Consider a dict like

mydict = {
  'Apple': {'American':'16', 'Mexican':10, 'Chinese':5},
  'Grapes':{'Arabian':'25','Indian':'20'} }

How do I access for instance a particular element of this dictionary? for instance, I would like to print the first element after some formatting the first element of Apple which in our case is ‘American’ only?

Additional information The above data structure was created by parsing an input file in a python function. Once created however it remains the same for that run.

I am using this data structure in my function.

So if the file changes, the next time this application is run the contents of the file are different and hence the contents of this data structure will be different but the format would be the same. So you see I in my function I don’t know that the first element in Apple is ‘American’ or anything else so I can’t directly use ‘American’ as a key.

Given that it is a dictionary you access it by using the keys. Getting the dictionary stored under “Apple”, do the following:

>>> mydict["Apple"]
{'American': '16', 'Mexican': 10, 'Chinese': 5}

And getting how many of them are American (16), do like this:

>>> mydict["Apple"]["American"]
'16'

If the questions is, if I know that I have a dict of dicts that contains ‘Apple’ as a fruit and ‘American’ as a type of apple, I would use:

myDict = {'Apple': {'American':'16', 'Mexican':10, 'Chinese':5},
          'Grapes':{'Arabian':'25','Indian':'20'} }


print myDict['Apple']['American']

as others suggested. If instead the questions is, you don’t know whether ‘Apple’ as a fruit and ‘American’ as a type of ‘Apple’ exist when you read an arbitrary file into your dict of dict data structure, you could do something like:

print [ftype['American'] for f,ftype in myDict.iteritems() if f == 'Apple' and 'American' in ftype]

or better yet so you don’t unnecessarily iterate over the entire dict of dicts if you know that only Apple has the type American:

if 'Apple' in myDict:
    if 'American' in myDict['Apple']:
        print myDict['Apple']['American']

In all of these cases it doesn’t matter what order the dictionaries actually store the entries. If you are really concerned about the order, then you might consider using an OrderedDict:

http://docs.python.org/dev/library/collections.html#collections.OrderedDict

As I noticed your description, you just know that your parser will give you a dictionary that its values are dictionary too like this:

sampleDict = {
              "key1": {"key10": "value10", "key11": "value11"},
              "key2": {"key20": "value20", "key21": "value21"}
              }

So you have to iterate over your parent dictionary. If you want to print out or access all first dictionary keys in sampleDict.values() list, you may use something like this:

for key, value in sampleDict.items():
    print value.keys()[0]

If you want to just access first key of the first item in sampleDict.values(), this may be useful:

print sampleDict.values()[0].keys()[0]

If you use the example you gave in the question, I mean:

sampleDict = {
              'Apple': {'American':'16', 'Mexican':10, 'Chinese':5},
              'Grapes':{'Arabian':'25','Indian':'20'}
              }

The output for the first code is:

American
Indian

And the output for the second code is:

American

EDIT 1:

Above code examples does not work for version 3 and above of python; since from version 3, python changed the type of output of methods keys and values from list to dict_values. Type dict_values is not accepting indexing, but it is iterable. So you need to change above codes as below:

First One:

for key, value in sampleDict.items():
    print(list(value.keys())[0])

Second One:

print(list(list(sampleDict.values())[0].keys())[0])

As a bonus, I’d like to offer kind of a different solution to your issue. You seem to be dealing with nested dictionaries, which is usually tedious, especially when you have to check for existence of an inner key.

There are some interesting libraries regarding this on pypi, here is a quick search for you.

In your specific case, dict_digger seems suited.

>>> import dict_digger
>>> d = {
  'Apple': {'American':'16', 'Mexican':10, 'Chinese':5},
  'Grapes':{'Arabian':'25','Indian':'20'} 
}

>>> print(dict_digger.dig(d, 'Apple','American'))
16
>>> print(dict_digger.dig(d, 'Grapes','American'))
None

You can use dict['Apple'].keys()[0] to get the first key in the Apple dictionary, but there’s no guarantee that it will be American. The order of keys in a dictionary can change depending on the contents of the dictionary and the order the keys were added.

I know this is 8 years old, but no one seems to have actually read and answered the question.

You can call .values() on a dict to get a list of the inner dicts and thus access them by index.

