How to get string objects instead of Unicode from JSON?

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Question :

How to get string objects instead of Unicode from JSON?

I’m using Python 2 to parse JSON from ASCII encoded text files.

When loading these files with either json or simplejson, all my string values are cast to Unicode objects instead of string objects. The problem is, I have to use the data with some libraries that only accept string objects. I can’t change the libraries nor update them.

Is it possible to get string objects instead of Unicode ones?


>>> import json
>>> original_list = ['a', 'b']
>>> json_list = json.dumps(original_list)
>>> json_list
'["a", "b"]'
>>> new_list = json.loads(json_list)
>>> new_list
[u'a', u'b']  # I want these to be of type `str`, not `unicode`


This question was asked a long time ago, when I was stuck with Python 2. One easy and clean solution for today is to use a recent version of Python — i.e. Python 3 and forward.

Asked By: Brutus


Answer #1:

A solution with object_hook

import json

def json_load_byteified(file_handle):
    return _byteify(
        json.load(file_handle, object_hook=_byteify),

def json_loads_byteified(json_text):
    return _byteify(
        json.loads(json_text, object_hook=_byteify),

def _byteify(data, ignore_dicts = False):
    # if this is a unicode string, return its string representation
    if isinstance(data, unicode):
        return data.encode('utf-8')
    # if this is a list of values, return list of byteified values
    if isinstance(data, list):
        return [ _byteify(item, ignore_dicts=True) for item in data ]
    # if this is a dictionary, return dictionary of byteified keys and values
    # but only if we haven't already byteified it
    if isinstance(data, dict) and not ignore_dicts:
        return {
            _byteify(key, ignore_dicts=True): _byteify(value, ignore_dicts=True)
            for key, value in data.iteritems()
    # if it's anything else, return it in its original form
    return data

Example usage:

>>> json_loads_byteified('{"Hello": "World"}')
{'Hello': 'World'}
>>> json_loads_byteified('"I am a top-level string"')
'I am a top-level string'
>>> json_loads_byteified('7')
>>> json_loads_byteified('["I am inside a list"]')
['I am inside a list']
>>> json_loads_byteified('[[[[[[[["I am inside a big nest of lists"]]]]]]]]')
[[[[[[[['I am inside a big nest of lists']]]]]]]]
>>> json_loads_byteified('{"foo": "bar", "things": [7, {"qux": "baz", "moo": {"cow": ["milk"]}}]}')
{'things': [7, {'qux': 'baz', 'moo': {'cow': ['milk']}}], 'foo': 'bar'}
>>> json_load_byteified(open('somefile.json'))
{'more json': 'from a file'}

How does this work and why would I use it?

Mark Amery’s function is shorter and clearer than these ones, so what’s the point of them? Why would you want to use them?

Purely for performance. Mark’s answer decodes the JSON text fully first with unicode strings, then recurses through the entire decoded value to convert all strings to byte strings. This has a couple of undesirable effects:

  • A copy of the entire decoded structure gets created in memory
  • If your JSON object is really deeply nested (500 levels or more) then you’ll hit Python’s maximum recursion depth

This answer mitigates both of those performance issues by using the object_hook parameter of json.load and json.loads. From the docs:

object_hook is an optional function that will be called with the result of any object literal decoded (a dict). The return value of object_hook will be used instead of the dict. This feature can be used to implement custom decoders

Since dictionaries nested many levels deep in other dictionaries get passed to object_hook as they’re decoded, we can byteify any strings or lists inside them at that point and avoid the need for deep recursion later.

Mark’s answer isn’t suitable for use as an object_hook as it stands, because it recurses into nested dictionaries. We prevent that recursion in this answer with the ignore_dicts parameter to _byteify, which gets passed to it at all times except when object_hook passes it a new dict to byteify. The ignore_dicts flag tells _byteify to ignore dicts since they already been byteified.

Finally, our implementations of json_load_byteified and json_loads_byteified call _byteify (with ignore_dicts=True) on the result returned from json.load or json.loads to handle the case where the JSON text being decoded doesn’t have a dict at the top level.

