Printing a particular subset of keys in a dictionary

I have a dictionary in Python where the keys are pathnames. For example:

dict["/A"] = 0
dict["/A/B"] = 1
dict["/A/C"] = 1

dict["/X"] = 10
dict["/X/Y"] = 11

I was wondering, what's a good way to print all "subpaths" given any key.

For example, given a function called "print_dict_path" that does this, something like

print_dict_path("/A")

or

print_dict_path("/A/B")

would print out something like:

"B" = 1
"C" = 1

The only method I开发者_运维技巧 can think of is something like using regex and going through the entire dictionary, but I'm not sure if that's the best method (nor am I that well versed in regex).

Thanks for any help.

One possibility without using regex is to just use startswith

top_path = '/A/B'
for p in d.iterkeys():
    if p.startswith(top_path):
        print d[p]

You can use str.find:

def print_dict_path(prefix, d):
    for k in d:
        if k.find(prefix) == 0:
            print "\"{0}\" = {1}".format(k,d[k])

Well, you'll definitely have to loop through the entire dict.

def filter_dict_path( d, sub ):
    for key, val in d.iteritems():
        if key.startswith(sub): ## or do you want `sub in key` ?
            yield key, val

print dict(filter_dict_path( old_dict, sub ))

You could speed this up by using the appropriate data structure: a Tree.

Is your dictionary structure fixed? It would be nicer to do this using nested dictionaries:

{
    "A": {
        "value": 0
        "dirs": {
            "B": {
                "value": 1
            }
            "C": {
                "value": 1
            }
        }
    "X": {
        "value": 10
        "dirs": {
            "Y": {
                "value": 11
            }
}

The underlying data structure here is a tree, but Python doesn't have that built in.

This removes one level of indenting, which may make the code in the body of the for loop more readable in some cases

top_path = '/A/B'
for p in (p for p in d.iterkeys() if p.startswith(top_path)):
    print d[p]

If you find performance to be a problem, consider using a trie instead of the dictionary