Multi-Level Defaultdict with Variable Depth

Multi-level defaultdict with variable depth?

You can do it without even defining a class:

from collections import defaultdict

nested_dict = lambda: defaultdict(nested_dict)
nest = nested_dict()

nest[0][1][2][3][4][5] = 6

Nested defaultdict of defaultdict

For an arbitrary number of levels:

def rec_dd():
    return defaultdict(rec_dd)

>>> x = rec_dd()
>>> x['a']['b']['c']['d']
defaultdict(<function rec_dd at 0x7f0dcef81500>, {})
>>> print json.dumps(x)
{"a": {"b": {"c": {"d": {}}}}}

Of course you could also do this with a lambda, but I find lambdas to be less readable. In any case it would look like this:

rec_dd = lambda: defaultdict(rec_dd)

Multiple levels of 'collection.defaultdict' in Python

Use:

from collections import defaultdict
d = defaultdict(lambda: defaultdict(int))

This will create a new defaultdict(int) whenever a new key is accessed in d.

Nested defaultdicts

The issue here is that defaultdict accepts a callable, which is used as a factory to generate the value when a key is missing. Once you understand that, the behaviour is clear:

x = defaultdict(dict)
x                    # it's a default dict
x['a']               # it's just a dict()
x['a']['b'] = 'c'    # it's just setting the 'b' item in the dict x['a']
x['a']['b']['z']     # oops, error, because x['a']['b'] is not a dict!

If you only require a finite level of nesting, using a plain old dict with tuple keys is usually a much easier data structure to work with. That will work fine for both the 2-d and 3-d examples shown in your question.

If you require arbitrary levels of nesting, however, you can consider the recursive defaultdict example shown here.

How can I set the position of my datagrid scrollbar in my winforms app?

You don't actually interact directly with the scrollbar, rather you set the FirstDisplayedScrollingRowIndex. So before it reloads, capture that index, once it's reloaded, reset it to that index.

EDIT: Good point in the comment. If you're using a DataGridView then this will work. If you're using the old DataGrid then the easiest way to do that is to inherit from it. See here: Linkage

The DataGrid has a protected GridVScrolled method that can be used to scroll the grid to a specific row. To use it, derive a new grid from the DataGrid and add a ScrollToRow method.

C# code

public void ScrollToRow(int theRow)
{
    //
    // Expose the protected GridVScrolled method allowing you
    // to programmatically scroll the grid to a particular row.
    //
    if (DataSource != null)
    {
        GridVScrolled(this, new ScrollEventArgs(ScrollEventType.LargeIncrement, theRow));
    }
}

Modify multi-level dictionaries

I was playing around with the idea of using multiple indexes, and a defaultdict. And this came out:

from collections import defaultdict

class LayeredDict(defaultdict):
    def __getitem__(self, key):
        if isinstance(key, (tuple, list)):
            if len(key) == 1:
                return self[key[0]]
            return self[key[0]][key[1:]]
        return super(LayeredDict, self).__getitem__(key)

    def __setitem__(self, key, value):
        if isinstance(key, (tuple, list)):
            if len(key) == 1:
                self[key[0]] = value
            else:
                self[key[0]][key[1:]] = value
        else:
            super(LayeredDict, self).__setitem__(key, value)

    def __init__(self, *args, **kwargs):
        super(LayeredDict, self).__init__(*args, **kwargs)
        self.default_factory = type(self)  # override default

I haven't fully tested it, but it should allow you to create any level of nested dictionaries, and index them with a tuple.

>>> x = LayeredDict()
>>> x['abc'] = 'blah'
>>> x['abc']
'blah'
>>> x[0, 8, 2] = 1.2345
>>> x[0, 8, 1] = 8.9
>>> x[0, 8, 'xyz'] = 10.1
>>> x[0, 8].keys()
[1, 2, 'xyz']
>>> x['abc', 1] = 5
*** TypeError: 'str' object does not support item assignment

Unfortunately expansion notation (or whatever it's called) isn't supported, but
you can just pass a list or tuple in as an index.

>>> keylist = (0, 8, 2)
>>> x[*keylist]
*** SyntaxError: invalid syntax (<stdin>, line 1)
>>> x[keylist]
1.2345

Also, the isinstance(key, (tuple, list)) condition means a tuple can't be used as a key.

How to update values in nested dictionary if keys are in a list?

Predefined dictionary structure: functools.reduce

You can define a function using functools.reduce to apply getitem repeatedly and then set a supplied value:

from functools import reduce
from operator import getitem

def set_nested_item(dataDict, mapList, val):
    """Set item in nested dictionary"""
    reduce(getitem, mapList[:-1], dataDict)[mapList[-1]] = val
    return dataDict

key_lst = ["key1", "key2", "key3"]
value = "my_value"
d = {"key1": {"key2": {"key3": "some_value"}}}

d = set_nested_item(d, key_lst, value)

print(d)
# {'key1': {'key2': {'key3': 'my_value'}}}

Note operator.getitem is used to access dict.__getitem__, or its more commonly used syntactic sugar dict[]. In this instance, functools.reduce calls getitem recursively on dataDict, successively using each value in mapList[:-1] as an argument. With [:-1], we intentionally leave out the last value, so we can use __setitem__ via dict[key] = value for the final key.

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Arbitrary dictionary nesting: collections.defaultdict

If you wish to add items at arbitrary branches not yet been defined, you can construct a defaultdict. For this, you can first defaultify your regular dictionary input, then use set_nested_item as before:

from collections import defaultdict

def dd_rec():
    return defaultdict(dd_rec)

def defaultify(d):
    if not isinstance(d, dict):
        return d
    return defaultdict(dd_rec, {k: defaultify(v) for k, v in d.items()})

dd = defaultify(d)

key_lst = ["key1", "key2", "key5", "key6"]
value = "my_value2"
dd = set_nested_item(dd, key_lst, value)

print(dd)

# defaultdict(<function __main__.<lambda>>,
#             {'key1': defaultdict(<function __main__.<lambda>>,
#                          {'key2': defaultdict(<function __main__.<lambda>>,
#                                       {'key3': 'my_value',
#                                        'key5': defaultdict(<function __main__.<lambda>>,
#                                                    {'key6': 'my_value2'})})})})

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