Answered By: Mirec Miskuf

Answer #2:

While there are some good answers here, I ended up using PyYAML to parse my JSON files, since it gives the keys and values as str type strings instead of unicode type. Because JSON is a subset of YAML it works nicely:

>>> import json
>>> import yaml
>>> list_org = ['a', 'b']
>>> list_dump = json.dumps(list_org)
>>> list_dump
'["a", "b"]'
>>> json.loads(list_dump)
[u'a', u'b']
>>> yaml.safe_load(list_dump)
['a', 'b']


Some things to note though:

  • I get string objects because all my entries are ASCII encoded. If I would use unicode encoded entries, I would get them back as unicode objects — there is no conversion!

  • You should (probably always) use PyYAML’s safe_load function; if you use it to load JSON files, you don’t need the “additional power” of the load function anyway.

  • If you want a YAML parser that has more support for the 1.2 version of the spec (and correctly parses very low numbers) try Ruamel YAML: pip install ruamel.yaml and import ruamel.yaml as yaml was all I needed in my tests.


As stated, there is no conversion! If you can’t be sure to only deal with ASCII values (and you can’t be sure most of the time), better use a conversion function:

I used the one from Mark Amery a couple of times now, it works great and is very easy to use. You can also use a similar function as an object_hook instead, as it might gain you a performance boost on big files. See the slightly more involved answer from Mirec Miskuf for that.

Answered By: Brutus

Answer #3:

There’s no built-in option to make the json module functions return byte strings instead of unicode strings. However, this short and simple recursive function will convert any decoded JSON object from using unicode strings to UTF-8-encoded byte strings:

def byteify(input):
    if isinstance(input, dict):
        return {byteify(key): byteify(value)
                for key, value in input.iteritems()}
    elif isinstance(input, list):
        return [byteify(element) for element in input]
    elif isinstance(input, unicode):
        return input.encode('utf-8')
        return input

Just call this on the output you get from a json.load or json.loads call.

A couple of notes:

  • To support Python 2.6 or earlier, replace return {byteify(key): byteify(value) for key, value in input.iteritems()} with return dict([(byteify(key), byteify(value)) for key, value in input.iteritems()]), since dictionary comprehensions weren’t supported until Python 2.7.
  • Since this answer recurses through the entire decoded object, it has a couple of undesirable performance characteristics that can be avoided with very careful use of the object_hook or object_pairs_hook parameters. Mirec Miskuf’s answer is so far the only one that manages to pull this off correctly, although as a consequence, it’s significantly more complicated than my approach.
Answered By: Mark Amery

Answer #4:

You can use the object_hook parameter for json.loads to pass in a converter. You don’t have to do the conversion after the fact. The json module will always pass the object_hook dicts only, and it will recursively pass in nested dicts, so you don’t have to recurse into nested dicts yourself. I don’t think I would convert unicode strings to numbers like Wells shows. If it’s a unicode string, it was quoted as a string in the JSON file, so it is supposed to be a string (or the file is bad).

Also, I’d try to avoid doing something like str(val) on a unicode object. You should use value.encode(encoding) with a valid encoding, depending on what your external lib expects.

So, for example:

def _decode_list(data):
    rv = []
    for item in data:
        if isinstance(item, unicode):
            item = item.encode('utf-8')
        elif isinstance(item, list):
            item = _decode_list(item)
        elif isinstance(item, dict):
            item = _decode_dict(item)
    return rv

def _decode_dict(data):
    rv = {}
    for key, value in data.iteritems():
        if isinstance(key, unicode):
            key = key.encode('utf-8')
        if isinstance(value, unicode):
            value = value.encode('utf-8')
        elif isinstance(value, list):
            value = _decode_list(value)
        elif isinstance(value, dict):
            value = _decode_dict(value)
        rv[key] = value
    return rv

obj = json.loads(s, object_hook=_decode_dict)
Answered By: Mike Brennan

Answer #5:

That’s because json has no difference between string objects and unicode objects. They’re all strings in javascript.

I think JSON is right to return unicode objects. In fact, I wouldn’t accept anything less, since javascript strings are in fact unicode objects (i.e. JSON (javascript) strings can store any kind of unicode character) so it makes sense to create unicode objects when translating strings from JSON. Plain strings just wouldn’t fit since the library would have to guess the encoding you want.

It’s better to use unicode string objects everywhere. So your best option is to update your libraries so they can deal with unicode objects.

But if you really want bytestrings, just encode the results to the encoding of your choice:

>>> nl = json.loads(js)
>>> nl
[u'a', u'b']
>>> nl = [s.encode('utf-8') for s in nl]
>>> nl
['a', 'b']
Answered By: nosklo

Answer #6:

There exists an easy work-around.

