I am trying to convert a list of dot-separated strings, e.g.
['one.two.three.four', 'one.six.seven.eight', 'five.nine.ten', 'twelve.zero']
into a tree (nested lists or dicts - anything that is easy to walk through). The real data happens to have 1 to 4 dot-separated parts of different length and has 2200 records in total. My actual goal is to fill in the set of 4 QComboBox'es with this data, in manner that the 1st QComboBox is filled with first set items ['one', 'five', 'twelve'] (no duplicates). Then depending on the chosen item, the 2nd QComboBox is filled with its related items: for 'one' it would be: ['two', 'six'], and so on, if there's another nested level.
So far I've got a working list -> nested dicts solution, but it's horribly slow, since I use regular dict(). And I seem to have a trouble to redesign it to a defaultdict in a way to easily work out filling the ComboBoxes properly.
My current code:
def list2tree(m):
tmp = {}
for i in range(len(m)):
if m.count('.') == 0:
return m
a = m.split('.', 1)
try:
tmp[a[0]].append(list2tree(a[1]))
except (KeyError, AttributeError):
tmp[a[0]] = list2tree(a[1])
return tmp
main_dict = {}
i = 0
for m in methods:
main_dict = list2tree(m)
i += 1
if (i % 100) == 0: print i, len(methods)
print main_dict, i, len(methods)
You can convert a Python list to a dictionary using a dictionary comprehension, dict. fromkeys(), or the zip() method. All three methods create a new dictionary. They do not modify the existing list.
In Python, a Nested dictionary can be created by placing the comma-separated dictionaries enclosed within braces.
Access Values using get() Another way to access value(s) in a nested dictionary ( employees ) is to use the dict. get() method. This method returns the value for a specified key. If the specified key does not exist, the get() method returns None (preventing a KeyError ).
You can make use of the eval function in python. nestq will be "nest['b']['v']['y']" where nest is the nested dictionary. The eval builtin function executes the given string. However, it is important to be careful about possible vulnerabilities that arise from use of eval function.
ls = ['one.two.three.four', 'one.six.seven.eight', 'five.nine.ten', 'twelve.zero']
tree = {}
for item in ls:
t = tree
for part in item.split('.'):
t = t.setdefault(part, {})
Result:
{
"twelve": {
"zero": {}
},
"five": {
"nine": {
"ten": {}
}
},
"one": {
"six": {
"seven": {
"eight": {}
}
},
"two": {
"three": {
"four": {}
}
}
}
}
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