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Python: Usage of anonymous list within dict of other values?

I am asked to generate and also later read back a json object which looks like:
{"name":"somename",
[{"id":123,"key1":"anydata"},
{"id":345,"key1":"x","key3":"yz"}]}
Normally, I would use a python dict and convert it to/from json. Here however is the problem, that both, list and sub-dict are anonymous.
I think it is not possible to make a dict like this, is it?

This is not a valid json object but you can just add a key for the list part. Like:
{"name":"somename","value" : [{"id":123,"key1":"anydata"},{"id":345,"key1":"x","key3":"yz"}]}
Now this is a valid json string.
>>> a = ast.literal_eval('{"name":"somename","value" : [{"id":123,"key1":"anydata"},{"id":345,"key1":"x","key3":"yz"}]}')
>>> print(a['name'])
>>> 'somename'
>>> print(a['value'][0]['id'])
>>> 123
For a variable list its is as simple as:
anon_list = [{"id":123,"key1":"anydata"},{"id":345,"key1":"x","key3":"yz"}]
a = {"name":"somename","value" : anon_list}
>>> print(a['name'])
>>> 'somename'
>>> print(a['value'][0]['id'])
>>> 123

how to create a dictionary from a set of properly formatted tuples in python

Is there a simple way to create a dictionary from a list of formatted tuples. e.g. if I do something like:
d={"responseStatus":"SUCCESS","sessionId":"01234","userId":2000004904}
This creates a dictionary called d. However, if I want to create a dictionary from a string which contains the same string, I can't do that
res=<some command that returns {"responseStatus":"SUCCESS","sessionId":"01234","userId":2000004904}>
print res
# returns {"responseStatus":"SUCCESS","sessionId":"01234","userId":2000004904}
d=dict(res)
This throws an error that says:
ValueError: dictionary update sequence element #0 has length 1; 2 is required

I strongly strongly suspect that you have json on your hands.
import json
d = json.loads('{"responseStatus":"SUCCESS","sessionId":"01234","userId":2000004904}')
would give you what you want.

Use dict(zip(tuples))
>>> u = ("foo", "bar")
>>> v = ("blah", "zoop")
>>> d = dict(zip(u, v))
>>> d
{'foo': 'blah', 'bar': 'zoop'}
Note, if you have an odd number of tuples this will not work.

Based on what you gave is, res is
# returns {"responseStatus":"SUCCESS","sessionId":"01234","userId":2000004904}
So the plan is to grab the string starting at the curly brace to the end and use json to decode it:
import json
# Discard the text before the curly brace
res = res[res.index('{'):]
# Turn that text into a dictionary
d = json.loads(res)

All you need to do in your particular case is
d = eval(res)
And please keep security in mind when using eval, especially if you're mixing it with ajax/json.
UPDATE
Since others pointed out you might be getting this data over the web and it isn't just a "how to make this work" question, use this:
import json
json.loads(res)

Why don't Python dictionaries treat keys independently in this script? [duplicate]

This question already has answers here:
How to copy a dictionary and only edit the copy
(23 answers)
Closed 8 years ago.
I'm expecting my frustration to be overridden with some enlightenment - here's a minimal version of the script to demonstrate the problem:
First I create a dictionary:
dic = {
'foo':{},
'bar':{}
}
Then we instantiate a template dictionary that can be iteratively appended
to keys of dic:
appendic= {
'is':'', # '' is a terminal value to be replaced later
}
So here we append appendic to each of the keys in dic:
dic['foo'] = appendic
dic['bar'] = appendic
Now we replace the terminal values, '', with something meaningful:
dic['foo']['is'] = 'foo'
dic['bar']['is'] = 'bar'
At this point, my intuition tells me that if we call:
print(dic['foo']['is']) we get 'foo'
But instead Python returns 'bar' ... to my un-trained mind that is counter-intuitive.
Questions:
How can I tell Python to keep the keys of dic independent?
Why is this the default behaviour? What use cases does this have?

