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I have multiple dicts (or sequences of key-value pairs) like this:
d1 = {key1: x1, key2: y1}
d2 = {key1: x2, key2: y2}
How can I efficiently get a result like this, as a new dict?
d = {key1: (x1, x2), key2: (y1, y2)}
See also: How can one make a dictionary with duplicate keys in Python?.
Here's a general solution that will handle an arbitrary amount of dictionaries, with cases when keys are in only some of the dictionaries:
from collections import defaultdict
d1 = {1: 2, 3: 4}
d2 = {1: 6, 3: 7}
dd = defaultdict(list)
for d in (d1, d2): # you can list as many input dicts as you want here
for key, value in d.items():
dd[key].append(value)
print(dd) # result: defaultdict(<type 'list'>, {1: [2, 6], 3: [4, 7]})
assuming all keys are always present in all dicts:
ds = [d1, d2]
d = {}
for k in d1.iterkeys():
d[k] = tuple(d[k] for d in ds)
Note: In Python 3.x use below code:
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = tuple(d[k] for d in ds)
and if the dic contain numpy arrays:
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = np.concatenate(list(d[k] for d in ds))
This function merges two dicts even if the keys in the two dictionaries are different:
def combine_dict(d1, d2):
return {
k: tuple(d[k] for d in (d1, d2) if k in d)
for k in set(d1.keys()) | set(d2.keys())
}
Example:
d1 = {
'a': 1,
'b': 2,
}
d2` = {
'b': 'boat',
'c': 'car',
}
combine_dict(d1, d2)
# Returns: {
# 'a': (1,),
# 'b': (2, 'boat'),
# 'c': ('car',)
# }
dict1 = {'m': 2, 'n': 4}
dict2 = {'n': 3, 'm': 1}
Making sure that the keys are in the same order:
dict2_sorted = {i:dict2[i] for i in dict1.keys()}
keys = dict1.keys()
values = zip(dict1.values(), dict2_sorted.values())
dictionary = dict(zip(keys, values))
gives:
{'m': (2, 1), 'n': (4, 3)}
If you only have d1 and d2,
from collections import defaultdict
d = defaultdict(list)
for a, b in d1.items() + d2.items():
d[a].append(b)
Here is one approach you can use which would work even if both dictonaries don't have same keys:
d1 = {'a':'test','b':'btest','d':'dreg'}
d2 = {'a':'cool','b':'main','c':'clear'}
d = {}
for key in set(d1.keys() + d2.keys()):
try:
d.setdefault(key,[]).append(d1[key])
except KeyError:
pass
try:
d.setdefault(key,[]).append(d2[key])
except KeyError:
pass
print d
This would generate below input:
{'a': ['test', 'cool'], 'c': ['clear'], 'b': ['btest', 'main'], 'd': ['dreg']}
Using precomputed keys
def merge(dicts):
# First, figure out which keys are present.
keys = set().union(*dicts)
# Build a dict with those keys, using a list comprehension to
# pull the values from the source dicts.
return {
k: [d[k] for d in dicts if k in d]
for k in keys
}
This is essentially Flux's answer, generalized for a list of input dicts.
The set().union trick works by making a set union of the keys in all the source dictionaries. The union method on a set (we start with an empty one) can accept an arbitrary number of arguments, and make a union of each input with the original set; and it can accept other iterables (it does not require other sets for the arguments) - it will iterate over them and look for all unique elements. Since iterating over a dict yields its keys, they can be passed directly to the union method.
In the case where the keys of all inputs are known to be the same, this can be simplified: the keys can be hard-coded (or inferred from one of the inputs), and the if check in the list comprehension becomes unnecessary:
def merge(dicts):
return {
k: [d[k] for d in dicts]
for k in dicts[0].keys()
}
This is analogous to blubb's answer, but using a dict comprehension rather than an explicit loop to build the final result.
We could also try something like Mahdi Ghelichi's answer:
def merge(dicts):
values = zip(*(d.values() for d in ds))
return dict(zip(dicts[0].keys(), values))
This should work in Python 3.5 and below: dicts with identical keys will store them in the same order, during the same run of the program (if you run the program again, you may get a different ordering, but still a consistent one).
In 3.6 and above, dictionaries preserve their insertion order (though they are only guaranteed to do so by the specification in 3.7 and above). Thus, input dicts could have the same keys in a different order, which would cause the first zip to combine the wrong values.
We can work around this by "sorting" the input dicts (re-creating them with keys in a consistent order, like [{k:d[k] for k in dicts[0].keys()} for d in dicts]. (In older versions, this would be extra work with no net effect.) However, this adds complexity, and this double-zip approach really doesn't offer any advantages over the previous one using a dict comprehension.
Building the result explicitly, discovering keys on the fly
As in Eli Bendersky's answer, but as a function:
from collections import defaultdict
def merge(dicts):
result = defaultdict(list)
for d in dicts:
for key, value in d.items():
result[key].append(value)
return result
This will produce a defaultdict, a subclass of dict defined by the standard library. The equivalent code using only built-in dicts might look like:
def merge(dicts):
result = {}
for d in dicts:
for key, value in d.items():
result.setdefault(key, []).append(value)
return result
Using other container types besides lists
The precomputed-key approach will work fine to make tuples; replace the list comprehension [d[k] for d in dicts if k in d] with tuple(d[k] for d in dicts if k in d). This passes a generator expression to the tuple constructor. (There is no "tuple comprehension".)
Since tuples are immutable and don't have an append method, the explicit loop approach should be modified by replacing .append(value) with += (value,). However, this may perform poorly if there is a lot of key duplication, since it must create a new tuple each time. It might be better to produce lists first and then convert the final result with something like {k: tuple(v) for (k, v) in merged.items()}.
