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I have two dictionaries as follows, I need to extract which strings in the tuple values are in one dictionary but not in other:
dict_a = {"s": ("mmmm", "iiiii", "p11"), "yyzz": ("oo", "i9")}
dict_b = {"s": ("mmmm",), "h": ("pp",), "g": ("rr",)}
The desired output:
{"s": ("iiiii", "p11"), "yyzz": ("oo", "i9")}
The order of the strings in the output doesn't matter.
One way that I tried to solve, but it doesn't produce the expected result:
>>> [item for item in dict_a.values() if item not in dict_b.values()]
[('mmmm', 'iiiii', 'p11'), ('oo', 'i9')]
If order doesn't matter, convert your dictionary values to sets, and subtract these:
{k: set(v) - set(dict_b.get(k, ())) for k, v in dict_a.items()}
The above takes all key-value pairs from dict_a, and for each such pair, outputs a new dictionary with those keys and a new value that's the set difference between the original value and the corresponding value from dict_b, if there is one:
>>> dict_a = {"s": ("mmmm", "iiiii", "p11"), "yyzz": ("oo", "i9")}
>>> dict_b = {"s": ("mmmm",), "h": ("pp",), "g": ("rr",)}
>>> {k: set(v) - set(dict_b.get(k, ())) for k, v in dict_a.items()}
{'s': {'p11', 'iiiii'}, 'yyzz': {'oo', 'i9'}}
The output will have sets, but these can be converted back to tuples if necessary:
{k: tuple(set(v) - set(dict_b.get(k, ()))) for k, v in dict_a.items()}
The dict_b.get(k, ()) call ensures there is always a tuple to give to set().
If you use the set.difference() method you don't even need to turn the dict_b value to a set:
{k: tuple(set(v).difference(dict_b.get(k, ()))) for k, v in dict_a.items()}
Demo of the latter two options:
>>> {k: tuple(set(v) - set(dict_b.get(k, ()))) for k, v in dict_a.items()}
{'s': ('p11', 'iiiii'), 'yyzz': ('oo', 'i9')}
>>> {k: tuple(set(v).difference(dict_b.get(k, ()))) for k, v in dict_a.items()}
{'s': ('p11', 'iiiii'), 'yyzz': ('oo', 'i9')}
{k: [v for v in vs if v not in dict_b.get(k, [])] for k,vs in dict_a.items()}
if you want to use tuples (or sets - just replace the cast)
{k: tuple(v for v in vs if v not in dict_b.get(k, [])) for k,vs in dict_a.items()}
Try this (see comments for explanations):
>>> out = {} # Initialise output dictionary
>>> for k, v in dict_a.items(): # Iterate through items of dict_a
... if k not in dict_b: # Check if the key is not in dict_b
... out[k] = v # If it isn't, add to out
... else: # Otherwise
... out[k] = tuple(set(v) - set(dict_b[k])) # Subtract sets to find the difference
...
>>> out
{'s': ('iiiii', 'p11'), 'yyzz': ('oo', 'i9')}
This can then be simplified using a dictionary comprehension:
>>> out = {k: tuple(set(v) - set(dict_b.get(k, ()))) for k, v in dict_a.items()}
>>> out
{'s': ('iiiii', 'p11'), 'yyzz': ('oo', 'i9')}
See this solution :
First iterate through all keys in dict_a
Check if the key is present or not in dict_b
Now, if present: take the tuples and iterate through dict_a's tuple(as per your question). now check weather that element is present or not in dict_b's tuple. If it is present just leave it. If it is not just add that element in tup_res.
Now after the for loop, add that key and tup value in the dict_res.
if the key is not present in dict_b simply add it in dict_res.
dict_a = {"s":("mmmm","iiiii","p11"), "yyzz":("oo","i9")}
dict_b = {"s":("mmmm"),"h":("pp",),"g":("rr",)}
dict_res = {}
for key in dict_a:
if key in dict_b:
tup_a = dict_a[key]
tup_b = dict_b[key]
tup_res = ()
for tup_ele in tup_a:
if tup_ele in tup_b:
pass
else:
tup_res = tup_res + (tup_ele,)
dict_res[key] = tup_res;
else:
dict_res[key] = dict_a[key];
print(dict_res)
It is giving correct output :)
I need to do a not "natural" operation on a dictionary so i wondering what is the best pythonic way to do this.