When you assign a appendic to two different keys, Python doesn't make a copy. It assigns a reference instead.
As a result, both dic['please_make_me_Foo'] and dic['dont_make_him_Bar'] refer to the same object. These are not separate dictionaries, they are both the same object, the one appendic also references to.
If you expected these to be separate dictionaries, create a copy of appendic instead. The dict.copy() method creates a shallow copy of a dictionary:
dic['please_make_me_Foo']= appendic.copy()
dic['dont_make_him_Bar'] = appendic.copy()
Shallow means that a new dictionary is created and all references to keys and values contained are copied over.
If appendic itself contains values that are also dictionaries, these would not be copied. The new copy and appendic would both refer to the same values. In most cases, that's not a problem because most primitive values (strings, integers, etc.) are immutable, and you never notice references are shared as you replace such values with new ones.

You make a dict:
appendic= {
'Python_made_me':''
}
Add it to your other dict twice
dic['please_make_me_Foo']= appendic
dic['dont_make_him_Bar'] = appendic
And set the single dict's Python_made_me value twice
dic['please_make_me_Foo']['Python_made_me'] = 'Foo'
dic['dont_make_him_Bar']['Python_made_me'] = 'Bar'
But because they're the same dict, the second line overwrites the first
If you need to copy it, you need to use the copy method:
dic['please_make_me_Foo']= appendic.copy()
dic['dont_make_him_Bar'] = appendic.copy()

ok, I'm just going to write this as a complement to the other answers. When you manipulate a dictionary, you manipulate the reference to an instance, which is the root cause of your mistake. Using hex(id(foo)) you get the memory address of foo, so let's show the address of d instance in the following example to make that tangible:
>>> hex(id(d))
'0x10bd95e60'
>>> hex(id(e[1]))
'0x10bd95e60'
>>> hex(id(f[1]))
'0x10bd95e60'
so if you add or remove values from e[1], you're actually changing the same instance as the one pointed by d, and as a dictionary is mutable, i.e. you can change values within.
Now you're wondering why that does not happen when you're handling integers? Because, in fact it does, it's just that integers are not mutable:
>>> i = 1
>>> hex(id(i))
'0x10ba51e90'
>>> j = i
>>> hex(id(j))
'0x10ba51e90'
>>> i = 2
>>> hex(id(i))
'0x10ba51eb0'
i.e. i is pointing to another place in the memory.
It's possible to create a mutable integer though, by using a class:
>>> class Integer:
... def __init__(self, i):
... self.i = i
...
>>> i = Integer(2)
>>> hex(id(i))
'0x10bd9b410'
>>> j = i
>>> hex(id(j))
'0x10bd9b410'
>>> j.i = 2
>>> i.i
2
>>> hex(id(i))
'0x10bd9b410'
In order to create a new instance of the same dictionary, you need to use the copy() member of a dict:
>>> hex(id(d))
'0x10bd95e60'
>>> w = d.copy()
>>> x = d.copy()
>>> y = d.copy()
>>> hex(id(w))
'0x10bd96128'
>>> hex(id(x))
'0x10bd95f80'
>>> hex(id(y))
'0x10bd96098'

dic['please_make_me_Foo']= appendic
dic['dont_make_him_Bar'] = appendic
appendic is an object - you are assigning a reference to the same object to both keys in dic. So when you change one, you change both.
Try this instead:
dic['please_make_me_Foo']= appendic.copy()
dic['dont_make_him_Bar'] = appendic.copy()

Equivalent of "genvarname" in Python

How can I generate a variable name from a string (say a concatenation of a letter and a number) ?
In Matlab, this task can be easily done using genvarname

