I have a deep nested object of various lists and dicts that I retrieve as json which I need to compare to another version of itself. The issue is that all lists are basically unsorted, therefore I need to sort before comparing them. Any deep diff library I've tried failed without proper sorting the dicts position in the lists, so here we go.
Sample object that requires sorting:
{
"main":{
"key1":"value1",
"key2":"value2",
"key3":[{
"sub1":"value2",
"sub2":{
"subsub":[{
"subsubsub1":10,
"subsubsub2":11,
"subsubsub3":[10,11,12]
},{
"subsubsub1":7,
"subsubsub2":8,
"subsubsub3":[9,7,8]
}]
}
},{
"sub1":"value1",
"sub2":{
"subsub":[{
"subsubsub1":1,
"subsubsub2":2,
"subsubsub3":[1,2,3]
},
{
"subsubsub1":4,
"subsubsub2":5,
"subsubsub3":[5,6,4]
}]
}
}]
}
}
Besides a few recursive loops I'm trying to sort the dicts by translating them with sorted lists into sorted tuples and hash them.
Edit:
The object is passed into unnest()
def unnest(d):
for k, v in d.items():
if isinstance(v, dict):
d.update({k: unnest(v)})
elif isinstance(v, list):
d.update({k: unsort(v)})
return d
def unsort(l):
for i, e in enumerate(l):
if isinstance(e, dict):
l[i] = unnest(e)
elif isinstance(e, list):
l[i] = unsort(e)
return sorted(l, key=lambda i: sort_hash(i))
def unnest_hash(d):
for k, v in d.items():
if isinstance(v, dict):
d.update({k: unnest_hash(v)})
elif isinstance(v, list):
d.update({k: sort_hash(v)})
return hash(tuple(sorted(d.items())))
def sort_hash(l):
if isinstance(l, list):
for i, e in enumerate(l):
if isinstance(e, dict):
l[i] = unnest_hash(e)
elif isinstance(e, list):
l[i] = sort_hash(e)
return hash(tuple(sorted(l)))
elif isinstance(l, dict):
return unnest_hash(l)
else:
return hash(l)
However for some reason the hash value gets written into the "sorted" list:
{'main': {'key1': 'value1', 'key2': 'value2', 'key3': [{'sub1': 'value2', 'sub2': -4046234112924644199}, {'sub1': 'value1', 'sub2': 4015568797712784641}]}}
How can I prevent the sort value in the lambda function to be written into the returned sorted list?
Thanks!
Your sort_hash function is mutating the value passed into it. That's why you see it in the original values are the sort:
l[i] = unnest_hash(e)
and
l[i] = sort_hash(e)
both modify the value you are trying to hash. unnest_hash also modifies the original values:
d.update({k: unnest_hash(v)})
A hash calculation for sorting must never modify the value it is hashing.
For example I have a dict:
{
"a" : 123,
"b" : {
"a" : 24324,
"c" : 9
},
"c" : {
"a" : 123,
"b" : 64
}
}
What is the best way to remove all a fields from this dictionary? I understand, that I can do an iteration over this dict, remove a field, that iterate on the keys, which are also dicts and so on.
But maybe there is more elegant way to do it?
EDIT
Thanks for your answers! But what about a situation, when keys can be lists, like "b" : [{"a" : 1, "c" : 2}]
Well, internally you have to process the items one by one, either by manual iteration or recursion.
Here's an attempt using recursion:
def remove_keys(d, to_remove):
if not isinstance(d, dict):
return d
return {k: remove_keys(v, to_remove)
for k, v in d.items() if k not in to_remove}
You would simply pass the dictionary to process as the first argument and the name of the keys to remove (as a set, preferably) as the second argument. In your case:
remove_keys(d, {"a"})
Whether it's more elegant or not, it depends on your taste, I would say :).