TL;DR – Use ast.literal_eval() instead of json.loads(). Both ast and json are in the standard library.

While not a ‘perfect’ answer, it gets one pretty far if your plan is to ignore Unicode altogether. In Python 2.7

import json, ast
d = { 'field' : 'value' }
print "JSON Fail: ", json.loads(json.dumps(d))
print "AST Win:", ast.literal_eval(json.dumps(d))


JSON Fail:  {u'field': u'value'}
AST Win: {'field': 'value'}

This gets more hairy when some objects are really Unicode strings. The full answer gets hairy quickly.

Answered By: Charles Merriam

Answer #7:

Mike Brennan’s answer is close, but there is no reason to re-traverse the entire structure. If you use the object_hook_pairs (Python 2.7+) parameter:

object_pairs_hook is an optional function that will be called with the result of any object literal decoded with an ordered list of pairs. The return value of object_pairs_hook will be used instead of the dict. This feature can be used to implement custom decoders that rely on the order that the key and value pairs are decoded (for example, collections.OrderedDict will remember the order of insertion). If object_hook is also defined, the object_pairs_hook takes priority.

With it, you get each JSON object handed to you, so you can do the decoding with no need for recursion:

def deunicodify_hook(pairs):
    new_pairs = []
    for key, value in pairs:
        if isinstance(value, unicode):
            value = value.encode('utf-8')
        if isinstance(key, unicode):
            key = key.encode('utf-8')
        new_pairs.append((key, value))
    return dict(new_pairs)

In [52]: open('test.json').read()
Out[52]: '{"1": "hello", "abc": [1, 2, 3], "def": {"hi": "mom"}, "boo": [1, "hi", "moo", {"5": "some"}]}'                                        

In [53]: json.load(open('test.json'))
{u'1': u'hello',
 u'abc': [1, 2, 3],
 u'boo': [1, u'hi', u'moo', {u'5': u'some'}],
 u'def': {u'hi': u'mom'}}

In [54]: json.load(open('test.json'), object_pairs_hook=deunicodify_hook)
{'1': 'hello',
 'abc': [1, 2, 3],
 'boo': [1, 'hi', 'moo', {'5': 'some'}],
 'def': {'hi': 'mom'}}

Notice that I never have to call the hook recursively since every object will get handed to the hook when you use the object_pairs_hook. You do have to care about lists, but as you can see, an object within a list will be properly converted, and you don’t have to recurse to make it happen.

EDIT: A coworker pointed out that Python2.6 doesn’t have object_hook_pairs. You can still use this will Python2.6 by making a very small change. In the hook above, change:

for key, value in pairs:


for key, value in pairs.iteritems():

Then use object_hook instead of object_pairs_hook:

In [66]: json.load(open('test.json'), object_hook=deunicodify_hook)
{'1': 'hello',
 'abc': [1, 2, 3],
 'boo': [1, 'hi', 'moo', {'5': 'some'}],
 'def': {'hi': 'mom'}}

Using object_pairs_hook results in one less dictionary being instantiated for each object in the JSON object, which, if you were parsing a huge document, might be worth while.

Answered By: Travis Jensen

Answer #8:

I’m afraid there’s no way to achieve this automatically within the simplejson library.

The scanner and decoder in simplejson are designed to produce unicode text. To do this, the library uses a function called c_scanstring (if it’s available, for speed), or py_scanstring if the C version is not available. The scanstring function is called several times by nearly every routine that simplejson has for decoding a structure that might contain text. You’d have to either monkeypatch the scanstring value in simplejson.decoder, or subclass JSONDecoder and provide pretty much your own entire implementation of anything that might contain text.

The reason that simplejson outputs unicode, however, is that the json spec specifically mentions that “A string is a collection of zero or more Unicode characters”… support for unicode is assumed as part of the format itself. Simplejson’s scanstring implementation goes so far as to scan and interpret unicode escapes (even error-checking for malformed multi-byte charset representations), so the only way it can reliably return the value to you is as unicode.

If you have an aged library that needs an str, I recommend you either laboriously search the nested data structure after parsing (which I acknowledge is what you explicitly said you wanted to avoid… sorry), or perhaps wrap your libraries in some sort of facade where you can massage the input parameters at a more granular level. The second approach might be more manageable than the first if your data structures are indeed deeply nested.

Answered By: Jarret Hardie

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