You can do this.
data[0]['f'] = var
You can do this.
data[0]['f'] = var
One possible issue I see is you set your JSON unconventionally within an array/list object. I would recommend using JSON in its most accepted form, i.e.:
test_json = { "a": 1, "b": 2}
Once you do this, adding a json element only involves the following line:
test_json["c"] = 3
This will result in:
{'a': 1, 'b': 2, 'c': 3}
Afterwards, you can add that json back into an array or a list of that is desired.
Unable to append data to Json array object with desired output
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I would do this:
data["list"].append({'b':'2'})
so simply you are adding an object to the list that is present in "data"
Elements are added to list using append():
>>> data = {'list': [{'a':'1'}]}
>>> data['list'].append({'b':'2'})
>>> data
{'list': [{'a': '1'}, {'b': '2'}]}
If you want to add element to a specific place in a list (i.e. to the beginning), use insert() instead:
>>> data['list'].insert(0, {'b':'2'})
>>> data
{'list': [{'b': '2'}, {'a': '1'}]}
After doing that, you can assemble JSON again from dictionary you modified:
>>> json.dumps(data)
'{"list": [{"b": "2"}, {"a": "1"}]}'
Hello I currently learning json in python i want to append a dictionary in a json file ontop of existing ones but every time i do this i get this error in VS-Code:
End of file expected.
Can somebody help me?
Here is the Code:
dict = {
"data1" : data3,
"data2" : data4
}
data = json.dumps(dict)
with open("index.json" , "a") as file:
json.dump(data , file)
def write(filename, new_data): # function that appends data to the specified file
with open(filename, "r") as file1:
data = json.load(file1)
data.append(new_data)
with open(filename, "w") as file1:
# Sets file's current position at offset.
file1.seek(0)
json.dump(data, file1, indent=4)I have this function that appends objects exactly how I want; however, it only does it to the bottom of the file rather than the top.
jsobj["a"]["b"]["e"].append({"f":var3, "g":var4, "h":var5})
jsobj["a"]["b"]["e"].append({"f":var6, "g":var7, "h":var8})
Just add the dictionary as a dictionary object not a string :
jsobj["a"]["b"]["e"].append(dict(f=var3))
Full source :
var1 = 11
var2 = 32
jsonobj = {"a":{"b":{"c": var1,
"d": var2,
"e": [],
},
},
}
var3 = 444
jsonobj["a"]["b"]["e"].append(dict(f=var3))
jsonobj will contain :
{'a': {'b': {'c': 11, 'd': 32, 'e': [{'f': 444}]}}}
First, accidents is a dictionary, and you can't write to a dictionary; you just set values in it.
So, what you want is:
for accident in accidents:
accident['Turn'] = 'right'
The thing you want to write out is the new JSON—after you've finished modifying the data, you can dump it back to a file.
Ideally you do this by writing to a new file, then moving it over the original:
with open('sanfrancisco_crashes_cp.json') as json_file:
json_data = json.load(json_file)
accidents = json_data['accidents']
for accident in accidents:
accident['Turn'] = 'right'
with tempfile.NamedTemporaryFile(dir='.', delete=False) as temp_file:
json.dump(temp_file, json_data)
os.replace(temp_file.name, 'sanfrancisco_crashes_cp.json')
But you can do it in-place if you really want to:
# notice r+, not rw, and notice that we have to keep the file open
# by moving everything into the with statement
with open('sanfrancisco_crashes_cp.json', 'r+') as json_file:
json_data = json.load(json_file)
accidents = json_data['accidents']
for accident in accidents:
accident['Turn'] = 'right'
# And we also have to move back to the start of the file to overwrite
json_file.seek(0, 0)
json.dump(json_file, json_data)
json_file.truncate()
If you're wondering why you got the specific error you did:
In Python—unlike many other languages—assignments aren't expressions, they're statements, which have to go on a line all by themselves.
But keyword arguments inside a function call have a very similar syntax. For example, see that tempfile.NamedTemporaryFile(dir='.', delete=False) in my example code above.
So, Python is trying to interpret your accident['Turn'] = 'right' as if it were a keyword argument, with the keyword accident['Turn']. But keywords can only be actual words (well, identifiers), not arbitrary expressions. So its attempt to interpret your code fails, and you get an error saying keyword can't be an expression.
I solved with that :
with open('sanfrancisco_crashes_cp.json') as json_file:
json_data = json.load(json_file)
accidents = json_data['accidents']
for accident in accidents:
accident['Turn'] = 'right'
with open('sanfrancisco_crashes_cp.json', "w") as f:
json.dump(json_data, f)
json might not be the best choice for on-disk formats; The trouble it has with appending data is a good example of why this might be. Specifically, json objects have a syntax that means the whole object must be read and parsed in order to understand any part of it.
Fortunately, there are lots of other options. A particularly simple one is CSV; which is supported well by python's standard library. The biggest downside is that it only works well for text; it requires additional action on the part of the programmer to convert the values to numbers or other formats, if needed.
Another option which does not have this limitation is to use a sqlite database, which also has built-in support in python. This would probably be a bigger departure from the code you already have, but it more naturally supports the 'modify a little bit' model you are apparently trying to build.
You probably want to use a JSON list instead of a dictionary as the toplevel element.
So, initialize the file with an empty list:
with open(DATA_FILENAME, mode='w', encoding='utf-8') as f:
json.dump([], f)
Then, you can append new entries to this list:
with open(DATA_FILENAME, mode='w', encoding='utf-8') as feedsjson:
entry = {'name': args.name, 'url': args.url}
feeds.append(entry)
json.dump(feeds, feedsjson)
Note that this will be slow to execute because you will rewrite the full contents of the file every time you call add. If you are calling it in a loop, consider adding all the feeds to a list in advance, then writing the list out in one go.