>>> mydict = {
...  'Apple': {'American':'16', 'Mexican':10, 'Chinese':5},
...  'Grapes':{'Arabian':'25','Indian':'20'} }

>>>mylist = list(mydict.values())
>>>mylist[0]
{'American':'16', 'Mexican':10, 'Chinese':5},
>>>mylist[1]
{'Arabian':'25','Indian':'20'}

>>>myInnerList1 = list(mylist[0].values())
>>>myInnerList1
['16', 10, 5]
>>>myInnerList2 = list(mylist[1].values())
>>>myInnerList2
['25', '20']

You can’t rely on order on dictionaries. But you may try this:

dict['Apple'].items()[0][0]

If you want the order to be preserved you may want to use this: http://www.python.org/dev/peps/pep-0372/#ordered-dict-api

Simple Example to understand how to access elements in the dictionary:-

Create a Dictionary

d = {'dog' : 'bark', 'cat' : 'meow' } 
print(d.get('cat'))
print(d.get('lion'))
print(d.get('lion', 'Not in the dictionary'))
print(d.get('lion', 'NA'))
print(d.get('dog', 'NA'))

Explore more about Python Dictionaries and learn interactively here…

Few people appear, despite the many answers to this question, to have pointed out that dictionaries are un-ordered mappings, and so (until the blessing of insertion order with Python 3.7) the idea of the “first” entry in a dictionary literally made no sense. And even an OrderedDict can only be accessed by numerical index using such uglinesses as mydict[mydict.keys()[0]] (Python 2 only, since in Python 3 keys() is a non-subscriptable iterator.)

From 3.7 onwards and in practice in 3,6 as well – the new behaviour was introduced then, but not included as part of the language specification until 3.7 – iteration over the keys, values or items of a dict (and, I believe, a set also) will yield the least-recently inserted objects first. There is still no simple way to access them by numerical index of insertion.

As to the question of selecting and “formatting” items, if you know the key you want to retrieve in the dictionary you would normally use the key as a subscript to retrieve it (my_var = mydict['Apple']).

If you really do want to be able to index the items by entry number (ignoring the fact that a particular entry’s number will change as insertions are made) then the appropriate structure would probably be a list of two-element tuples. Instead of

mydict = {
  'Apple': {'American':'16', 'Mexican':10, 'Chinese':5},
  'Grapes':{'Arabian':'25','Indian':'20'} }

you might use:

mylist = [
    ('Apple', {'American':'16', 'Mexican':10, 'Chinese':5}),
    ('Grapes', {'Arabian': '25', 'Indian': '20'}
]

Under this regime the first entry is mylist[0] in classic list-endexed form, and its value is ('Apple', {'American':'16', 'Mexican':10, 'Chinese':5}). You could iterate over the whole list as follows:

for (key, value) in mylist:  # unpacks to avoid tuple indexing
    if key == 'Apple':
        if 'American' in value:
            print(value['American'])

but if you know you are looking for the key “Apple”, why wouldn’t you just use a dict instead?

You could introduce an additional level of indirection by cacheing the list of keys, but the complexities of keeping two data structures in synchronisation would inevitably add to the complexity of your code.

With the following small function, digging into a tree-shaped dictionary becomes quite easy:

def dig(tree, path):
    for key in path.split("."):
        if isinstance(tree, dict) and tree.get(key):
            tree = tree[key]
        else:
            return None
    return tree

Now, dig(mydict, "Apple.Mexican") returns 10, while dig(mydict, "Grape") yields the subtree {'Arabian':'25','Indian':'20'}. If a key is not contained in the dictionary, dig returns None.

Note that you can easily change (or even parameterize) the separator char from ‘.’ to ‘/’, ‘|’ etc.

I have a dict and would like to remove all the keys for which there are empty value strings.

metadata = {u'Composite:PreviewImage': u'(Binary data 101973 bytes)',
            u'EXIF:CFAPattern2': u''}

What is the best way to do this?

Python 2.X

dict((k, v) for k, v in metadata.iteritems() if v)

Python 2.7 – 3.X

{k: v for k, v in metadata.items() if v is not None}

Note that all of your keys have values. It’s just that some of those values are the empty string. There’s no such thing as a key in a dict without a value; if it didn’t have a value, it wouldn’t be in the dict.

It can get even shorter than BrenBarn’s solution (and more readable I think)

{k: v for k, v in metadata.items() if v}

Tested with Python 2.7.3.

If you really need to modify the original dictionary:

empty_keys = [k for k,v in metadata.iteritems() if not v]
for k in empty_keys:
    del metadata[k]

Note that we have to make a list of the empty keys because we can’t modify a dictionary while iterating through it (as you may have noticed). This is less expensive (memory-wise) than creating a brand-new dictionary, though, unless there are a lot of entries with empty values.