If you want to actually mutate the dictionary:
def remove_keys_m(d, to_remove):
if not isinstance(d, dict):
return d
for k in to_remove:
if k in d:
del d[k]
for k in d:
d[k] = remove_keys(d[k], to_remove)
return d
def remove_entries(d, k):
if k in d:
del d[k]
for value in d.values():
if isinstance(value, dict):
remove_entries(value, k)
Here's a pretty basic recursive function for doing it
Edit:
If you also want to handle lists nested in dicts, nested in lists, etc, something like the below will owrk.
def remove_from_dict_in_list(l, k):
for i in l:
if isinstance(i, list):
remove_from_dict_in_list(i, k)
elif isinstance(i, dict):
remove_entries(i, k)
def remove_entries(d, k):
if k in d:
del d[k]
for value in d.values():
if isinstance(value, dict):
remove_entries(value, k)
elif isinstance(value, list):
remove_from_dict_in_list(value, k)
While a recursive solution is more robust, here is a simpler solution that will work for the posted example and any other dictionary with only one nested layer:
s = {
"a" : 123,
"b" : {
"a" : 24324,
"c" : 9
},
"c" : {
"a" : 123,
"b" : 64
}
}
final_data = {a:{c:d for c, d in b.items() if c != "a"} for a, b in s.items() if a != "a"}
Output:
{'c': {'b': 64}, 'b': {'c': 9}}
I think this example will answer your question.
dic = {"a": 1,
"b": {"a": 2,
"c": {"a": 3}},
"d": 4}
key_to_del = "a"
def get_key(dic):
for key, value in dic.items():
if key == key_to_del:
del dic[key]
if isinstance(value, dict):
get_key(value)
return dic
print "BEFOR", dic
print "AFTER", get_key(dic)
enter image description here
original dictionary keys are all integers. How can I convert all the integer keys to strings using a shorter approach?
original = {1:{},2:{101:"OneZeroOne",202:"TwoZeroTwo"}}
result = {}
for key in original:
if not key in result:
result[str(key)]={}
for i, value in original[key].items():
result[str(key)][str(i)] = value
print result
prints:
{'1': {}, '2': {'202': 'TwoZeroTwo', '101': 'OneZeroOne'}}
Depending on what types of data you have:
original = {1:{},2:{101:"OneZeroOne",202:"TwoZeroTwo"}}
result= json.loads(json.dumps(original))
print(result)
prints:
{'2': {'101': 'OneZeroOne', '202': 'TwoZeroTwo'}, '1': {}}
def f(d):
new = {}
for k,v in d.items():
if isinstance(v, dict):
v = f(v)
new[str(k)] = v
return new
import json
original = {1:{},2:{101:"OneZeroOne",202:"TwoZeroTwo"}}
text = json.dumps(original)
json.loads(text)
out:
{'1': {}, '2': {'101': 'OneZeroOne', '202': 'TwoZeroTwo'}}
If you don't know the number of levels, then a recursive solution is probably best:
def convert_dict(d):
return {str(k): convert_value(v) for k,v in d.items()}
def convert_list(lst):
return [convert_value(item) for item in lst]
def convert_value(v):
if isinstance(v, dict):
return convert_dict(v)
elif isinstance(v, list):
return convert_list(v)
# more elifs..
else:
return v
if you know that all values are either dicts or simple values, then you can remove all the elifs and the convert_list function
for k, v in d.iteritems():
if type(v) is dict:
for t, c in v.iteritems():
print "{0} : {1}".format(t, c)
I'm trying to loop through a dictionary and print out all key value pairs where the value is not a nested dictionary. If the value is a dictionary I want to go into it and print out its key value pairs...etc. Any help?
EDIT
How about this? It still only prints one thing.
def printDict(d):
for k, v in d.iteritems():
if type(v) is dict:
printDict(v)
else:
print "{0} : {1}".format(k, v)
Full Test Case
Dictionary:
{u'xml': {u'config': {u'portstatus': {u'status': u'good'}, u'target': u'1'},
u'port': u'11'}}
Result:
xml : {u'config': {u'portstatus': {u'status': u'good'}, u'target': u'1'}, u'port': u'11'}
As said by Niklas, you need recursion, i.e. you want to define a function to print your dict, and if the value is a dict, you want to call your print function using this new dict.
