.load_all() , which seems to work but my main issue still exists.

Answer 1

I think your yaml file should look like (or at least something like, so it's structured correctly anyway):

instance:
     Id: i-aaaaaaaa
     environment: us-east
     serverId: someServer
     awsHostname: ip-someip
     serverName: somewebsite.com
     ipAddr: 192.168.0.1
     roles: [webserver,php]

Then, yaml.load(...) returns:

{'instance': {'environment': 'us-east', 'roles': ['webserver', 'php'], 'awsHostname': 'ip-someip', 'serverName': 'somewebsite.com', 'ipAddr': '192.168.0.1', 'serverId': 'someServer', 'Id': 'i-aaaaaaaa'}}

And you can go from there...

---

So used like:

>>> for key, value in yaml.load(open('test.txt'))['instance'].iteritems():
    print key, value


environment us-east
roles ['webserver', 'php']
awsHostname ip-someip
serverName somewebsite.com
ipAddr 192.168.0.1
serverId someServer
Id i-aaaaaaaa

Answer 2

An additional bug in your code, that doesn't have to do with YAML:

for key in dict:
    if key in dict == "instanceId": # This doesn't do what you want
        print key, dict[key]

in is an operator that works on sequence types , and also on maps. This is why this isn't a syntax error... but it doesn't do what you want.

key in dict will always evaluate to True , because all the keys you're iterating through are in the dict. So your code boils down to True == "instanceId" , which will always evaluate to False , because the boolean value True is never equal to that string.

You might have noticed that the print statement doesn't produce any output; this is because it never gets called.

Answer 3

Just use python-benedict , it's a dict subclass that provides I/O support for most common formats, including yaml .

from benedict import benedict

# path can be a yaml string, a filepath or a remote url
path = 'path/to/data.yml'

d = benedict.from_yaml(path)

# do stuff with your dict
# ...

# write it back to disk
d.to_yaml(filepath=path)

It's well tested and documented, check the README to see all the features: https://github.com/fabiocaccamo/python-benedict

Install using pip: pip install python-benedict

Note: I am the author of this project

Answer 4

You can use the bios package for Python3 like below:

import bios

my_dict = bios.read('data.yml')

bios is reading the raw data from a file and converting a python dict object. According to the extension of the file, it can figure out the file type.

Answer 5

If you have reached this question trying to figure out how to get a python dictionary from a yaml file using the pyyaml library try the safe_load option as shown below.

import yaml
dict = yaml.safe_load("data.yml")

Answer 6

I love to use Path, makes a beautiful one-liner

import yaml
from pathlib import Path
conf = yaml.safe_load(Path('data.yml').read_text())