In Python 3, to load json previously saved like this:
json.dumps(dictionary)
the output is something like
{\"(\'Hello\',)\": 6, \"
import json
data = json.load(open('/Users/laxmanjeergal/Desktop/json.json'))
jtopy=json.dumps(data) #json.dumps take a dictionary as input and returns a string as output.
dict_json=json.loads(jtopy) # json.loads take a string as input and returns a dictionary as output.
print(dict_json["shipments"])
You are passing a dictionary to a function that expects a string.
This syntax:
{"('Hello',)": 6, "('Hi',)": 5}
is both a valid Python dictionary literal and a valid JSON object literal. But loads
doesn't take a dictionary; it takes a string, which it then interprets as JSON and returns the result as a dictionary (or string or array or number, depending on the JSON, but usually a dictionary).
If you pass this string to loads
:
'''{"('Hello',)": 6, "('Hi',)": 5}'''
then it will return a dictionary that looks a lot like the one you are trying to pass to it.
You could also exploit the similarity of JSON object literals to Python dictionary literals by doing this:
json.loads(str({"('Hello',)": 6, "('Hi',)": 5}))
But in either case you would just get back the dictionary that you're passing in, so I'm not sure what it would accomplish. What's your goal?
json.loads
take a string as input and returns a dictionary as output.
json.dumps
take a dictionary as input and returns a string as output.
With json.loads({"('Hello',)": 6, "('Hi',)": 5})
,
You are calling json.loads
with a dictionary as input.
You can fix it as follows (though I'm not quite sure what's the point of that):
d1 = {"('Hello',)": 6, "('Hi',)": 5}
s1 = json.dumps(d1)
d2 = json.loads(s1)