I'm using Python 3 and I'm trying to merge two dictionaries of dictionaries into one dictionary. I'm successful, however, when I add a new key/value pair to the new dictionary, it also adds it to both of the original dictionaries. I would like the original dictionaries to remain unchanged.
dict1 = {'key_val_1': {'a': '1', 'b': '2', 'c': '3'}}
dict2 = {'key_val_2': {'d': '4', 'e': '5', 'f': '6'}}
dict3 = dict1 | dict2
for x in dict3:
dict3[x]['g'] = '7'
The above code will append 'g': '7' to all 3 dictionaries and I only want to alter dict3. I have to assume that this is the intended behavior, but for the life of me I can't understand why (or how to get the desired results).
I believe the root of your problem is your assumption that when concatenating the two dictionaries dict1 and dict2 that python makes a copy of the dictionaries before concatenating them. In fact Python simply creates a new object with pointers to each of the parts. With this in mind, when you change the contents of a part of dict3 you are in reality changing the underlying dictionaries dict1 and dict2. To remedy this condition, you need to make copies of the underlying dictionaries before concatenating them or merge them rather than concatenating them.
Using the copy function:
from copy import deepcopy
dict3 = deepcopy(dict1) | deepcopy(dict2)
Now dict3 contains independent copies of dict1 and dict2
To merge the dicts:
from copy import copy
def merge(d1, d2):
rslt = dict()
for k in d1.keys():
rslt[k] = d1[k].copy() #Note still necessary to copy underlying dict
for k in d2.keys():
rslt[k] = d2[k].copy()
return rslt
then use:
dict3 = merge(dict1, dict2)
Issue you are having is because dict3 consists reference to the sub-dicts in dict1 and dict2. And the dict objects are mutable. So, when you change a dict in one place, it effects all the place where it is referenced. You can verify it by using the id() function. Example:
>>> print(id(dict1['key_val_1']))
>>> 140294429633472
>>> print(id(dict3['key_val_1']))
>>> 140294429633472
>>> print(id(dict2['key_val_2']))
>>> 140294429633728
>>> print(id(dict3['key_val_2']))
>>> 140294429633728
From above example you can verify that, the sub-dict in dict1 and dict2 are referenced in dict3. So, when you modify them in dict3 the orginial dicts are also modified, as dict are mutable.
So, to solve your issue, you need to make a deep copy of each sub-dict before merging them.
d = {'A': ['A11117',
'33465'
'17160144',
'A11-33465',
'3040',
'A11-33465 W1',
'nor'], 'B': ['maD', 'vern', 'first', 'A2lRights']}
I have a dictionary d and I would like to sort the values based on length of characters. For instance, for key A the value A11-33465 W1 would be first because it contains 12 characters followed by 'A11-33465' because it contains 9 characters etc. I would like this output:
d = {'A': ['A11-33465 W1',
' A11-33465',
'17160144',
'A11117',
'33465',
'3040',
'nor'],
'B': ['A2lRights',
'first',
'vern',
'maD']}
(I understand that dictionaries are not able to be sorted but I have examples below that didn't work for me but the answer contains a dictionary that was sorted)
I have tried the following
python sorting dictionary by length of values
print(' '.join(sorted(d, key=lambda k: len(d[k]), reverse=True)))
Sort a dictionary by length of the value
sorted_items = sorted(d.items(), key = lambda item : len(item[1]))
newd = dict(sorted_items[-2:])
How do I sort a dictionary by value?
import operator
sorted_x = sorted(d.items(), key=operator.itemgetter(1))
But they both do not give me what I am looking for.
How do I get my desired output?
You are not sorting the dict, you are sorting the lists inside it. The simplest will be a loop that sorts the lists in-place:
for k, lst in d.items():
lst.sort(key=len, reverse=True)
This will turn d into:
{'A': ['3346517160144', 'A11-33465 W1', 'A11-33465', 'A11117', '3040', 'nor'],
'B': ['A2lRights', 'first', 'vern', 'maD']}
If you want to keep the original data intact, use a comprehension like:
sorted_d = {k: sorted(lst, key=len, reverse=True) for k, lst in d.items()}
d = {'x': 1, 'y': 2, 'z': 3}
for key in d:
print(key, 'corresponds to', d[key])
How does Python recognize that it needs only to read the key from the dictionary? Is key a special keyword, or is it simply a variable?
key is just a variable name.
for key in d:
will simply loop over the keys in the dictionary, rather than the keys and values. To loop over both key and value you can use the following:
For Python 3.x:
for key, value in d.items():
For Python 2.x:
for key, value in d.iteritems():
To test for yourself, change the word key to poop.
