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First solution in Uncategorized category for YAML. More Types by Jaime
def yaml(a):
d={}
[d.update(z) for z in [{y[0]:y[1][1:].replace('\\','"')} for y in [x.split(':') for x in a.replace('n\'','\n').replace('"','').split('\n') if ':' in x]]]
veo= dict(map(lambda y: (y[0],int(y[1]) if y[1].isdigit() else y[1]), d.items()))
veo= dict(map(lambda y: (y[0],False if y[1]=='false' else y[1]), veo.items()))
veo= dict(map(lambda y: (y[0],None if y[1]=='' else y[1]), veo.items()))
veo= dict(map(lambda y: (y[0],None if y[1]=='null' else y[1]), veo.items()))
veo= dict(map(lambda y: (y[0],'null' if y[1]=='null ' else y[1]), veo.items()))
return (veo)
April 4, 2020