Similar modifications can be made to get sets (although there is a set comprehension, using {}), Numpy arrays etc. For example, we can generalize both approaches with a container type like so:
def merge(dicts, value_type=list):
# First, figure out which keys are present.
keys = set().union(*dicts)
# Build a dict with those keys, using a list comprehension to
# pull the values from the source dicts.
return {
k: value_type(d[k] for d in dicts if k in d)
for k in keys
}
and
from collections import defaultdict
def merge(dicts, value_type=list):
# We stick with hard-coded `list` for the first part,
# because even other mutable types will offer different interfaces.
result = defaultdict(list)
for d in dicts:
for key, value in d.items():
result[key].append(value)
# This is redundant for the default case, of course.
return {k:value_type(v) for (k, v) in result}
If the input values are already sequences
Rather than wrapping the values from the source in a new list, often people want to take inputs where the values are all already lists, and concatenate those lists in the output (or concatenate tuples or 1-dimensional Numpy arrays, combine sets, etc.).
This is still a trivial modification. For precomputed keys, use a nested list comprehension, ordered to get a flat result:
def merge(dicts):
keys = set().union(*dicts)
return {
k: [v for d in dicts if k in d for v in d[k]]
# Alternately:
# k: [v for d in dicts for v in d.get(k, [])]
for k in keys
}
One might instead think of using sum to concatenate results from the original list comprehension. Don't do this - it will perform poorly when there are a lot of duplicate keys. The built-in sum isn't optimized for sequences (and will explicitly disallow "summing" strings) and will try to create a new list with each addition internally.
With the explicit loop approach, use .extend instead of .append:
from collections import defaultdict
def merge(dicts):
result = defaultdict(list)
for d in dicts:
for key, value in d.items():
result[key].extend(value)
return result
The extend method of lists accepts any iterable, so this will work with inputs that have tuples for the values - of course, it still uses lists in the output; and of course, those can be converted back as shown previously.
If the inputs have one item each
A common version of this problem involves input dicts that each have a single key-value pair. Alternately, the input might be (key, value) tuples (or lists).
The above approaches will still work, of course. For tuple inputs, converting them to dicts first, like [{k:v} for (k, v) in tuples], allows for using the directly. Alternately, the explicit iteration approach can be modified to accept the tuples directly, like in Victoria Stuart's answer:
from collections import defaultdict
def merge(pairs):
result = defaultdict(list)
for key, value in pairs:
result[key].extend(value)
return result
(The code was simplified because there is no need to iterate over key-value pairs when there is only one of them and it has been provided directly.)
However, for these single-item cases it may work better to sort the values by key and then use itertools.groupby. In this case, it will be easier to work with the tuples. That looks like:
from itertools import groupby
def merge(tuples):
grouped = groupby(tuples, key=lambda t: t[0])
return {k: [kv[1] for kv in ts] for k, ts in grouped}
Here, t is used as a name for one of the tuples from the input. The grouped iterator will provide pairs of a "key" value k (the first element that was common to the tuples being grouped) and an iterator ts over the tuples in that group. Then we extract the values from the key-value pairs kv in the ts, make a list from those, and use that as the value for the k key in the resulting dict.
To merge one-item dicts this way, of course, convert them to tuples first. One simple way to do this, for a list of one-item dicts, is [next(iter(d.items())) for d in dicts].
Assuming there are two dictionaries with exact same keys, below is the most succinct way of doing it (python3 should be used for both the solution).
d1 = {'a': 1, 'b': 2, 'c':3}
d2 = {'a': 5, 'b': 6, 'c':7}
# get keys from one of the dictionary
ks = [k for k in d1.keys()]
print(ks)
['a', 'b', 'c']
# call values from each dictionary on available keys
d_merged = {k: (d1[k], d2[k]) for k in ks}
print(d_merged)
{'a': (1, 5), 'b': (2, 6), 'c': (3, 7)}
# to merge values as list
d_merged = {k: [d1[k], d2[k]] for k in ks}
print(d_merged)
{'a': [1, 5], 'b': [2, 6], 'c': [3, 7]}
If there are two dictionaries with some common keys, but a few different keys, a list of all the keys should be prepared.
d1 = {'a': 1, 'b': 2, 'c':3, 'd': 9}
d2 = {'a': 5, 'b': 6, 'c':7, 'e': 4}
# get keys from one of the dictionary
d1_ks = [k for k in d1.keys()]
d2_ks = [k for k in d2.keys()]
all_ks = set(d1_ks + d2_ks)
print(all_ks)
['a', 'b', 'c', 'd', 'e']
# call values from each dictionary on available keys
d_merged = {k: [d1.get(k), d2.get(k)] for k in all_ks}
print(d_merged)
{'d': [9, None], 'a': [1, 5], 'b': [2, 6], 'c': [3, 7], 'e': [None, 4]}
There is a great library funcy doing what you need in a just one, short line.
from funcy import join_with
from pprint import pprint
d1 = {"key1": "x1", "key2": "y1"}
d2 = {"key1": "x2", "key2": "y2"}
list_of_dicts = [d1, d2]
merged_dict = join_with(tuple, list_of_dicts)
pprint(merged_dict)
Output:
{'key1': ('x1', 'x2'), 'key2': ('y1', 'y2')}
More info here: funcy -> join_with.
