I have two or more dictionary, I like to merge it as one with retaining multiple values of the same key as list. I would not able to share the original code, so please help me with the following example.
Input:
a= {'a':1, 'b': 2}
b= {'aa':4, 'b': 6}
c= {'aa':3, 'c': 8}
Output:
c= {'a':1,'aa':[3,4],'b': [2,6], 'c': 8}
I suggest you read up on the defaultdict: it lets you provide a factory method that initializes missing keys, i.e. if a key is looked up but not found, it creates a value by calling factory_method(missing_key). See this example, it might make things clearer:
from collections import defaultdict
a = {'a': 1, 'b': 2}
b = {'aa': 4, 'b': 6}
c = {'aa': 3, 'c': 8}
stuff = [a, b, c]
# our factory method is the list-constructor `list`,
# so whenever we look up a value that doesn't exist, a list is created;
# we can always be sure that we have list-values
store = defaultdict(list)
for s in stuff:
for k, v in s.items():
# since we know that our value is always a list, we can safely append
store[k].append(v)
print(store)
This has the "downside" of creating one-element lists for single occurences of values, but maybe you are able to work around that.
Please find below to resolve your issue. I hope this would work for you.
from collections import defaultdict
a = {'a':1, 'b': 2}
b = {'aa':4, 'b': 6}
c={'aa':3, 'c': 8}
dd = defaultdict(list)
for d in (a,b,c):
for key, value in d.items():
dd[key].append(value)
print(dd)
Use defaultdict to automatically create a dictionary entry with an empty list.
To process all source dictionaries in a single loop, use itertools.chain.
The main loop just adds a value from the current item, to the list under
the current key.
As you wrote, for cases when under some key there is only one item,
you have to generate a work dictionary (using dictonary comprehension),
limited to items with value (list) containing only one item.
The value of such item shoud contain only the first (and only) number
from the source list.
Then use this dictionary to update d.
So the whole script can be surprisingly short, as below:
from collections import defaultdict
from itertools import chain
a = {'a':1, 'b': 2}
b = {'aa':4, 'b': 6}
c = {'aa':3, 'c': 8}
d = defaultdict(list)
for k, v in chain(a.items(), b.items(), c.items()):
d[k].append(v)
d.update({ k: v[0] for k, v in d.items() if len(v) == 1 })
As you can see, the actual processing code is contained in only 4 (last) lines.
If you print d, the result is:
defaultdict(list, {'a': 1, 'b': [2, 6], 'aa': [4, 3], 'c': 8})
I have:
myDict = {'a': [1,2,3], 'b':[4,5,6], 'c':[7,8,9]}
I want:
myDict = {'a': set([1,2,3]), 'b':set([4,5,6]), 'c':set([7,8,9])}
Is there a one-liner I can use to do this rather than looping through it and converting the type of the values?
You'll have to loop anyway:
{key: set(value) for key, value in yourData.items()}
If you're using Python 3.6+, you can also do this:
dict(zip(myDict.keys(), map(set, myDict.values())))
This can be done with map by mapping the values to type set
myDict = dict(map(lambda x: (x[0], set(x[1])), myDict.items()))
Or with either version of dictionary comprehension as well
myDict = {k: set(v) for k, v in myDict.items()}
myDict = {k: set(myDict[k]) for k in myDict}
You can use comprehension for it:
Basically, loop through the key-value pairs and create set out of each value for the corresponding key.
>>> myDict = {'a': [1,2,3], 'b':[4,5,6], 'c':[7,8,9]}
>>> myDict = {k: set(v) for k, v in myDict.items()}
>>> myDict
{'a': {1, 2, 3}, 'b': {4, 5, 6}, 'c': {8, 9, 7}}
You can't do it without looping anyway, but you can have the looping done in one line, with the following code:
myDict = {k:set(v) for k, v in myDict.items()}
This is basically traversing each item in your dictionary and converting the lists to sets and combining the key(str):value(set) pairs to a new dictionary and assigning it back to myDict variable.