Something like :
def myprint(d):
for k, v in d.items():
if isinstance(v, dict):
myprint(v)
else:
print("{0} : {1}".format(k, v))
There are potential problems if you write your own recursive implementation or the iterative equivalent with stack. See this example:
dic = {}
dic["key1"] = {}
dic["key1"]["key1.1"] = "value1"
dic["key2"] = {}
dic["key2"]["key2.1"] = "value2"
dic["key2"]["key2.2"] = dic["key1"]
dic["key2"]["key2.3"] = dic
In the normal sense, nested dictionary will be a n-nary tree like data structure. But the definition doesn't exclude the possibility of a cross edge or even a back edge (thus no longer a tree). For instance, here key2.2 holds to the dictionary from key1, key2.3 points to the entire dictionary(back edge/cycle). When there is a back edge(cycle), the stack/recursion will run infinitely.
root<-------back edge
/ \ |
_key1 __key2__ |
/ / \ \ |
|->key1.1 key2.1 key2.2 key2.3
| / | |
| value1 value2 |
| |
cross edge----------|
If you print this dictionary with this implementation from Scharron
def myprint(d):
for k, v in d.items():
if isinstance(v, dict):
myprint(v)
else:
print "{0} : {1}".format(k, v)
You would see this error:
> RuntimeError: maximum recursion depth exceeded while calling a Python object
The same goes with the implementation from senderle.
Similarly, you get an infinite loop with this implementation from Fred Foo:
def myprint(d):
stack = list(d.items())
while stack:
k, v = stack.pop()
if isinstance(v, dict):
stack.extend(v.items())
else:
print("%s: %s" % (k, v))
However, Python actually detects cycles in nested dictionary:
print dic
{'key2': {'key2.1': 'value2', 'key2.3': {...},
'key2.2': {'key1.1': 'value1'}}, 'key1': {'key1.1': 'value1'}}
"{...}" is where a cycle is detected.
As requested by Moondra this is a way to avoid cycles (DFS):
def myprint(d):
stack = list(d.items())
visited = set()
while stack:
k, v = stack.pop()
if isinstance(v, dict):
if k not in visited:
stack.extend(v.items())
else:
print("%s: %s" % (k, v))
visited.add(k)
Since a dict is iterable, you can apply the classic nested container iterable formula to this problem with only a couple of minor changes. Here's a Python 2 version (see below for 3):
import collections
def nested_dict_iter(nested):
for key, value in nested.iteritems():
if isinstance(value, collections.Mapping):
for inner_key, inner_value in nested_dict_iter(value):
yield inner_key, inner_value
else:
yield key, value
Test:
list(nested_dict_iter({'a':{'b':{'c':1, 'd':2},
'e':{'f':3, 'g':4}},
'h':{'i':5, 'j':6}}))
# output: [('g', 4), ('f', 3), ('c', 1), ('d', 2), ('i', 5), ('j', 6)]
In Python 2, It might be possible to create a custom Mapping that qualifies as a Mapping but doesn't contain iteritems, in which case this will fail. The docs don't indicate that iteritems is required for a Mapping; on the other hand, the source gives Mapping types an iteritems method. So for custom Mappings, inherit from collections.Mapping explicitly just in case.
In Python 3, there are a number of improvements to be made. As of Python 3.3, abstract base classes live in collections.abc. They remain in collections too for backwards compatibility, but it's nicer having our abstract base classes together in one namespace. So this imports abc from collections. Python 3.3 also adds yield from, which is designed for just these sorts of situations. This is not empty syntactic sugar; it may lead to faster code and more sensible interactions with coroutines.
from collections import abc
def nested_dict_iter(nested):
for key, value in nested.items():
if isinstance(value, abc.Mapping):
yield from nested_dict_iter(value)
else:
yield key, value
Alternative iterative solution:
def myprint(d):
stack = d.items()
while stack:
k, v = stack.pop()
if isinstance(v, dict):
stack.extend(v.iteritems())
else:
print("%s: %s" % (k, v))
Slightly different version I wrote that keeps track of the keys along the way to get there
def print_dict(v, prefix=''):
if isinstance(v, dict):
for k, v2 in v.items():
p2 = "{}['{}']".format(prefix, k)
print_dict(v2, p2)
elif isinstance(v, list):
for i, v2 in enumerate(v):
p2 = "{}[{}]".format(prefix, i)
print_dict(v2, p2)
else:
print('{} = {}'.format(prefix, repr(v)))
On your data, it'll print
data['xml']['config']['portstatus']['status'] = u'good'
data['xml']['config']['target'] = u'1'
data['xml']['port'] = u'11'
It's also easy to modify it to track the prefix as a tuple of keys rather than a string if you need it that way.