def merge(d1, d2, merge):
result = dict(d1)
for k,v in d2.iteritems():
if k in result:
result[k] = merge(result[k], v)
else:
result[k] = v
return result
d1 = {'a': 1, 'b': 2}
d2 = {'a': 1, 'b': 3, 'c': 2}
print merge(d1, d2, lambda x, y:(x,y))
{'a': (1, 1), 'c': 2, 'b': (2, 3)}
If keys are nested:
d1 = { 'key1': { 'nkey1': 'x1' }, 'key2': { 'nkey2': 'y1' } }
d2 = { 'key1': { 'nkey1': 'x2' }, 'key2': { 'nkey2': 'y2' } }
ds = [d1, d2]
d = {}
for k in d1.keys():
for k2 in d1[k].keys():
d.setdefault(k, {})
d[k].setdefault(k2, [])
d[k][k2] = tuple(d[k][k2] for d in ds)
yields:
{'key1': {'nkey1': ('x1', 'x2')}, 'key2': {'nkey2': ('y1', 'y2')}}
Modifying this answer to create a dictionary of tuples (what the OP asked for), instead of a dictionary of lists:
from collections import defaultdict
d1 = {1: 2, 3: 4}
d2 = {1: 6, 3: 7}
dd = defaultdict(tuple)
for d in (d1, d2): # you can list as many input dicts as you want here
for key, value in d.items():
dd[key] += (value,)
print(dd)
The above prints the following:
defaultdict(<class 'tuple'>, {1: (2, 6), 3: (4, 7)})
d1 ={'B': 10, 'C ': 7, 'A': 20}
d2 ={'B': 101, 'Y ': 7, 'X': 8}
d3 ={'A': 201, 'Y ': 77, 'Z': 8}
def CreateNewDictionaryAssemblingAllValues1(d1,d2,d3):
aa = {
k :[d[k] for d in (d1,d2,d3) if k in d ] for k in set(d1.keys() | d2.keys() | d3.keys() )
}
aap = print(aa)
return aap
CreateNewDictionaryAssemblingAllValues1(d1, d2, d3)
"""
Output :
{'X': [8], 'C ': [7], 'Y ': [7, 77], 'Z': [8], 'B': [10, 101], 'A': [20, 201]}
"""
From blubb answer:
You can also directly form the tuple using values from each list
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = (d1[k], d2[k])
This might be useful if you had a specific ordering for your tuples
ds = [d1, d2, d3, d4]
d = {}
for k in d1.keys():
d[k] = (d3[k], d1[k], d4[k], d2[k]) #if you wanted tuple in order of d3, d1, d4, d2
Using below method we can merge two dictionaries having same keys.
def update_dict(dict1: dict, dict2: dict) -> dict:
output_dict = {}
for key in dict1.keys():
output_dict.update({key: []})
if type(dict1[key]) != str:
for value in dict1[key]:
output_dict[key].append(value)
else:
output_dict[key].append(dict1[key])
if type(dict2[key]) != str:
for value in dict2[key]:
output_dict[key].append(value)
else:
output_dict[key].append(dict2[key])
return output_dict
Input: d1 = {key1: x1, key2: y1} d2 = {key1: x2, key2: y2}
Output: {'key1': ['x1', 'x2'], 'key2': ['y1', 'y2']}
dicts = [dict1,dict2,dict3]
out = dict(zip(dicts[0].keys(),[[dic[list(dic.keys())[key]] for dic in dicts] for key in range(0,len(dicts[0]))]))
A compact possibility
d1={'a':1,'b':2}
d2={'c':3,'d':4}
context={**d1, **d2}
context
{'b': 2, 'c': 3, 'd': 4, 'a': 1}
I have multiple dicts (or sequences of key-value pairs) like this:
d1 = {key1: x1, key2: y1}
d2 = {key1: x2, key2: y2}
How can I efficiently get a result like this, as a new dict?
d = {key1: (x1, x2), key2: (y1, y2)}
See also: How can one make a dictionary with duplicate keys in Python?.
Here's a general solution that will handle an arbitrary amount of dictionaries, with cases when keys are in only some of the dictionaries:
from collections import defaultdict
d1 = {1: 2, 3: 4}
d2 = {1: 6, 3: 7}
dd = defaultdict(list)
for d in (d1, d2): # you can list as many input dicts as you want here
for key, value in d.items():
dd[key].append(value)
print(dd) # result: defaultdict(<type 'list'>, {1: [2, 6], 3: [4, 7]})
assuming all keys are always present in all dicts:
ds = [d1, d2]
d = {}
for k in d1.iterkeys():
d[k] = tuple(d[k] for d in ds)
Note: In Python 3.x use below code:
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = tuple(d[k] for d in ds)
and if the dic contain numpy arrays:
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = np.concatenate(list(d[k] for d in ds))
This function merges two dicts even if the keys in the two dictionaries are different:
def combine_dict(d1, d2):
return {
k: tuple(d[k] for d in (d1, d2) if k in d)
for k in set(d1.keys()) | set(d2.keys())
}
Example:
d1 = {
'a': 1,
'b': 2,
}
d2` = {
'b': 'boat',
'c': 'car',
}
combine_dict(d1, d2)
# Returns: {
# 'a': (1,),
# 'b': (2, 'boat'),
# 'c': ('car',)
# }
dict1 = {'m': 2, 'n': 4}
dict2 = {'n': 3, 'm': 1}
Making sure that the keys are in the same order:
dict2_sorted = {i:dict2[i] for i in dict1.keys()}
keys = dict1.keys()
values = zip(dict1.values(), dict2_sorted.values())
dictionary = dict(zip(keys, values))
gives:
{'m': (2, 1), 'n': (4, 3)}
If you only have d1 and d2,
from collections import defaultdict
d = defaultdict(list)
for a, b in d1.items() + d2.items():
d[a].append(b)
Here is one approach you can use which would work even if both dictonaries don't have same keys:
d1 = {'a':'test','b':'btest','d':'dreg'}
d2 = {'a':'cool','b':'main','c':'clear'}
d = {}
for key in set(d1.keys() + d2.keys()):
try:
d.setdefault(key,[]).append(d1[key])
except KeyError:
pass
try:
d.setdefault(key,[]).append(d2[key])
except KeyError:
pass
print d
This would generate below input:
{'a': ['test', 'cool'], 'c': ['clear'], 'b': ['btest', 'main'], 'd': ['dreg']}
Using precomputed keys
def merge(dicts):
# First, figure out which keys are present.
keys = set().union(*dicts)
# Build a dict with those keys, using a list comprehension to
# pull the values from the source dicts.
return {
k: [d[k] for d in dicts if k in d]
for k in keys
}
This is essentially Flux's answer, generalized for a list of input dicts.