This question already has answers here:
switching keys and values in a dictionary in python [duplicate]
(10 answers)
Closed 5 years ago.
I'm sorry if this is a foolish question or a duplicate - I looked but didn't really find anything on this specific question:
I wrote a little cyphering tool for practise, now I'm working on the deciphering part and wondering, how I can use the dictionary I used as a lookuptable to revert this... I can't access the dictionary's key via its value, right?
So I thought I'd turn it around like this:
for x, y in cyphertable.items():
DEcyphertable = dict(zip(x, y))
But this doesn't seem to work.
A: What am I doing wrong?
and B: How else could I make each "i" in a string look up the value and replace it with the corresponding key?
Using Dict comprehension
reverted_dict = {value: key for key, value in cyphertable.items()}
Using zip
reverted_dict = dict(zip(cyphertable.values(), cyphertable.keys()))
You can do that simply by:
new_dict = {v: k for k, v in old_dict.items()}
But, note that you may loose some items if you have duplicated values in old_dict (values become keys and keys are unique in a dict).
Outputs:
>>> old_dict = {'a': 1, 'b': 2}
>>> new_dict = {v: k for k, v in old_dict.items()}
>>> new_dict
{1: 'a', 2: 'b'}
If old_dict contains duplicated values:
>>> old_dict = {'a': 1, 'b': 2, 'c': 1}
>>> new_dict = {v: k for k, v in old_dict.items()}
>>> new_dict
{1: 'c', 2: 'b'}
Here it is,
In [38]: a = {'a':1,'b':2,'c':3}
In [39]: {j:i for i,j in a.items()}
Out[39]: {1: 'a', 2: 'b', 3: 'c'}
Or
In [40]: dict(zip(a.values(),a.keys()))
Out[40]: {1: 'a', 2: 'b', 3: 'c'}
I am trying to find a way of deleting duplicate shaders in Maya using Python Dictionaries.
Here is what I'm doing:
I want to put all maya shaders into a dictionary as keys and put the corresponding texture file as the value. Then I want the script to run through the dictionary and find any keys that share the same value and stuff them into an array or another dictionary.
This is basically what I have right now:
shaders_dict = {'a': somePath, 'b': somePath,
'c': differentPath, 'd': differentPath}
duplicate_shaders_dict = {}`
how can I now run through that dictionary to compile another dictionary that looks something like this:
duplicate_shaders_dict = {'b':somePath, 'd':differentPath }
And the tricky part being since there are duplicates I want the script to skip the original key so it doesn't also get stuffed in to duplicate shaders dictionary.
I would probably do something like this. First, make the inverse dictionary:
>>> from collections import defaultdict
>>>
>>> shaders_dict = {'a':'somePath', 'b':'somePath', 'c':'differentPath', 'd':'differentPath'}
>>>
>>> inverse_dict = defaultdict(list)
>>> for k,v in shaders_dict.iteritems():
... inverse_dict[v].append(k)
...
>>> inverse_dict
defaultdict(<type 'list'>, {'differentPath': ['c', 'd'], 'somePath': ['a', 'b']})
This basically inverts the dictionary by looping over every key, value pair and appending the key to a list associated with the value.
Then split this:
>>> first_shaders_dict = {}
>>> duplicate_shaders_dict = {}
>>> for v, ks in inverse_dict.iteritems():
... first, rest = ks[0], ks[1:]
... first_shaders_dict[first] = v
... for r in rest:
... duplicate_shaders_dict[r] = v
...
>>> first_shaders_dict
{'a': 'somePath', 'c': 'differentPath'}
>>> duplicate_shaders_dict
{'b': 'somePath', 'd': 'differentPath'}
Hmm. This assumes that the texture files are hashable and so can serve as dictionary keys. If they're not, then I'd have to work around that. Also, since as #freespace notes there's no ordering here, if you wanted a particular order we'd have to iterate over sorted keys or the like.