I need to simplify a dictionary by removing all the keys on it with the same value (keys are differents, values are the same)
For example:
Input:
dict = {key1 : [1,2,3], key2: [1,2,6], key3: [1,2,3]}
expected output:
{key1 : [1,2,3], key2:[1,2,6]}
I dont care about which key is delete (on the example: key1 or key3)
Exchange keys and values; duplicated key-value pairs will be removed as a side effect (because dictionary does not allow duplicated keys). Exchange keys and values again.
>>> d = {'key1': [1,2,3], 'key2': [1,2,6], 'key3': [1,2,3]}
>>> d2 = {tuple(v): k for k, v in d.items()} # exchange keys, values
>>> d = {v: list(k) for k, v in d2.items()} # exchange again
>>> d
{'key2': [1, 2, 6], 'key1': [1, 2, 3]}
NOTE: tuple(v) was used because list is not hashable; cannot be used as key directly.
BTW, don't use dict as a variable name. It will shadow builtin function/type dict.
This solution deletes the keys with same values without creating a new dictionary.
seen = set()
for key in mydict.keys():
value = tuple(mydict[key])
if value in seen:
del mydict[key]
else:
seen.add(value)
I think you can do it this way also. But I don't say as there seems to be more efficient ways. It is in-line.
for i in dictionary.keys():
if dictionary.values().count(dictionary[i]) > 1:
del dictionary[i]
You can iterate over your dict items and use a set to check what we have seen so far, deleting a key if we have already seen the value:
d = {"key1" : [1,2,3], "key2": [1,2,6], "key3": [1,2,3]}
seen = set()
for k, v in d.items(): # list(items) for python3
temp = tuple(v)
if temp in seen:
del d[k]
seen.add(temp)
print(d)
{'key1': [1, 2, 3], 'key2': [1, 2, 6]}
This will be more efficient that using creating a dict and reversing the values as you only have to cast to tuple once not from a tuple back to a list.
this worked for me:
seen = set()
for key in mydict.copy():
value = tuple(mydict[key])
if value in seen:
del mydict[key]
else:
seen.add(value)
What is a good 1-liner (other than putting this into a function) to achieve this:
# like dict1.update(dict2) but does not add any keys from dict2
# that are not already in dict1
for k in dict1:
if k in dict2:
dict1[k]=dict2[k]
I guess that this would work, but uses a "protected" function:
[dict1.__setitem__(k, dict2[k]) for k in dict1 if k in dict2]
dict1.update((k, dict2[k]) for k in set(dict2).intersection(dict1))
is how I'd do it in Python 2.6 or below (see further on how to do this in later versions).
Next to another mapping, dict.update() can also take an iterable of (key, value) tuples, which we generate based on the set intersection of the two dictionaries (so all keys they have in common).
Demo:
>>> dict1 = {'foo':'bar', 'ham': 'eggs'}
>>> dict2 = {'ham': 'spam', 'bar': 'baz'}
>>> dict1.update((k, dict2[k]) for k in set(dict2).intersection(dict1))
>>> dict1
{'foo': 'bar', 'ham': 'spam'}
In python 2.7 you can use the new Dict views to achieve the same without casting to sets:
dict1.update((k, dict2[k]) for k in dict1.viewkeys() & dict2.viewkeys())
In Python 3, dict views are the default, so you can instead spell this as:
dict1.update((k, dict2[k]) for k in dict1.keys() & dict2.keys())
Non-destructively:
dict((k, dict2.get(k, v)) for k, v in dict1.items())
Modifying dict1:
dict1.update((k, v) for k, v in dict2.items() if k in dict1)
Using map/zip
As two lines for readability:
match_keys = dict1.keys() & dict2.keys()
dict2.update(**dict(zip(match_keys, map(dict2.get, match_keys))))
Or as a one-liner:
dict2.update(**dict(zip(dict1.keys() & dict2.keys(), map(dict2.get, dict1.keys() & dict2.keys()))))
non-destrctively:
new_dict = {**dict2, **dict(zip(dict1.keys() & dict2.keys(), map(dict2.get, dict1.keys() & dict2.keys())))}
Code goes below:
d = {'a':0, 'b':0, 'c':0, 'd':0} #at the beginning, all the values are 0.
s = 'cbad' #a string
indices = map(s.index, d.keys()) #get every key's index in s, i.e., a-2, b-1, c-0, d-3
#then set the values to keys' index
d = dict(zip(d.keys(), indices)) #this is how I do it, any better way?
print d #{'a':2, 'c':0, 'b':1, 'd':3}
Any other way to do that?