BrenBarn’s solution is ideal (and pythonic, I might add). Here is another (fp) solution, however:

from operator import itemgetter
dict(filter(itemgetter(1), metadata.items()))

If you want a full-featured, yet succinct approach to handling real-world data structures which are often nested, and can even contain cycles, I recommend looking at the remap utility from the boltons utility package.

After pip install boltons or copying iterutils.py into your project, just do:

from boltons.iterutils import remap

drop_falsey = lambda path, key, value: bool(value)
clean = remap(metadata, visit=drop_falsey)

This page has many more examples, including ones working with much larger objects from Github’s API.

It’s pure-Python, so it works everywhere, and is fully tested in Python 2.7 and 3.3+. Best of all, I wrote it for exactly cases like this, so if you find a case it doesn’t handle, you can bug me to fix it right here.

Based on Ryan’s solution, if you also have lists and nested dictionaries:

For Python 2:

def remove_empty_from_dict(d):
    if type(d) is dict:
        return dict((k, remove_empty_from_dict(v)) for k, v in d.iteritems() if v and remove_empty_from_dict(v))
    elif type(d) is list:
        return [remove_empty_from_dict(v) for v in d if v and remove_empty_from_dict(v)]
    else:
        return d

For Python 3:

def remove_empty_from_dict(d):
    if type(d) is dict:
        return dict((k, remove_empty_from_dict(v)) for k, v in d.items() if v and remove_empty_from_dict(v))
    elif type(d) is list:
        return [remove_empty_from_dict(v) for v in d if v and remove_empty_from_dict(v)]
    else:
        return d

If you have a nested dictionary, and you want this to work even for empty sub-elements, you can use a recursive variant of BrenBarn’s suggestion:

def scrub_dict(d):
    if type(d) is dict:
        return dict((k, scrub_dict(v)) for k, v in d.iteritems() if v and scrub_dict(v))
    else:
        return d

Quick Answer (TL;DR)

Example01

### example01 -------------------

mydict  =   { "alpha":0,
              "bravo":"0",
              "charlie":"three",
              "delta":[],
              "echo":False,
              "foxy":"False",
              "golf":"",
              "hotel":"   ",                        
            }
newdict =   dict([(vkey, vdata) for vkey, vdata in mydict.iteritems() if(vdata) ])
print newdict

### result01 -------------------
result01 ='''
{'foxy': 'False', 'charlie': 'three', 'bravo': '0'}
'''

Detailed Answer

Problem

• Context: Python 2.x
• Scenario: Developer wishes modify a dictionary to exclude blank values
  • aka remove empty values from a dictionary
  • aka delete keys with blank values
  • aka filter dictionary for non-blank values over each key-value pair

Solution

• example01 use python list-comprehension syntax with simple conditional to remove “empty” values

Pitfalls

• example01 only operates on a copy of the original dictionary (does not modify in place)
• example01 may produce unexpected results depending on what developer means by “empty”
  • Does developer mean to keep values that are falsy?
  • If the values in the dictionary are not gauranteed to be strings, developer may have unexpected data loss.
  • result01 shows that only three key-value pairs were preserved from the original set

Alternate example

• example02 helps deal with potential pitfalls
• The approach is to use a more precise definition of “empty” by changing the conditional.
• Here we only want to filter out values that evaluate to blank strings.
• Here we also use .strip() to filter out values that consist of only whitespace.

Example02

### example02 -------------------

mydict  =   { "alpha":0,
              "bravo":"0",
              "charlie":"three",
              "delta":[],
              "echo":False,
              "foxy":"False",
              "golf":"",
              "hotel":"   ",
            }
newdict =   dict([(vkey, vdata) for vkey, vdata in mydict.iteritems() if(str(vdata).strip()) ])
print newdict

### result02 -------------------
result02 ='''
{'alpha': 0,
  'bravo': '0', 
  'charlie': 'three', 
  'delta': [],
  'echo': False,
  'foxy': 'False'
  }
'''

See also

• list-comprehension
• falsy
• checking for empty string
• modifying original dictionary in place
• dictionary comprehensions
• pitfalls of checking for empty string

For python 3

dict((k, v) for k, v in metadata.items() if v)

Building on the answers from patriciasz and nneonneo, and accounting for the possibility that you might want to delete keys that have only certain falsy things (e.g. '') but not others (e.g. 0), or perhaps you even want to include some truthy things (e.g. 'SPAM'), then you could make a highly specific hitlist:

unwanted = ['', u'', None, False, [], 'SPAM']

Unfortunately, this doesn’t quite work, because for example 0 in unwanted evaluates to True. We need to discriminate between 0 and other falsy things, so we have to use is:

any([0 is i for i in unwanted])

…evaluates to False.