--
Update: I didn't like the above much. Shorter itertools-based version:
>>> import itertools
>>> shaders_dict = {'a':'somePath', 'b':'somePath', 'c':'differentPath', 'd':'differentPath'}
>>> keys = sorted(sorted(shaders_dict),key=shaders_dict.get)
>>> by_val = [(v, list(ks)) for v, ks in itertools.groupby(keys, shaders_dict.get)]
>>> first_dict = dict((ks[0],v) for v,ks in by_val)
>>> duplicate_dict = dict((k,v) for v,ks in by_val for k in ks[1:])
>>> first_dict
{'a': 'somePath', 'c': 'differentPath'}
>>> duplicate_dict
{'b': 'somePath', 'd': 'differentPath'}
One simple solution is to reverse the dictionary. Given:
>>> d = {'a': 'somePath', 'b': 'somePath',
... 'c': 'differentPath', 'd': 'differentPath'}
You can reverse it like this:
>>> r = dict((v,k) for k,v in d.iteritems())
Which gives you:
>>> r
{'differentPath': 'd', 'somePath': 'b'}
And if you reverse that, you have the original dictionary with duplicates removed:
>>> d = dict((v,k) for k,v in r.iteritems())
>>> d
{'b': 'somePath', 'd': 'differentPath'}
This question already has answers here:
How to merge dicts, collecting values from matching keys?
(17 answers)
Closed 3 months ago.
I have to merge list of python dictionary. For eg:
dicts[0] = {'a':1, 'b':2, 'c':3}
dicts[1] = {'a':1, 'd':2, 'c':'foo'}
dicts[2] = {'e':57,'c':3}
super_dict = {'a':[1], 'b':[2], 'c':[3,'foo'], 'd':[2], 'e':[57]}
I wrote the following code:
super_dict = {}
for d in dicts:
for k, v in d.items():
if super_dict.get(k) is None:
super_dict[k] = []
if v not in super_dict.get(k):
super_dict[k].append(v)
Can it be presented more elegantly / optimized?
Note
I found another question on SO but its about merging exactly 2 dictionaries.
You can iterate over the dictionaries directly -- no need to use range. The setdefault method of dict looks up a key, and returns the value if found. If not found, it returns a default, and also assigns that default to the key.
super_dict = {}
for d in dicts:
for k, v in d.iteritems(): # d.items() in Python 3+
super_dict.setdefault(k, []).append(v)
Also, you might consider using a defaultdict. This just automates setdefault by calling a function to return a default value when a key isn't found.
import collections
super_dict = collections.defaultdict(list)
for d in dicts:
for k, v in d.iteritems(): # d.items() in Python 3+
super_dict[k].append(v)
Also, as Sven Marnach astutely observed, you seem to want no duplication of values in your lists. In that case, set gets you what you want:
import collections
super_dict = collections.defaultdict(set)
for d in dicts:
for k, v in d.iteritems(): # d.items() in Python 3+
super_dict[k].add(v)
from collections import defaultdict
dicts = [{'a':1, 'b':2, 'c':3},
{'a':1, 'd':2, 'c':'foo'},
{'e':57, 'c':3} ]
super_dict = defaultdict(set) # uses set to avoid duplicates
for d in dicts:
for k, v in d.items(): # use d.iteritems() in python 2
super_dict[k].add(v)
you can use this behaviour of dict. (a bit elegant)
a = {'a':1, 'b':2, 'c':3}
b = {'d':1, 'e':2, 'f':3}
c = {1:1, 2:2, 3:3}
merge = {**a, **b, **c}
print(merge) # {'a': 1, 'b': 2, 'c': 3, 'd': 1, 'e': 2, 'f': 3, 1: 1, 2: 2, 3: 3}
and you are good to go :)
Merge the keys of all dicts, and for each key assemble the list of values:
super_dict = {}
for k in set(k for d in dicts for k in d):
super_dict[k] = [d[k] for d in dicts if k in d]
The expression set(k for d in dicts for k in d) builds a set of all unique keys of all dictionaries. For each of these unique keys, we use the list comprehension [d[k] for d in dicts if k in d] to build the list of values from all dicts for this key.