The set().union trick works by making a set union of the keys in all the source dictionaries. The union method on a set (we start with an empty one) can accept an arbitrary number of arguments, and make a union of each input with the original set; and it can accept other iterables (it does not require other sets for the arguments) - it will iterate over them and look for all unique elements. Since iterating over a dict yields its keys, they can be passed directly to the union method.
In the case where the keys of all inputs are known to be the same, this can be simplified: the keys can be hard-coded (or inferred from one of the inputs), and the if check in the list comprehension becomes unnecessary:
def merge(dicts):
return {
k: [d[k] for d in dicts]
for k in dicts[0].keys()
}
This is analogous to blubb's answer, but using a dict comprehension rather than an explicit loop to build the final result.
We could also try something like Mahdi Ghelichi's answer:
def merge(dicts):
values = zip(*(d.values() for d in ds))
return dict(zip(dicts[0].keys(), values))
This should work in Python 3.5 and below: dicts with identical keys will store them in the same order, during the same run of the program (if you run the program again, you may get a different ordering, but still a consistent one).
In 3.6 and above, dictionaries preserve their insertion order (though they are only guaranteed to do so by the specification in 3.7 and above). Thus, input dicts could have the same keys in a different order, which would cause the first zip to combine the wrong values.
We can work around this by "sorting" the input dicts (re-creating them with keys in a consistent order, like [{k:d[k] for k in dicts[0].keys()} for d in dicts]. (In older versions, this would be extra work with no net effect.) However, this adds complexity, and this double-zip approach really doesn't offer any advantages over the previous one using a dict comprehension.
Building the result explicitly, discovering keys on the fly
As in Eli Bendersky's answer, but as a function:
from collections import defaultdict
def merge(dicts):
result = defaultdict(list)
for d in dicts:
for key, value in d.items():
result[key].append(value)
return result
This will produce a defaultdict, a subclass of dict defined by the standard library. The equivalent code using only built-in dicts might look like:
def merge(dicts):
result = {}
for d in dicts:
for key, value in d.items():
result.setdefault(key, []).append(value)
return result
Using other container types besides lists
The precomputed-key approach will work fine to make tuples; replace the list comprehension [d[k] for d in dicts if k in d] with tuple(d[k] for d in dicts if k in d). This passes a generator expression to the tuple constructor. (There is no "tuple comprehension".)
Since tuples are immutable and don't have an append method, the explicit loop approach should be modified by replacing .append(value) with += (value,). However, this may perform poorly if there is a lot of key duplication, since it must create a new tuple each time. It might be better to produce lists first and then convert the final result with something like {k: tuple(v) for (k, v) in merged.items()}.
Similar modifications can be made to get sets (although there is a set comprehension, using {}), Numpy arrays etc. For example, we can generalize both approaches with a container type like so:
def merge(dicts, value_type=list):
# First, figure out which keys are present.
keys = set().union(*dicts)
# Build a dict with those keys, using a list comprehension to
# pull the values from the source dicts.
return {
k: value_type(d[k] for d in dicts if k in d)
for k in keys
}
and
from collections import defaultdict
def merge(dicts, value_type=list):
# We stick with hard-coded `list` for the first part,
# because even other mutable types will offer different interfaces.
result = defaultdict(list)
for d in dicts:
for key, value in d.items():
result[key].append(value)
# This is redundant for the default case, of course.
return {k:value_type(v) for (k, v) in result}
If the input values are already sequences
Rather than wrapping the values from the source in a new list, often people want to take inputs where the values are all already lists, and concatenate those lists in the output (or concatenate tuples or 1-dimensional Numpy arrays, combine sets, etc.).
This is still a trivial modification. For precomputed keys, use a nested list comprehension, ordered to get a flat result:
def merge(dicts):
keys = set().union(*dicts)
return {
k: [v for d in dicts if k in d for v in d[k]]
# Alternately:
# k: [v for d in dicts for v in d.get(k, [])]
for k in keys
}
One might instead think of using sum to concatenate results from the original list comprehension. Don't do this - it will perform poorly when there are a lot of duplicate keys. The built-in sum isn't optimized for sequences (and will explicitly disallow "summing" strings) and will try to create a new list with each addition internally.
With the explicit loop approach, use .extend instead of .append:
from collections import defaultdict
def merge(dicts):
result = defaultdict(list)
for d in dicts:
for key, value in d.items():
result[key].extend(value)
return result
The extend method of lists accepts any iterable, so this will work with inputs that have tuples for the values - of course, it still uses lists in the output; and of course, those can be converted back as shown previously.
If the inputs have one item each
A common version of this problem involves input dicts that each have a single key-value pair. Alternately, the input might be (key, value) tuples (or lists).
The above approaches will still work, of course. For tuple inputs, converting them to dicts first, like [{k:v} for (k, v) in tuples], allows for using the directly. Alternately, the explicit iteration approach can be modified to accept the tuples directly, like in Victoria Stuart's answer:
from collections import defaultdict
def merge(pairs):
result = defaultdict(list)
for key, value in pairs:
result[key].extend(value)
return result
(The code was simplified because there is no need to iterate over key-value pairs when there is only one of them and it has been provided directly.)