PS. the code above is just a simple one to demonstrate my question.
Something like this might make your code more readable:
dict([(x,y) for y,x in enumerate('cbad')])
But you should give more details what you really want to do. Your code will probably fail if the characters in s do not fit the keys of d. So d is just a container for the keys and the values are not important. Why not start with a list in that case?
use update() method of dict:
d.update((k,s.index(k)) for k in d.iterkeys())
What about
d = {'a':0, 'b':0, 'c':0, 'd':0}
s = 'cbad'
for k in d.iterkeys():
d[k] = s.index(k)
? It's no functional programming anymore but should be more performant and more pythonic, perhaps :-).
EDIT: A function variant using python dict-comprehensions (needs Python 2.7+ or 3+):
d.update({k : s.index(k) for k in d.iterkeys()})
or even
{k : s.index(k) for k in d.iterkeys()}
if a new dict is okay!
for k in d.iterkeys():
d[k] = s.index[k]
Or, if you don't already know the letters in the string:
d = {}
for i in range(len(s)):
d[s[i]]=i
another one liner:
dict([(k,s.index(k)) for (k,v) in d.items()])
You don't need to pass a list of tuples to dict. Instead, you can use a dictionary comprehension with enumerate:
s = 'cbad'
d = {v: k for k, v in enumerate(s)}
If you need to process the intermediary steps, including initial setting of values, you can use:
d = dict.fromkeys('abcd', 0)
s = 'cbad'
indices = {v: k for k, v in enumerate(s)}
d = {k: indices[k] for k in d} # dictionary comprehension
d = dict(zip(d, map(indices.get, d))) # dict + zip alternative
print(d)
# {'a': 2, 'b': 1, 'c': 0, 'd': 3}
You choose the right way but think that no need to create dict and then modify it if you have ability to do this in the same time:
keys = ['a','b','c','d']
strK = 'bcad'
res = dict(zip(keys, (strK.index(i) for i in keys)))
Dict comprehension for python 2.7 and above
{key : indice for key, indice in zip(d.keys(), map(s.index, d.keys()))}
>>> d = {'a':0, 'b':0, 'c':0, 'd':0}
>>> s = 'cbad'
>>> for x in d:
d[x]=s.find(x)
>>> d
{'a': 2, 'c': 0, 'b': 1, 'd': 3}
This question already has answers here:
Reverse / invert a dictionary mapping
(32 answers)
Closed 10 months ago.
I receive a dictionary as input, and would like to to return a dictionary whose keys will be the input's values and whose value will be the corresponding input keys. Values are unique.
For example, say my input is:
a = dict()
a['one']=1
a['two']=2
I would like my output to be:
{1: 'one', 2: 'two'}
To clarify I would like my result to be the equivalent of the following:
res = dict()
res[1] = 'one'
res[2] = 'two'
Any neat Pythonic way to achieve this?
Python 2:
res = dict((v,k) for k,v in a.iteritems())
Python 3 (thanks to #erik):
res = dict((v,k) for k,v in a.items())
new_dict = dict(zip(my_dict.values(), my_dict.keys()))
From Python 2.7 on, including 3.0+, there's an arguably shorter, more readable version:
>>> my_dict = {'x':1, 'y':2, 'z':3}
>>> {v: k for k, v in my_dict.items()}
{1: 'x', 2: 'y', 3: 'z'}
You can make use of dict comprehensions:
Python 3
res = {v: k for k, v in a.items()}
Python 2
res = {v: k for k, v in a.iteritems()}
Edited: For Python 3, use a.items() instead of a.iteritems(). Discussions about the differences between them can be found in iteritems in Python on SO.
In [1]: my_dict = {'x':1, 'y':2, 'z':3}
Python 3
In [2]: dict((value, key) for key, value in my_dict.items())
Out[2]: {1: 'x', 2: 'y', 3: 'z'}
Python 2
In [2]: dict((value, key) for key, value in my_dict.iteritems())
Out[2]: {1: 'x', 2: 'y', 3: 'z'}
The current leading answer assumes values are unique which is not always the case. What if values are not unique? You will loose information!