Since you only seem to one the unique value of each key, you might want to use sets instead:
super_dict = {}
for k in set(k for d in dicts for k in d):
super_dict[k] = set(d[k] for d in dicts if k in d)
It seems like most of the answers using comprehensions are not all that readable. In case any gets lost in the mess of answers above this might be helpful (although extremely late...). Just loop over the items of each dict and place them in a separate one.
super_dict = {key:val for d in dicts for key,val in d.items()}
When the value of the keys are in list:
from collections import defaultdict
dicts = [{'a':[1], 'b':[2], 'c':[3]},
{'a':[11], 'd':[2], 'c':['foo']},
{'e':[57], 'c':[3], "a": [1]} ]
super_dict = defaultdict(list) # uses set to avoid duplicates
for d in dicts:
for k, v in d.items(): # use d.iteritems() in python 2
super_dict[k] = list(set(super_dict[k] + v))
combined_dict = {}
for elem in super_dict.keys():
combined_dict[elem] = super_dict[elem]
combined_dict
## output: {'a': [1, 11], 'b': [2], 'c': [3, 'foo'], 'd': [2], 'e': [57]}
I have a very easy to go solution without any imports.
I use the dict.update() method.
But sadly it will overwrite, if same key appears in more than one dictionary, then the most recently merged dict's value will appear in the output.
dict1 = {'Name': 'Zara', 'Age': 7}
dict2 = {'Sex': 'female' }
dict3 = {'Status': 'single', 'Age': 27}
dict4 = {'Occupation':'nurse', 'Wage': 3000}
def mergedict(*args):
output = {}
for arg in args:
output.update(arg)
return output
print(mergedict(dict1, dict2, dict3, dict4))
The output is this:
{'Name': 'Zara', 'Age': 27, 'Sex': 'female', 'Status': 'single', 'Occupation': 'nurse', 'Wage': 3000}
Perhaps a more modern and concise approach for those who use python 3.3 or later versions is the use of ChainMap from the collections module.
from collections import ChainMap
d1 = {'a': 1, 'b': 3}
d2 = {'c': 2}
d3 = {'d': 7, 'a': 9}
d4 = {}
combo = dict(ChainMap(d1, d2, d3, d4))
# {'d': 7, 'a': 1, 'c': 2, 'b': 3}
For a larger collection of dict objects then star operator works
dict(ChainMap(*dict_collection))
Note that the resulting dictionary seems to only keep the value of the first key it encounters in the ordered collection and ignores any further duplicates.
This may be a bit more elegant:
super_dict = {}
for d in dicts:
for k, v in d.iteritems():
l=super_dict.setdefault(k,[])
if v not in l:
l.append(v)
UPDATE: made change suggested by Sven
UPDATE: changed to avoid duplicates (thanks Marcin and Steven)
Never forget that the standard libraries have a wealth of tools for dealing with dicts and iteration:
from itertools import chain
from collections import defaultdict
super_dict = defaultdict(list)
for k,v in chain.from_iterable(d.iteritems() for d in dicts):
if v not in super_dict[k]: super_dict[k].append(v)
Note that the if v not in super_dict[k] can be avoided by using defaultdict(set) as per Steven Rumbalski's answer.
If you assume that the keys in which you are interested are at the same nested level, you can recursively traverse each dictionary and create a new dictionary using that key, effectively merging them.
merged = {}
for d in dicts:
def walk(d,merge):
for key, item in d.items():
if isinstance(item, dict):
merge.setdefault(key, {})
walk(item, merge[key])
else:
merge.setdefault(key, [])
merge[key].append(item)
walk(d,merged)
For example, say you have the following dictionaries you want to merge.