Now use it to del the unwanted things:

unwanted_keys = [k for k, v in metadata.items() if any([v is i for i in unwanted])]
for k in unwanted_keys: del metadata[k]

If you want a new dictionary, instead of modifying metadata in place:

newdict = {k: v for k, v in metadata.items() if not any([v is i for i in unwanted])}

I read all replies in this thread and some referred also to this thread: Remove empty dicts in nested dictionary with recursive function

I originally used solution here and it worked great:

Attempt 1: Too Hot (not performant or future-proof):

def scrub_dict(d):
    if type(d) is dict:
        return dict((k, scrub_dict(v)) for k, v in d.iteritems() if v and scrub_dict(v))
    else:
        return d

But some performance and compatibility concerns were raised in Python 2.7 world:

1. use isinstance instead of type
2. unroll the list comp into for loop for efficiency
3. use python3 safe items instead of iteritems

Attempt 2: Too Cold (Lacks Memoization):

def scrub_dict(d):
    new_dict = {}
    for k, v in d.items():
        if isinstance(v,dict):
            v = scrub_dict(v)
        if not v in (u'', None, {}):
            new_dict[k] = v
    return new_dict

DOH! This is not recursive and not at all memoizant.

Attempt 3: Just Right (so far):

def scrub_dict(d):
    new_dict = {}
    for k, v in d.items():
        if isinstance(v,dict):
            v = scrub_dict(v)
        if not v in (u'', None, {}):
            new_dict[k] = v
    return new_dict

Dicts mixed with Arrays

• The answer at Attempt 3: Just Right (so far) from BlissRage’s answer does not properly handle arrays elements. I’m including a patch in case anyone needs it. The method is handles list with the statement block of if isinstance(v, list):, which scrubs the list using the original scrub_dict(d) implementation.

    @staticmethod
    def scrub_dict(d):
        new_dict = {}
        for k, v in d.items():
            if isinstance(v, dict):
                v = scrub_dict(v)
            if isinstance(v, list):
                v = scrub_list(v)
            if not v in (u'', None, {}):
                new_dict[k] = v
        return new_dict

    @staticmethod
    def scrub_list(d):
        scrubbed_list = []
        for i in d:
            if isinstance(i, dict):
                i = scrub_dict(i)
            scrubbed_list.append(i)
        return scrubbed_list

An alternative way you can do this, is using dictionary comprehension. This should be compatible with 2.7+

result = {
    key: value for key, value in
    {"foo": "bar", "lorem": None}.items()
    if value
}

Here is an option if you are using pandas:

import pandas as pd

d = dict.fromkeys(['a', 'b', 'c', 'd'])
d['b'] = 'not null'
d['c'] = ''  # empty string

print(d)

# convert `dict` to `Series` and replace any blank strings with `None`;
# use the `.dropna()` method and
# then convert back to a `dict`
d_ = pd.Series(d).replace('', None).dropna().to_dict()

print(d_)

Some of Methods mentioned above ignores if there are any integers and float with values 0 & 0.0

If someone wants to avoid the above can use below code(removes empty strings and None values from nested dictionary and nested list):

def remove_empty_from_dict(d):
    if type(d) is dict:
        _temp = {}
        for k,v in d.items():
            if v == None or v == "":
                pass
            elif type(v) is int or type(v) is float:
                _temp[k] = remove_empty_from_dict(v)
            elif (v or remove_empty_from_dict(v)):
                _temp[k] = remove_empty_from_dict(v)
        return _temp
    elif type(d) is list:
        return [remove_empty_from_dict(v) for v in d if( (str(v).strip() or str(remove_empty_from_dict(v)).strip()) and (v != None or remove_empty_from_dict(v) != None))]
    else:
        return d

“As I also currently write a desktop application for my work with Python, I found in data-entry application when there is lots of entry and which some are not mandatory thus user can left it blank, for validation purpose, it is easy to grab all entries and then discard empty key or value of a dictionary. So my code above a show how we can easy take them out, using dictionary comprehension and keep dictionary value element which is not blank. I use Python 3.8.3

data = {'':'', '20':'', '50':'', '100':'1.1', '200':'1.2'}

dic = {key:value for key,value in data.items() if value != ''}

print(dic)

{'100': '1.1', '200': '1.2'}

Some benchmarking:

1. List comprehension recreate dict

In [7]: %%timeit dic = {str(i):i for i in xrange(10)}; dic['10'] = None; dic['5'] = None
   ...: dic = {k: v for k, v in dic.items() if v is not None} 
   1000000 loops, best of 7: 375 ns per loop