Here is pythonic way to do it. This function will allow you to loop through key-value pair in all the levels. It does not save the whole thing to the memory but rather walks through the dict as you loop through it
def recursive_items(dictionary):
for key, value in dictionary.items():
if type(value) is dict:
yield (key, value)
yield from recursive_items(value)
else:
yield (key, value)
a = {'a': {1: {1: 2, 3: 4}, 2: {5: 6}}}
for key, value in recursive_items(a):
print(key, value)
Prints
a {1: {1: 2, 3: 4}, 2: {5: 6}}
1 {1: 2, 3: 4}
1 2
3 4
2 {5: 6}
5 6
A alternative solution to work with lists based on Scharron's solution
def myprint(d):
my_list = d.iteritems() if isinstance(d, dict) else enumerate(d)
for k, v in my_list:
if isinstance(v, dict) or isinstance(v, list):
myprint(v)
else:
print u"{0} : {1}".format(k, v)
I am using the following code to print all the values of a nested dictionary, taking into account where the value could be a list containing dictionaries. This was useful to me when parsing a JSON file into a dictionary and needing to quickly check whether any of its values are None.
d = {
"user": 10,
"time": "2017-03-15T14:02:49.301000",
"metadata": [
{"foo": "bar"},
"some_string"
]
}
def print_nested(d):
if isinstance(d, dict):
for k, v in d.items():
print_nested(v)
elif hasattr(d, '__iter__') and not isinstance(d, str):
for item in d:
print_nested(item)
elif isinstance(d, str):
print(d)
else:
print(d)
print_nested(d)
Output:
10
2017-03-15T14:02:49.301000
bar
some_string
Your question already has been answered well, but I recommend using isinstance(d, collections.Mapping) instead of isinstance(d, dict). It works for dict(), collections.OrderedDict(), and collections.UserDict().
The generally correct version is:
def myprint(d):
for k, v in d.items():
if isinstance(v, collections.Mapping):
myprint(v)
else:
print("{0} : {1}".format(k, v))
Iterative solution as an alternative:
def traverse_nested_dict(d):
iters = [d.iteritems()]
while iters:
it = iters.pop()
try:
k, v = it.next()
except StopIteration:
continue
iters.append(it)
if isinstance(v, dict):
iters.append(v.iteritems())
else:
yield k, v
d = {"a": 1, "b": 2, "c": {"d": 3, "e": {"f": 4}}}
for k, v in traverse_nested_dict(d):
print k, v
Here's a modified version of Fred Foo's answer for Python 2. In the original response, only the deepest level of nesting is output. If you output the keys as lists, you can keep the keys for all levels, although to reference them you need to reference a list of lists.
Here's the function:
def NestIter(nested):
for key, value in nested.iteritems():
if isinstance(value, collections.Mapping):
for inner_key, inner_value in NestIter(value):
yield [key, inner_key], inner_value
else:
yield [key],value
To reference the keys:
for keys, vals in mynested:
print(mynested[keys[0]][keys[1][0]][keys[1][1][0]])
for a three-level dictionary.
You need to know the number of levels before to access multiple keys and the number of levels should be constant (it may be possible to add a small bit of script to check the number of nesting levels when iterating through values, but I haven't yet looked at this).
I find this approach a bit more flexible, here you just providing generator function that emits key, value pairs and can be easily extended to also iterate over lists.
def traverse(value, key=None):
if isinstance(value, dict):
for k, v in value.items():
yield from traverse(v, k)
else:
yield key, value
Then you can write your own myprint function, then would print those key value pairs.
def myprint(d):
for k, v in traverse(d):
print(f"{k} : {v}")
A test:
myprint({
'xml': {
'config': {
'portstatus': {
'status': 'good',
},
'target': '1',
},
'port': '11',
},
})
Output:
status : good
target : 1
port : 11
I tested this on Python 3.6.
Nested dictionaries looping using isinstance() and yield function.
**isinstance is afunction that returns the given input and reference is true or false as in below case dict is true so it go for iteration.