dicts = [{'A': {'A1': {'FOO': [1,2,3]}}},
{'A': {'A1': {'A2': {'BOO': [4,5,6]}}}},
{'A': {'A1': {'FOO': [7,8]}}},
{'B': {'B1': {'COO': [9]}}},
{'B': {'B2': {'DOO': [10,11,12]}}},
{'C': {'C1': {'C2': {'POO':[13,14,15]}}}},
{'C': {'C1': {'ROO': [16,17]}}}]
Using the key at each level, you should get something like this:
{'A': {'A1': {'FOO': [[1, 2, 3], [7, 8]],
'A2': {'BOO': [[4, 5, 6]]}}},
'B': {'B1': {'COO': [[9]]},
'B2': {'DOO': [[10, 11, 12]]}},
'C': {'C1': {'C2': {'POO': [[13, 14, 15]]},
'ROO': [[16, 17]]}}}
Note: I assume the leaf at each branch is a list of some kind, but you can obviously change the logic to do whatever is necessary for your situation.
This is a more recent enhancement over the prior answer by ElbowPipe, using newer syntax introduced in Python 3.9 for merging dictionaries. Note that this answer does not merge conflicting values into a list!
> import functools
> import operator
> functools.reduce(operator.or_, [{0:1}, {2:3, 4:5}, {2:6}])
{0: 1, 2: 6, 4: 5}
For a oneliner, the following could be used:
{key: {d[key] for d in dicts if key in d} for key in {key for d in dicts for key in d}}
although readibility would benefit from naming the combined key set:
combined_key_set = {key for d in dicts for key in d}
super_dict = {key: {d[key] for d in dicts if key in d} for key in combined_key_set}
Elegance can be debated but personally I prefer comprehensions over for loops. :)
(The dictionary and set comprehensions are available in Python 2.7/3.1 and newer.)
python 3.x (reduce is builtin for python 2.x, so no need to import if in 2.x)
import operator
from functools import operator.add
a = [{'a': 1}, {'b': 2}, {'c': 3, 'd': 4}]
dict(reduce(operator.add, map(list,(map(dict.items, a))))
map(dict.items, a) # converts to list of key, value iterators
map(list, ... # converts to iterator equivalent of [[[a, 1]], [[b, 2]], [[c, 3],[d,4]]]
reduce(operator.add, ... # reduces the multiple list down to a single list
My solution is similar to #senderle proposed, but instead of for loop I used map
super_dict = defaultdict(set)
map(lambda y: map(lambda x: super_dict[x].add(y[x]), y), dicts)
The use of defaultdict is good, this also can be done with the use of itertools.groupby.
import itertools
# output all dict items, and sort them by key
dicts_ele = sorted( ( item for d in dicts for item in d.items() ), key = lambda x: x[0] )
# groups items by key
ele_groups = itertools.groupby( dicts_ele, key = lambda x: x[0] )
# iterates over groups and get item value
merged = { k: set( v[1] for v in grouped ) for k, grouped in ele_groups }
and obviously, you can merge this block of code into one-line style
merged = {
k: set( v[1] for v in grouped )
for k, grouped in (
itertools.groupby(
sorted(
( item for d in dicts for item in d.items() ),
key = lambda x: x[0]
),
key = lambda x: x[0]
)
)
}
I'm a bit late to the game but I did it in 2 lines with no dependencies beyond python itself:
flatten = lambda *c: (b for a in c for b in (flatten(*a) if isinstance(a, (tuple, list)) else (a,)))
o = reduce(lambda d1,d2: dict((k, list(flatten([d1.get(k), d2.get(k)]))) for k in set(d1.keys() + d2.keys())), dicts)
# output:
# {'a': [1, 1, None], 'c': [3, 'foo', 3], 'b': [2, None, None], 'e': [None, 57], 'd': [None, 2, None]}
Though if you don't care about nested lists, then:
o2 = reduce(lambda d1,d2: dict((k, [d1.get(k), d2.get(k)]) for k in set(d1.keys() + d2.keys())), dicts)
# output:
# {'a': [[1, 1], None], 'c': [[3, 'foo'], 3], 'b': [[2, None], None], 'e': [None, 57], 'd': [[None, 2], None]}