For example:
d = {'a':3, 'b': 2, 'c': 2}
{v:k for k,v in d.iteritems()}
returns {2: 'b', 3: 'a'}.
The information about 'c' was completely ignored.
Ideally it should had be something like {2: ['b','c'], 3: ['a']}. This is what the bottom implementation does.
Python 2.x
def reverse_non_unique_mapping(d):
dinv = {}
for k, v in d.iteritems():
if v in dinv:
dinv[v].append(k)
else:
dinv[v] = [k]
return dinv
Python 3.x
def reverse_non_unique_mapping(d):
dinv = {}
for k, v in d.items():
if v in dinv:
dinv[v].append(k)
else:
dinv[v] = [k]
return dinv
You could try:
Python 3
d={'one':1,'two':2}
d2=dict((value,key) for key,value in d.items())
d2
{'two': 2, 'one': 1}
Python 2
d={'one':1,'two':2}
d2=dict((value,key) for key,value in d.iteritems())
d2
{'two': 2, 'one': 1}
Beware that you cannot 'reverse' a dictionary if
More than one key shares the same value. For example {'one':1,'two':1}. The new dictionary can only have one item with key 1.
One or more of the values is unhashable. For example {'one':[1]}. [1] is a valid value but not a valid key.
See this thread on the python mailing list for a discussion on the subject.
res = dict(zip(a.values(), a.keys()))
new_dict = dict( (my_dict[k], k) for k in my_dict)
or even better, but only works in Python 3:
new_dict = { my_dict[k]: k for k in my_dict}
Another way to expand on Ilya Prokin's response is to actually use the reversed function.
dict(map(reversed, my_dict.items()))
In essence, your dictionary is iterated through (using .items()) where each item is a key/value pair, and those items are swapped with the reversed function. When this is passed to the dict constructor, it turns them into value/key pairs which is what you want.
Suggestion for an improvement for Javier answer :
dict(zip(d.values(),d))
Instead of d.keys() you can write just d, because if you go through dictionary with an iterator, it will return the keys of the relevant dictionary.
Ex. for this behavior :
d = {'a':1,'b':2}
for k in d:
k
'a'
'b'
Can be done easily with dictionary comprehension:
{d[i]:i for i in d}
dict(map(lambda x: x[::-1], YourDict.items()))
.items() returns a list of tuples of (key, value). map() goes through elements of the list and applies lambda x:[::-1] to each its element (tuple) to reverse it, so each tuple becomes (value, key) in the new list spitted out of map. Finally, dict() makes a dict from the new list.
Hanan's answer is the correct one as it covers more general case (the other answers are kind of misleading for someone unaware of the duplicate situation). An improvement to Hanan's answer is using setdefault:
mydict = {1:a, 2:a, 3:b}
result = {}
for i in mydict:
result.setdefault(mydict[i],[]).append(i)
print(result)
>>> result = {a:[1,2], b:[3]}
Using loop:-
newdict = {} #Will contain reversed key:value pairs.
for key, value in zip(my_dict.keys(), my_dict.values()):
# Operations on key/value can also be performed.
newdict[value] = key
If you're using Python3, it's slightly different:
res = dict((v,k) for k,v in a.items())
Adding an in-place solution:
>>> d = {1: 'one', 2: 'two', 3: 'three', 4: 'four'}
>>> for k in list(d.keys()):
... d[d.pop(k)] = k
...
>>> d
{'two': 2, 'one': 1, 'four': 4, 'three': 3}
In Python3, it is critical that you use list(d.keys()) because dict.keys returns a view of the keys. If you are using Python2, d.keys() is enough.
I find this version the most comprehensive one:
a = {1: 'one', 2: 'two'}
swapped_a = {value : key for key, value in a.items()}
print(swapped_a)
output :
{'one': 1, 'two': 2}
An alternative that is not quite as readable (in my opinion) as some of the other answers:
new_dict = dict(zip(*list(zip(*old_dict.items()))[::-1]))
where list(zip(*old_dict.items()))[::-1] gives a list of 2 tuples, old_dict's values and keys, respectively.