However, for these single-item cases it may work better to sort the values by key and then use itertools.groupby. In this case, it will be easier to work with the tuples. That looks like:
from itertools import groupby
def merge(tuples):
grouped = groupby(tuples, key=lambda t: t[0])
return {k: [kv[1] for kv in ts] for k, ts in grouped}
Here, t is used as a name for one of the tuples from the input. The grouped iterator will provide pairs of a "key" value k (the first element that was common to the tuples being grouped) and an iterator ts over the tuples in that group. Then we extract the values from the key-value pairs kv in the ts, make a list from those, and use that as the value for the k key in the resulting dict.
To merge one-item dicts this way, of course, convert them to tuples first. One simple way to do this, for a list of one-item dicts, is [next(iter(d.items())) for d in dicts].
Assuming there are two dictionaries with exact same keys, below is the most succinct way of doing it (python3 should be used for both the solution).
d1 = {'a': 1, 'b': 2, 'c':3}
d2 = {'a': 5, 'b': 6, 'c':7}
# get keys from one of the dictionary
ks = [k for k in d1.keys()]
print(ks)
['a', 'b', 'c']
# call values from each dictionary on available keys
d_merged = {k: (d1[k], d2[k]) for k in ks}
print(d_merged)
{'a': (1, 5), 'b': (2, 6), 'c': (3, 7)}
# to merge values as list
d_merged = {k: [d1[k], d2[k]] for k in ks}
print(d_merged)
{'a': [1, 5], 'b': [2, 6], 'c': [3, 7]}
If there are two dictionaries with some common keys, but a few different keys, a list of all the keys should be prepared.
d1 = {'a': 1, 'b': 2, 'c':3, 'd': 9}
d2 = {'a': 5, 'b': 6, 'c':7, 'e': 4}
# get keys from one of the dictionary
d1_ks = [k for k in d1.keys()]
d2_ks = [k for k in d2.keys()]
all_ks = set(d1_ks + d2_ks)
print(all_ks)
['a', 'b', 'c', 'd', 'e']
# call values from each dictionary on available keys
d_merged = {k: [d1.get(k), d2.get(k)] for k in all_ks}
print(d_merged)
{'d': [9, None], 'a': [1, 5], 'b': [2, 6], 'c': [3, 7], 'e': [None, 4]}
There is a great library funcy doing what you need in a just one, short line.
from funcy import join_with
from pprint import pprint
d1 = {"key1": "x1", "key2": "y1"}
d2 = {"key1": "x2", "key2": "y2"}
list_of_dicts = [d1, d2]
merged_dict = join_with(tuple, list_of_dicts)
pprint(merged_dict)
Output:
{'key1': ('x1', 'x2'), 'key2': ('y1', 'y2')}
More info here: funcy -> join_with.
def merge(d1, d2, merge):
result = dict(d1)
for k,v in d2.iteritems():
if k in result:
result[k] = merge(result[k], v)
else:
result[k] = v
return result
d1 = {'a': 1, 'b': 2}
d2 = {'a': 1, 'b': 3, 'c': 2}
print merge(d1, d2, lambda x, y:(x,y))
{'a': (1, 1), 'c': 2, 'b': (2, 3)}
If keys are nested:
d1 = { 'key1': { 'nkey1': 'x1' }, 'key2': { 'nkey2': 'y1' } }
d2 = { 'key1': { 'nkey1': 'x2' }, 'key2': { 'nkey2': 'y2' } }
ds = [d1, d2]
d = {}
for k in d1.keys():
for k2 in d1[k].keys():
d.setdefault(k, {})
d[k].setdefault(k2, [])
d[k][k2] = tuple(d[k][k2] for d in ds)
yields:
{'key1': {'nkey1': ('x1', 'x2')}, 'key2': {'nkey2': ('y1', 'y2')}}
Modifying this answer to create a dictionary of tuples (what the OP asked for), instead of a dictionary of lists:
from collections import defaultdict
d1 = {1: 2, 3: 4}
d2 = {1: 6, 3: 7}
dd = defaultdict(tuple)
for d in (d1, d2): # you can list as many input dicts as you want here
for key, value in d.items():
dd[key] += (value,)
print(dd)
The above prints the following:
defaultdict(<class 'tuple'>, {1: (2, 6), 3: (4, 7)})
d1 ={'B': 10, 'C ': 7, 'A': 20}
d2 ={'B': 101, 'Y ': 7, 'X': 8}
d3 ={'A': 201, 'Y ': 77, 'Z': 8}
def CreateNewDictionaryAssemblingAllValues1(d1,d2,d3):
aa = {
k :[d[k] for d in (d1,d2,d3) if k in d ] for k in set(d1.keys() | d2.keys() | d3.keys() )
}
aap = print(aa)
return aap
CreateNewDictionaryAssemblingAllValues1(d1, d2, d3)
"""
Output :
{'X': [8], 'C ': [7], 'Y ': [7, 77], 'Z': [8], 'B': [10, 101], 'A': [20, 201]}
"""
From blubb answer:
You can also directly form the tuple using values from each list
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = (d1[k], d2[k])
This might be useful if you had a specific ordering for your tuples
ds = [d1, d2, d3, d4]
d = {}
for k in d1.keys():
d[k] = (d3[k], d1[k], d4[k], d2[k]) #if you wanted tuple in order of d3, d1, d4, d2
Using below method we can merge two dictionaries having same keys.
def update_dict(dict1: dict, dict2: dict) -> dict:
output_dict = {}
for key in dict1.keys():
output_dict.update({key: []})
if type(dict1[key]) != str:
for value in dict1[key]:
output_dict[key].append(value)
else:
output_dict[key].append(dict1[key])
if type(dict2[key]) != str:
for value in dict2[key]:
output_dict[key].append(value)
else:
output_dict[key].append(dict2[key])
return output_dict
Input: d1 = {key1: x1, key2: y1} d2 = {key1: x2, key2: y2}
Output: {'key1': ['x1', 'x2'], 'key2': ['y1', 'y2']}
dicts = [dict1,dict2,dict3]
out = dict(zip(dicts[0].keys(),[[dic[list(dic.keys())[key]] for dic in dicts] for key in range(0,len(dicts[0]))]))
A compact possibility
d1={'a':1,'b':2}
d2={'c':3,'d':4}
context={**d1, **d2}
context
{'b': 2, 'c': 3, 'd': 4, 'a': 1}
I would like to get a new dictionary with keys only if both dictionaries have those keys in them, and then get the values of the second one.