**Yield is used to return from a function without destroying the states of its local variable and when the function is called, the execution starts from the last yield statement. Any function that contains a yield keyword is termed a generator.
students= {'emp1': {'name': 'Bob', 'job': 'Mgr'},
'emp2': {'name': 'Kim', 'job': 'Dev','emp3': {'namee': 'Saam', 'j0ob': 'Deev'}},
'emp4': {'name': 'Sam', 'job': 'Dev'}}
def nested_dict_pairs_iterator(dict_obj):
for key, value in dict_obj.items():
# Check if value is of dict type
if isinstance(value, dict):
# If value is dict then iterate over all its values
for pair in nested_dict_pairs_iterator(value):
yield (key, *pair)
else:
# If value is not dict type then yield the value
yield (key, value)
for pair in nested_dict_pairs_iterator(students):
print(pair)
For a ready-made solution install ndicts
pip install ndicts
Import a NestedDict in your script
from ndicts.ndicts import NestedDict
Initialize
dictionary = {
u'xml': {
u'config': {
u'portstatus': {u'status': u'good'},
u'target': u'1'
},
u'port': u'11'
}
}
nd = NestedDict(dictionary)
Iterate
for key, value in nd.items():
print(key, value)
While the original solution from #Scharron is beautiful and simple, it cannot handle the list very well:
def myprint(d):
for k, v in d.items():
if isinstance(v, dict):
myprint(v)
else:
print("{0} : {1}".format(k, v))
So this code can be slightly modified like this to handle list in elements:
def myprint(d):
for k, v in d.items():
if isinstance(v, dict):
myprint(v)
elif isinstance(v, list):
for i in v:
myprint(i)
else:
print("{0} : {1}".format(k, v))
These answers work for only 2 levels of sub-dictionaries. For more try this:
nested_dict = {'dictA': {'key_1': 'value_1', 'key_1A': 'value_1A','key_1Asub1': {'Asub1': 'Asub1_val', 'sub_subA1': {'sub_subA1_key':'sub_subA1_val'}}},
'dictB': {'key_2': 'value_2'},
1: {'key_3': 'value_3', 'key_3A': 'value_3A'}}
def print_dict(dictionary):
dictionary_array = [dictionary]
for sub_dictionary in dictionary_array:
if type(sub_dictionary) is dict:
for key, value in sub_dictionary.items():
print("key=", key)
print("value", value)
if type(value) is dict:
dictionary_array.append(value)
print_dict(nested_dict)
You can print recursively with a dictionary comprehension:
def print_key_pairs(d):
{k: print_key_pairs(v) if isinstance(v, dict) else print(f'{k}: {v}') for k, v in d.items()}
For your test case this is the output:
>>> print_key_pairs({u'xml': {u'config': {u'portstatus': {u'status': u'good'}, u'target': u'1'}, u'port': u'11'}})
status: good
target: 1
port: 11
Returns a tuple of each key and value and the key contains the full path
from typing import Mapping, Tuple, Iterator
def traverse_dict(nested: Mapping, parent_key="", keys_to_not_traverse_further=tuple()) -> Iterator[Tuple[str, str]]:
"""Each key is joined with it's parent using dot as a separator.
Once a `parent_key` matches `keys_to_not_traverse_further`
it will no longer find its child dicts.
"""
for key, value in nested.items():
if isinstance(value, abc.Mapping) and key not in keys_to_not_traverse_further:
yield from traverse_dict(value, f"{parent_key}.{key}", keys_to_not_traverse_further)
else:
yield f"{parent_key}.{key}", value
Let's test it
my_dict = {
"isbn": "123-456-222",
"author": {"lastname": "Doe", "firstname": "Jane"},
"editor": {"lastname": "Smith", "firstname": "Jane"},
"title": "The Ultimate Database Study Guide",
"category": ["Non-Fiction", "Technology"],
"first": {
"second": {"third": {"fourth": {"blah": "yadda"}}},
"fifth": {"sixth": "seventh"},
},
}
for k, v in traverse_dict(my_dict):
print(k, v)
Returns
.isbn 123-456-222
.author.lastname Doe
.author.firstname Jane
.editor.lastname Smith
.editor.firstname Jane
.title The Ultimate Database Study Guide
.category ['Non-Fiction', 'Technology']
.first.second.third.fourth.blah yadda
.first.fifth.sixth seventh
If you don't care about some child dicts e.g names in this case then
use the keys_to_not_traverse_further
for k, v in traverse_dict(my_dict, parent_key="", keys_to_not_traverse_further=("author","editor")):
print(k, v)
Returns
.isbn 123-456-222
.author {'lastname': 'Doe', 'firstname': 'Jane'}
.editor {'lastname': 'Smith', 'firstname': 'Jane'}
.title The Ultimate Database Study Guide
.category ['Non-Fiction', 'Technology']
.first.second.third.fourth.blah yadda
.first.fifth.sixth seventh