In Python 3.x, iteritems() was replaced with simply items(), which returns a set-like view backed by the dict, like iteritems() but even better.
This is also available in 2.7 as viewitems().
The operation items() will work for both 2 and 3, but in 2 it will return a list of the dictionary's (key, value) pairs, which will not reflect changes to the dict that happen after the items() call. If you want the 2.x behavior in 3.x, you can call list(d.items()).
It's not that key is a special word, but that dictionaries implement the iterator protocol. You could do this in your class, e.g. see this question for how to build class iterators.
In the case of dictionaries, it's implemented at the C level. The details are available in PEP 234. In particular, the section titled "Dictionary Iterators":
Dictionaries implement a tp_iter slot that returns an efficient
iterator that iterates over the keys of the dictionary. [...] This
means that we can write
for k in dict: ...
which is equivalent to, but much faster than
for k in dict.keys(): ...
as long as the restriction on modifications to the dictionary
(either by the loop or by another thread) are not violated.
Add methods to dictionaries that return different kinds of
iterators explicitly:
for key in dict.iterkeys(): ...
for value in dict.itervalues(): ...
for key, value in dict.iteritems(): ...
This means that for x in dict is shorthand for for x in
dict.iterkeys().
In Python 3, dict.iterkeys(), dict.itervalues() and dict.iteritems() are no longer supported. Use dict.keys(), dict.values() and dict.items() instead.
Iterating over a dict iterates through its keys in no particular order, as you can see here:
(This is no longer the case in Python 3.6, but note that it's not guaranteed behaviour yet.)
>>> d = {'x': 1, 'y': 2, 'z': 3}
>>> list(d)
['y', 'x', 'z']
>>> d.keys()
['y', 'x', 'z']
For your example, it is a better idea to use dict.items():
>>> d.items()
[('y', 2), ('x', 1), ('z', 3)]
This gives you a list of tuples. When you loop over them like this, each tuple is unpacked into k and v automatically:
for k,v in d.items():
print(k, 'corresponds to', v)
Using k and v as variable names when looping over a dict is quite common if the body of the loop is only a few lines. For more complicated loops it may be a good idea to use more descriptive names:
for letter, number in d.items():
print(letter, 'corresponds to', number)
It's a good idea to get into the habit of using format strings:
for letter, number in d.items():
print('{0} corresponds to {1}'.format(letter, number))
key is simply a variable.
For Python2.X:
>>> d = {'x': 1, 'y': 2, 'z': 3}
>>> for my_var in d:
>>> print my_var, 'corresponds to', d[my_var]
x corresponds to 1
y corresponds to 2
z corresponds to 3
... or better,
d = {'x': 1, 'y': 2, 'z': 3}
for the_key, the_value in d.iteritems():
print the_key, 'corresponds to', the_value
For Python3.X:
d = {'x': 1, 'y': 2, 'z': 3}
for the_key, the_value in d.items():
print(the_key, 'corresponds to', the_value)
When you iterate through dictionaries using the for .. in ..-syntax, it always iterates over the keys (the values are accessible using dictionary[key]).
To iterate over key-value pairs, use the following:
for k,v in dict.iteritems() in Python 2
for k,v in dict.items() in Python 3
This is a very common looping idiom. in is an operator. For when to use for key in dict and when it must be for key in dict.keys() see David Goodger's Idiomatic Python article (archived copy).
I have a use case where I have to iterate through the dict to get the key, value pair, also the index indicating where I am. This is how I do it:
d = {'x': 1, 'y': 2, 'z': 3}
for i, (key, value) in enumerate(d.items()):
print(i, key, value)
Note that the parentheses around the key, value are important, without them, you'd get an ValueError "not enough values to unpack".
Iterating over dictionaries using 'for' loops
d = {'x': 1, 'y': 2, 'z': 3}
for key in d:
...
How does Python recognize that it needs only to read the key from the
dictionary? Is key a special word in Python? Or is it simply a
variable?
It's not just for loops. The important word here is "iterating".