# example:
Dict1 = {'A':3, 'B':5, 'C':2, 'D':5}
Dict2 = {'B':3, 'C':1, 'K':5}
# result--> {'B':3, 'C':1}
As a dictionary comprehension:
>>> {k:v for k, v in Dict2.items() if k in Dict1}
{'B': 3, 'C': 1}
Or use filter:
>>> dict(filter(lambda x: x[0] in Dict1, Dict2.items()))
{'B': 3, 'C': 1}
>>>
Just another solution, doesn't use comprehension. This function loops through the keys k in Dict1 and tries to add Dict2[k] to a new dict which is returned at the end. I think the try-except approach is "pythonic".
def shared_keys(a, b):
"""
returns dict of all KVs in b which are also in a
"""
shared = {}
for k in a.keys():
try:
shared[k] = b[k]
except:
pass
return shared
Dict1 = {'A':3, 'B':5, 'C':2, 'D':5}
Dict2 = {'B':3, 'C':1, 'K':5}
print(shared_keys(Dict1, Dict2))
# >>> {'B': 3, 'C': 1}
I have multiple dicts (or sequences of key-value pairs) like this:
d1 = {key1: x1, key2: y1}
d2 = {key1: x2, key2: y2}
How can I efficiently get a result like this, as a new dict?
d = {key1: (x1, x2), key2: (y1, y2)}
See also: How can one make a dictionary with duplicate keys in Python?.
Here's a general solution that will handle an arbitrary amount of dictionaries, with cases when keys are in only some of the dictionaries:
from collections import defaultdict
d1 = {1: 2, 3: 4}
d2 = {1: 6, 3: 7}
dd = defaultdict(list)
for d in (d1, d2): # you can list as many input dicts as you want here
for key, value in d.items():
dd[key].append(value)
print(dd) # result: defaultdict(<type 'list'>, {1: [2, 6], 3: [4, 7]})
assuming all keys are always present in all dicts:
ds = [d1, d2]
d = {}
for k in d1.iterkeys():
d[k] = tuple(d[k] for d in ds)
Note: In Python 3.x use below code:
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = tuple(d[k] for d in ds)
and if the dic contain numpy arrays:
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = np.concatenate(list(d[k] for d in ds))
This function merges two dicts even if the keys in the two dictionaries are different:
def combine_dict(d1, d2):
return {
k: tuple(d[k] for d in (d1, d2) if k in d)
for k in set(d1.keys()) | set(d2.keys())
}
Example:
d1 = {
'a': 1,
'b': 2,
}
d2` = {
'b': 'boat',
'c': 'car',
}
combine_dict(d1, d2)
# Returns: {
# 'a': (1,),
# 'b': (2, 'boat'),
# 'c': ('car',)
# }
dict1 = {'m': 2, 'n': 4}
dict2 = {'n': 3, 'm': 1}
Making sure that the keys are in the same order:
dict2_sorted = {i:dict2[i] for i in dict1.keys()}
keys = dict1.keys()
values = zip(dict1.values(), dict2_sorted.values())
dictionary = dict(zip(keys, values))
gives:
{'m': (2, 1), 'n': (4, 3)}
If you only have d1 and d2,
from collections import defaultdict
d = defaultdict(list)
for a, b in d1.items() + d2.items():
d[a].append(b)
Here is one approach you can use which would work even if both dictonaries don't have same keys:
d1 = {'a':'test','b':'btest','d':'dreg'}
d2 = {'a':'cool','b':'main','c':'clear'}
d = {}
for key in set(d1.keys() + d2.keys()):
try:
d.setdefault(key,[]).append(d1[key])
except KeyError:
pass
try:
d.setdefault(key,[]).append(d2[key])
except KeyError:
pass
print d
This would generate below input:
{'a': ['test', 'cool'], 'c': ['clear'], 'b': ['btest', 'main'], 'd': ['dreg']}
Using precomputed keys
def merge(dicts):
# First, figure out which keys are present.
keys = set().union(*dicts)
# Build a dict with those keys, using a list comprehension to
# pull the values from the source dicts.
return {
k: [d[k] for d in dicts if k in d]
for k in keys
}
This is essentially Flux's answer, generalized for a list of input dicts.
The set().union trick works by making a set union of the keys in all the source dictionaries. The union method on a set (we start with an empty one) can accept an arbitrary number of arguments, and make a union of each input with the original set; and it can accept other iterables (it does not require other sets for the arguments) - it will iterate over them and look for all unique elements. Since iterating over a dict yields its keys, they can be passed directly to the union method.
In the case where the keys of all inputs are known to be the same, this can be simplified: the keys can be hard-coded (or inferred from one of the inputs), and the if check in the list comprehension becomes unnecessary:
def merge(dicts):
return {
k: [d[k] for d in dicts]
for k in dicts[0].keys()
}
This is analogous to blubb's answer, but using a dict comprehension rather than an explicit loop to build the final result.
We could also try something like Mahdi Ghelichi's answer:
def merge(dicts):
values = zip(*(d.values() for d in ds))
return dict(zip(dicts[0].keys(), values))
This should work in Python 3.5 and below: dicts with identical keys will store them in the same order, during the same run of the program (if you run the program again, you may get a different ordering, but still a consistent one).
