Convert dictionary into "=" seperated string-CodePudding

I am learning python and I am stuck on something.

I am trying to convert list of dictionaries into = seperated list. I have tried many times but it is either showing an error or showing unexpected results.

python_program.py

list_of_dictionary = [{"name": "Cool", "id": 0}, {"name": "Good", "id": 1}, {"name": "Bad", "id": 3}]

s = ",".join([str(i) for i in list_of_dictionary])

print(s)
// {"name": "Cool", "id": 0}, {"name": "Good", "id": 1}, {"name": "Bad", "id": 3}
// this is converted as string

// converting the string to dict
d = ast.literal_eval(s)


e = ", ".join(["=".join([key, str(val)]) for key, val in d.items()])

but this is showing

AttributeError: 'tuple' object has no attribute 'items'

I am trying to get it like

0=Cool, 1=Good, 3=Bad

Then I tried

s = ", ".join(["=".join([key, str(val)]) for key, val in list_of_dictionary[0].items()])

print(s)

But is showed

name=Cool, id=0

not

0=Cool

I am not including key in string just values of the dictionary.

CodePudding user response：

Why .items? you are looking for 2 specific keys. Also, using an f-string (or .format for that matter) removes the need to call str(..._).

list_of_dictionary = [
    {"name": "Cool", "id": 0},
    {"name": "Good", "id": 1},
    {"name": "Bad", "id": 3}
]

print(', '.join(f'{d["id"]}={d["name"]}' for d in list_of_dictionary))

outputs

0=Cool, 1=Good, 3=Bad

CodePudding user response：

Try this:

', '.join(['='.join([str(v['id']), v['name']]) for v in list_of_dictionary])