A dictionary is a mapping of keys to values:
d = {'x': 1, 'y': 2, 'z': 3}
Any time we iterate over it, we iterate over the keys. The variable name key is only intended to be descriptive - and it is quite apt for the purpose.
This happens in a list comprehension:
>>> [k for k in d]
['x', 'y', 'z']
It happens when we pass the dictionary to list (or any other collection type object):
>>> list(d)
['x', 'y', 'z']
The way Python iterates is, in a context where it needs to, it calls the __iter__ method of the object (in this case the dictionary) which returns an iterator (in this case, a keyiterator object):
>>> d.__iter__()
<dict_keyiterator object at 0x7fb1747bee08>
We shouldn't use these special methods ourselves, instead, use the respective builtin function to call it, iter:
>>> key_iterator = iter(d)
>>> key_iterator
<dict_keyiterator object at 0x7fb172fa9188>
Iterators have a __next__ method - but we call it with the builtin function, next:
>>> next(key_iterator)
'x'
>>> next(key_iterator)
'y'
>>> next(key_iterator)
'z'
>>> next(key_iterator)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
When an iterator is exhausted, it raises StopIteration. This is how Python knows to exit a for loop, or a list comprehension, or a generator expression, or any other iterative context. Once an iterator raises StopIteration it will always raise it - if you want to iterate again, you need a new one.
>>> list(key_iterator)
[]
>>> new_key_iterator = iter(d)
>>> list(new_key_iterator)
['x', 'y', 'z']
Returning to dicts
We've seen dicts iterating in many contexts. What we've seen is that any time we iterate over a dict, we get the keys. Back to the original example:
d = {'x': 1, 'y': 2, 'z': 3}
for key in d:
If we change the variable name, we still get the keys. Let's try it:
>>> for each_key in d:
... print(each_key, '=>', d[each_key])
...
x => 1
y => 2
z => 3
If we want to iterate over the values, we need to use the .values method of dicts, or for both together, .items:
>>> list(d.values())
[1, 2, 3]
>>> list(d.items())
[('x', 1), ('y', 2), ('z', 3)]
In the example given, it would be more efficient to iterate over the items like this:
for a_key, corresponding_value in d.items():
print(a_key, corresponding_value)
But for academic purposes, the question's example is just fine.
For Iterating through dictionaries, The below code can be used.
dictionary= {1:"a", 2:"b", 3:"c"}
#To iterate over the keys
for key in dictionary.keys():
print(key)
#To Iterate over the values
for value in dictionary.values():
print(value)
#To Iterate both the keys and values
for key, value in dictionary.items():
print(key,'\t', value)
You can check the implementation of CPython's dicttype on GitHub. This is the signature of method that implements the dict iterator:
_PyDict_Next(PyObject *op, Py_ssize_t *ppos, PyObject **pkey,
PyObject **pvalue, Py_hash_t *phash)
CPython dictobject.c
To iterate over keys, it is slower but better to use my_dict.keys(). If you tried to do something like this:
for key in my_dict:
my_dict[key+"-1"] = my_dict[key]-1
it would create a runtime error because you are changing the keys while the program is running. If you are absolutely set on reducing time, use the for key in my_dict way, but you have been warned.
If you are looking for a clear and visual example:
cat = {'name': 'Snowy', 'color': 'White' ,'age': 14}
for key , value in cat.items():
print(key, ': ', value)
Result:
name: Snowy
color: White
age: 14
This will print the output in sorted order by values in ascending order.
d = {'x': 3, 'y': 1, 'z': 2}
def by_value(item):
return item[1]
for key, value in sorted(d.items(), key=by_value):
print(key, '->', value)
Output:
y -> 1
z -> 2
x -> 3
Let's get straight to the point. If the word key is just a variable, as you have mentioned then the main thing to note is that when you run a 'FOR LOOP' over a dictionary it runs through only the 'keys' and ignores the 'values'.
d = {'x': 1, 'y': 2, 'z': 3}
for key in d:
print (key, 'corresponds to', d[key])
rather try this:
d = {'x': 1, 'y': 2, 'z': 3}
for i in d:
print (i, 'corresponds to', d[i])
but if you use a function like:
d = {'x': 1, 'y': 2, 'z': 3}
print(d.keys())
in the above case 'keys' is just not a variable, its a function.
A dictionary in Python is a collection of key-value pairs. Each key is connected to a value, and you can use a key to access the value associated with that key. A key's value can be a number, a string, a list, or even another dictionary. In this case, threat each "key-value pair" as a separate row in the table: d is your table with two columns. the key is the first column, key[value] is your second column. Your for loop is a standard way to iterate over a table.