In 3.6 and above, dictionaries preserve their insertion order (though they are only guaranteed to do so by the specification in 3.7 and above). Thus, input dicts could have the same keys in a different order, which would cause the first zip to combine the wrong values.
We can work around this by "sorting" the input dicts (re-creating them with keys in a consistent order, like [{k:d[k] for k in dicts[0].keys()} for d in dicts]. (In older versions, this would be extra work with no net effect.) However, this adds complexity, and this double-zip approach really doesn't offer any advantages over the previous one using a dict comprehension.
Building the result explicitly, discovering keys on the fly
As in Eli Bendersky's answer, but as a function:
from collections import defaultdict
def merge(dicts):
result = defaultdict(list)
for d in dicts:
for key, value in d.items():
result[key].append(value)
return result
This will produce a defaultdict, a subclass of dict defined by the standard library. The equivalent code using only built-in dicts might look like:
def merge(dicts):
result = {}
for d in dicts:
for key, value in d.items():
result.setdefault(key, []).append(value)
return result
Using other container types besides lists
The precomputed-key approach will work fine to make tuples; replace the list comprehension [d[k] for d in dicts if k in d] with tuple(d[k] for d in dicts if k in d). This passes a generator expression to the tuple constructor. (There is no "tuple comprehension".)
Since tuples are immutable and don't have an append method, the explicit loop approach should be modified by replacing .append(value) with += (value,). However, this may perform poorly if there is a lot of key duplication, since it must create a new tuple each time. It might be better to produce lists first and then convert the final result with something like {k: tuple(v) for (k, v) in merged.items()}.
Similar modifications can be made to get sets (although there is a set comprehension, using {}), Numpy arrays etc. For example, we can generalize both approaches with a container type like so:
def merge(dicts, value_type=list):
# First, figure out which keys are present.
keys = set().union(*dicts)
# Build a dict with those keys, using a list comprehension to
# pull the values from the source dicts.
return {
k: value_type(d[k] for d in dicts if k in d)
for k in keys
}
and
from collections import defaultdict
def merge(dicts, value_type=list):
# We stick with hard-coded `list` for the first part,
# because even other mutable types will offer different interfaces.
result = defaultdict(list)
for d in dicts:
for key, value in d.items():
result[key].append(value)
# This is redundant for the default case, of course.
return {k:value_type(v) for (k, v) in result}
If the input values are already sequences
Rather than wrapping the values from the source in a new list, often people want to take inputs where the values are all already lists, and concatenate those lists in the output (or concatenate tuples or 1-dimensional Numpy arrays, combine sets, etc.).
This is still a trivial modification. For precomputed keys, use a nested list comprehension, ordered to get a flat result:
def merge(dicts):
keys = set().union(*dicts)
return {
k: [v for d in dicts if k in d for v in d[k]]
# Alternately:
# k: [v for d in dicts for v in d.get(k, [])]
for k in keys
}
One might instead think of using sum to concatenate results from the original list comprehension. Don't do this - it will perform poorly when there are a lot of duplicate keys. The built-in sum isn't optimized for sequences (and will explicitly disallow "summing" strings) and will try to create a new list with each addition internally.
With the explicit loop approach, use .extend instead of .append:
from collections import defaultdict
def merge(dicts):
result = defaultdict(list)
for d in dicts:
for key, value in d.items():
result[key].extend(value)
return result
The extend method of lists accepts any iterable, so this will work with inputs that have tuples for the values - of course, it still uses lists in the output; and of course, those can be converted back as shown previously.
If the inputs have one item each
A common version of this problem involves input dicts that each have a single key-value pair. Alternately, the input might be (key, value) tuples (or lists).
The above approaches will still work, of course. For tuple inputs, converting them to dicts first, like [{k:v} for (k, v) in tuples], allows for using the directly. Alternately, the explicit iteration approach can be modified to accept the tuples directly, like in Victoria Stuart's answer:
from collections import defaultdict
def merge(pairs):
result = defaultdict(list)
for key, value in pairs:
result[key].extend(value)
return result
(The code was simplified because there is no need to iterate over key-value pairs when there is only one of them and it has been provided directly.)
However, for these single-item cases it may work better to sort the values by key and then use itertools.groupby. In this case, it will be easier to work with the tuples. That looks like:
from itertools import groupby
def merge(tuples):
grouped = groupby(tuples, key=lambda t: t[0])
return {k: [kv[1] for kv in ts] for k, ts in grouped}
Here, t is used as a name for one of the tuples from the input. The grouped iterator will provide pairs of a "key" value k (the first element that was common to the tuples being grouped) and an iterator ts over the tuples in that group. Then we extract the values from the key-value pairs kv in the ts, make a list from those, and use that as the value for the k key in the resulting dict.
To merge one-item dicts this way, of course, convert them to tuples first. One simple way to do this, for a list of one-item dicts, is [next(iter(d.items())) for d in dicts].