Here's a really bad way (undefined behavior), but I think it shows the path to a more reasonable technique.
Your current namespace is really a dictionary under the covers:
>>> local_namespace = locals()
>>> name = "myVarName"
>>> local_namespace[name] = 'VarData'
>>> myVarName
'VarData'
But that's not very DRY - you have to write the name of the variable twice! It would be nice to use a variable that stored the name of our dynamically created variable so we didn't have to type it twice:
>>> name
'myVarName'
obviously doesn't work for this. But we can use our dictionary again:
>>> local_namespace[name]
'VarData'
So now we can store and recall the value associated with our variable. But wait - there's no need to use the special locals() dictionary for this - an ordinary dictionary will do!
>>> d = {}
>>> d[name] = 'VarData'
>>> d[name]
'VarData'
And now we have all these added benefits, like being able to keep track of the names of several of these variables in a list:
>>> l = []
>>> l.append('myVarName')
>>> l.append('anotherVarName')
Dictionaries even do this for us:
>>> d['anotherVarName'] = 123
>>> d.keys()
['myVarName', 'anotherVarName']
Unless you're doing terrifically wacky things, it's hard to imagine how constructing variable names could be more useful than using a dictionary.

You can use exec("").
But you really(!!!) don't want to.
>>> name="myVarName"
>>> exec(name+"='VarData'")
>>> myVarName
'VarData'

why does updating a dict that was appended to a list change the list?

My code will be more clear I think-
someList = list()
foo = {'a':'b'}
someList.append(foo)
print someList
>>> [{'a':'b'}]
defaultbazz = {'a':2, 'b':'t', 'c':'gg'}
for k, v in defaultbazz.iteritems():
foo[k] = v
print someList
>>> [{'a': 2, 'c': 'gg', 'b': 't'}]
Shouldn't the last print be [{'a':'b'}]? I didn't updated the someList, I want it as is..
It's seems to me uninterpreted behavior..
But if that's how python works, how can I find workaround? Even setting a new dict updates the original one dict.. I mean:
someList = list()
foo = {'a':'b'}
someList.append(foo)
print someList
>>> [{'a':'b'}]
bar = foo
defaultbazz = {'a':2, 'b':'t', 'c':'gg'}
for k, v in defaultbazz.iteritems():
bar[k] = v
print someList
>>> [{'a': 2, 'c': 'gg', 'b': 't'}]
I'll be thankful if someone can maybe explain me why it's happen..

It looks like you are expecting your dict to be copied when you add it to a list or assign it to a new variable, but that is not how Python operates. If you assign a dict -- actually, if you assign any object -- you are not creating a new object, but instead you are simply giving your object a new name. (An object can have multiple names.)
So, when you edit your object under the new name, the single instance of that object changes, and that change is visible when you access the object through any name.
If you want to copy your object, then you can do this:
bar = dict(foo)
or
bar = foo.copy()

To simplify:
a = {2: 3}
b = [a]
b contains a "reference" to a (and is a dict which is mutable) - so if a is modified then accessing a via the list b, will display the modified a.
You have to explicitly create a copy of a, which can be done in this case as:
b = [dict(a)]
But you should look at the copy module for copy.copy() and copy.deepcopy()

Dictionaries are mutable objects, hence the result of your script.
I guess you want a new object, i.e. a copy of the original one:
import copy
someList.append(copy.copy(foo))

Variables in Python are just names of objects. If you change the object from any name "attached" to it, you will see the changes from every other name. Python never creates copies automatically for you, in particular:
someList.append(foo)
doesn't create a copy of foo and put it on someList, it appends the object that the name foo refers to onto the list.
You can create a second name for this object
bar = foo
but this does not create a copy either. In particular
foo['x'] = 42
and
bar['x'] = 42
will then operate on exactly the same object. You can verify this by printing the memory address of the object:
print id(foo), id(bar)
and see that they are the same.
If you need a copy in Python, you'll need to create one explicitly. Depending on what you need, the copy module -- either copy.copy() or copy.deepcopy() -- will do what you want:
import copy
bar = copy.copy(foo)
print id(foo), id(bar)
should now print different memory locations.

Dicts are mutable, which means that they can change. It's because foo is inside of someList and you're changing foo in the for-loop. Take a look at this simple example:
a_dict = {'a':'b'}
a_list = [a_dict]
print a_list # [{'a':'b'}]
#change the dict
a_dict['a'] = 'c'
print a_list # [{'a':'c'}]