I am trying to write some code that involves creating a default dictionary of dictionaries. However, I have no idea how to initialise/create such a thing. My current attempt looks something like this:
from collections import defaultdict
inner_dict = {}
dict_of_dicts = defaultdict(inner_dict(int))
The use of this default dict of dictionaries is to for each pair of words that I produce from a file I open (e.g. [['M UH M', 'm oo m']] ), to set each segment of the first word delimited by empty space as a key in the outer dictionary, and then for each segment in the second word delimited by empty space count the frequency of that segment.
For example
[['M UH M', 'm oo m']]
(<class 'dict'>, {'M': {'m': 2}, 'UH': {'oo': 1}})
Having just run this now it doesn't seem to have output any errors, however I was just wondering if something like this will actually produce a default dictionary of dictionaries.
Apologies if this is a duplicate, however previous answers to these questions have been confusing and in a different context.
To initialise a defaultdict that creates dictionaries as its default value you would use:
d = defaultdict(dict)
For this particular problem, a collections.Counter would be more suitable
>>> from collections import defaultdict, Counter
>>> d = defaultdict(Counter)
>>> for a, b in zip(*[x.split() for x in ['M UH M', 'm oo m']]):
... d[a][b] += 1
>>> print(d)
defaultdict(collections.Counter,
{'M': Counter({'m': 2}), 'UH': Counter({'oo': 1})})
Edit
You expressed interest in a comment about the equivalent without a Counter. Here is the equivalent using a plain dict
>>> from collections import defaultdict
>>> d = defaultdict(dict)
>>> for a, b in zip(*[x.split() for x in ['M UH M', 'm oo m']]):
... d[a][b] = d[a].get(b, 0) + 1
>>> print(d)
defaultdict(dict, {'M': {'m': 2}, 'UH': {'oo': 1}})
You also could a use a normal dictionary and its setdefault method.
my_dict.setdefault(key, default) will look up my_dict[key] and ...
... if the key already exists, return its current value without modifying it, or ...
... assign the default value (my_dict[key] = default) and then return that.
So you can call my_dict.setdefault(key, {}) always when you want to get a value from your outer dictionary instead of the normal my_dict[key] to retrieve either the real value assigned with this key if it#s present, or to get a new empty dictionary as default value which gets automatically stored into your outer dictionary as well.
Example:
outer_dict = {"M": {"m": 2}}
inner_dict = d.setdefault("UH", {})
# outer_dict = {"M": {"m": 2}, "UH": {}}
# inner_dict = {}
inner_dict["oo"] = 1
# outer_dict = {"M": {"m": 2}, "UH": {"oo": 1}}
# inner_dict = {"oo": 1}
inner_dict = d.setdefault("UH", {})
# outer_dict = {"M": {"m": 2}, "UH": {"oo": 1}}
# inner_dict = {"oo": 1}
inner_dict["xy"] = 3
# outer_dict = {"M": {"m": 2}, "UH": {"oo": 1, "xy": 3}}
# inner_dict = {"oo": 1, "xy": 3}
This way you always get a valid inner_dict, either an empty default one or the one that's already present for the given key. As dictionaries are mutable data types, modifying the returned inner_dict will also modify the dictionary inside outer_dict.
The other answers propose alternative solutions or show you can make a default dictionary of dictionaries using d = defaultdict(dict)
but the question asked how to make a default dictionary of default dictionaries, my navie first attempt was this:
from collections import defaultdict
my_dict = defaultdict(defaultdict(list))
however this throw an error: *** TypeError: first argument must be callable or None
so my second attempt which works is to make a callable using the lambda key word to make an anonymous function:
from collections import defaultdict
my_dict = defaultdict(lambda: defaultdict(list))
which is more concise than the alternative method using a regular function:
from collections import defaultdict
def default_dict_maker():
return defaultdict(list)
my_dict = defaultdict(default_dict_maker)
you can check it works by assigning:
my_dict[2][3] = 5
my_dict[2][3]
>>> 5
or by trying to return a value:
my_dict[0][0]
>>> []
my_dict[5]
>>> defaultdict(<class 'list'>, {})
tl;dr
this is your oneline answer my_dict = defaultdict(lambda: defaultdict(list))