Assuming there are two dictionaries with exact same keys, below is the most succinct way of doing it (python3 should be used for both the solution).
d1 = {'a': 1, 'b': 2, 'c':3}
d2 = {'a': 5, 'b': 6, 'c':7}
# get keys from one of the dictionary
ks = [k for k in d1.keys()]
print(ks)
['a', 'b', 'c']
# call values from each dictionary on available keys
d_merged = {k: (d1[k], d2[k]) for k in ks}
print(d_merged)
{'a': (1, 5), 'b': (2, 6), 'c': (3, 7)}
# to merge values as list
d_merged = {k: [d1[k], d2[k]] for k in ks}
print(d_merged)
{'a': [1, 5], 'b': [2, 6], 'c': [3, 7]}
If there are two dictionaries with some common keys, but a few different keys, a list of all the keys should be prepared.
d1 = {'a': 1, 'b': 2, 'c':3, 'd': 9}
d2 = {'a': 5, 'b': 6, 'c':7, 'e': 4}
# get keys from one of the dictionary
d1_ks = [k for k in d1.keys()]
d2_ks = [k for k in d2.keys()]
all_ks = set(d1_ks + d2_ks)
print(all_ks)
['a', 'b', 'c', 'd', 'e']
# call values from each dictionary on available keys
d_merged = {k: [d1.get(k), d2.get(k)] for k in all_ks}
print(d_merged)
{'d': [9, None], 'a': [1, 5], 'b': [2, 6], 'c': [3, 7], 'e': [None, 4]}
There is a great library funcy doing what you need in a just one, short line.
from funcy import join_with
from pprint import pprint
d1 = {"key1": "x1", "key2": "y1"}
d2 = {"key1": "x2", "key2": "y2"}
list_of_dicts = [d1, d2]
merged_dict = join_with(tuple, list_of_dicts)
pprint(merged_dict)
Output:
{'key1': ('x1', 'x2'), 'key2': ('y1', 'y2')}
More info here: funcy -> join_with.
def merge(d1, d2, merge):
result = dict(d1)
for k,v in d2.iteritems():
if k in result:
result[k] = merge(result[k], v)
else:
result[k] = v
return result
d1 = {'a': 1, 'b': 2}
d2 = {'a': 1, 'b': 3, 'c': 2}
print merge(d1, d2, lambda x, y:(x,y))
{'a': (1, 1), 'c': 2, 'b': (2, 3)}
If keys are nested:
d1 = { 'key1': { 'nkey1': 'x1' }, 'key2': { 'nkey2': 'y1' } }
d2 = { 'key1': { 'nkey1': 'x2' }, 'key2': { 'nkey2': 'y2' } }
ds = [d1, d2]
d = {}
for k in d1.keys():
for k2 in d1[k].keys():
d.setdefault(k, {})
d[k].setdefault(k2, [])
d[k][k2] = tuple(d[k][k2] for d in ds)
yields:
{'key1': {'nkey1': ('x1', 'x2')}, 'key2': {'nkey2': ('y1', 'y2')}}
Modifying this answer to create a dictionary of tuples (what the OP asked for), instead of a dictionary of lists:
from collections import defaultdict
d1 = {1: 2, 3: 4}
d2 = {1: 6, 3: 7}
dd = defaultdict(tuple)
for d in (d1, d2): # you can list as many input dicts as you want here
for key, value in d.items():
dd[key] += (value,)
print(dd)
The above prints the following:
defaultdict(<class 'tuple'>, {1: (2, 6), 3: (4, 7)})
d1 ={'B': 10, 'C ': 7, 'A': 20}
d2 ={'B': 101, 'Y ': 7, 'X': 8}
d3 ={'A': 201, 'Y ': 77, 'Z': 8}
def CreateNewDictionaryAssemblingAllValues1(d1,d2,d3):
aa = {
k :[d[k] for d in (d1,d2,d3) if k in d ] for k in set(d1.keys() | d2.keys() | d3.keys() )
}
aap = print(aa)
return aap
CreateNewDictionaryAssemblingAllValues1(d1, d2, d3)
"""
Output :
{'X': [8], 'C ': [7], 'Y ': [7, 77], 'Z': [8], 'B': [10, 101], 'A': [20, 201]}
"""
From blubb answer:
You can also directly form the tuple using values from each list
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = (d1[k], d2[k])
This might be useful if you had a specific ordering for your tuples
ds = [d1, d2, d3, d4]
d = {}
for k in d1.keys():
d[k] = (d3[k], d1[k], d4[k], d2[k]) #if you wanted tuple in order of d3, d1, d4, d2
Using below method we can merge two dictionaries having same keys.
def update_dict(dict1: dict, dict2: dict) -> dict:
output_dict = {}
for key in dict1.keys():
output_dict.update({key: []})
if type(dict1[key]) != str:
for value in dict1[key]:
output_dict[key].append(value)
else:
output_dict[key].append(dict1[key])
if type(dict2[key]) != str:
for value in dict2[key]:
output_dict[key].append(value)
else:
output_dict[key].append(dict2[key])
return output_dict
Input: d1 = {key1: x1, key2: y1} d2 = {key1: x2, key2: y2}
Output: {'key1': ['x1', 'x2'], 'key2': ['y1', 'y2']}
dicts = [dict1,dict2,dict3]
out = dict(zip(dicts[0].keys(),[[dic[list(dic.keys())[key]] for dic in dicts] for key in range(0,len(dicts[0]))]))
A compact possibility
d1={'a':1,'b':2}
d2={'c':3,'d':4}
context={**d1, **d2}
context
{'b': 2, 'c': 3, 'd': 4, 'a': 1}
Consider two dictionaries:
dict1 = {'a': 35, 'b': 39, 'c': 20} # (with the values as integers)
dict2 = {'a': 23, 'c': 12}
I want to obtain the following:
dict_new = {'a': 0.657, 'c': 0.6} # (with the values as floats, as values of dict2/dict1)
You can get the common keys using dict2.keys() & dict1 and then just do the division:
dict1 = {'a':35, 'b': 39, 'c':20} #(with the values as integers)
dict2 = {'a':23, 'c':12}
d3 = {k: dict2[k] / dict1[k] for k in dict2.keys() & dict1}
If you want the values rounded to three decimal places use round(dict2[k] / dict1[k],3), if the keys from dict2 should always be in dict1 then you can simply iterate over the items of dict2:
d = {k:v / dict1[k] for k,v in dict2.items()}
dic_new = {}
for key in dic2.keys():
dic_new[key]=float(dict2[key])/dict1[key]