2. List comprehension recreate dict using dict()

In [8]: %%timeit dic = {str(i):i for i in xrange(10)}; dic['10'] = None; dic['5'] = None
   ...: dic = dict((k, v) for k, v in dic.items() if v is not None)
1000000 loops, best of 7: 681 ns per loop

3. Loop and delete key if v is None

In [10]: %%timeit dic = {str(i):i for i in xrange(10)}; dic['10'] = None; dic['5'] = None
    ...: for k, v in dic.items():
    ...:   if v is None:
    ...:     del dic[k]
    ...: 
10000000 loops, best of 7: 160 ns per loop

so loop and delete is the fastest at 160ns, list comprehension is half as slow at ~375ns and with a call to dict() is half as slow again ~680ns.

Wrapping 3 into a function brings it back down again to about 275ns. Also for me PyPy was about twice as fast as neet python.

I want to write a custom class that behaves like dict – so, I am inheriting from dict.

My question, though, is: Do I need to create a private dict member in my __init__() method?. I don’t see the point of this, since I already have the dict behavior if I simply inherit from dict.

Can anyone point out why most of the inheritance snippets look like the one below?

class CustomDictOne(dict):
   def __init__(self):
      self._mydict = {} 

   # other methods follow

Instead of the simpler…

class CustomDictTwo(dict):
   def __init__(self):
      # initialize my other stuff here ...

   # other methods follow

Actually, I think I suspect the answer to the question is so that users cannot directly access your dictionary (i.e. they have to use the access methods that you have provided).

However, what about the array access operator []? How would one implement that? So far, I have not seen an example that shows how to override the [] operator.

So if a [] access function is not provided in the custom class, the inherited base methods will be operating on a different dictionary?

I tried the following snippet to test out my understanding of Python inheritance:

class myDict(dict):
    def __init__(self):
        self._dict = {}

    def add(self, id, val):
        self._dict[id] = val


md = myDict()
md.add('id', 123)
print md[id]

I got the following error:

KeyError: < built-in function id>

What is wrong with the code above?

How do I correct the class myDict so that I can write code like this?

md = myDict()
md['id'] = 123

[Edit]

I have edited the code sample above to get rid of the silly error I made before I dashed away from my desk. It was a typo (I should have spotted it from the error message).

Check the documentation on emulating container types. In your case, the first parameter to add should be self.

class Mapping(dict):

    def __setitem__(self, key, item):
        self.__dict__[key] = item

    def __getitem__(self, key):
        return self.__dict__[key]

    def __repr__(self):
        return repr(self.__dict__)

    def __len__(self):
        return len(self.__dict__)

    def __delitem__(self, key):
        del self.__dict__[key]

    def clear(self):
        return self.__dict__.clear()

    def copy(self):
        return self.__dict__.copy()

    def has_key(self, k):
        return k in self.__dict__

    def update(self, *args, **kwargs):
        return self.__dict__.update(*args, **kwargs)

    def keys(self):
        return self.__dict__.keys()

    def values(self):
        return self.__dict__.values()

    def items(self):
        return self.__dict__.items()

    def pop(self, *args):
        return self.__dict__.pop(*args)

    def __cmp__(self, dict_):
        return self.__cmp__(self.__dict__, dict_)

    def __contains__(self, item):
        return item in self.__dict__

    def __iter__(self):
        return iter(self.__dict__)

    def __unicode__(self):
        return unicode(repr(self.__dict__))


o = Mapping()
o.foo = "bar"
o['lumberjack'] = 'foo'
o.update({'a': 'b'}, c=44)
print 'lumberjack' in o
print o

In [187]: run mapping.py
True
{'a': 'b', 'lumberjack': 'foo', 'foo': 'bar', 'c': 44}

Like this

class CustomDictOne(dict):
   def __init__(self,*arg,**kw):
      super(CustomDictOne, self).__init__(*arg, **kw)

Now you can use the built-in functions, like dict.get() as self.get().

You do not need to wrap a hidden self._dict. Your class already is a dict.

For the sake of completeness, here is the link to the documentation mentioned by @björn-pollex for the latest Python 2.x (2.7.7 as of the time of writing):

Emulating Container Types

(Sorry for not using the comments function, I’m just not allowed to do so by stackoverflow.)

The problem with this chunk of code:

class myDict(dict):
    def __init__(self):
        self._dict = {}

    def add(id, val):
        self._dict[id] = val


md = myDict()
md.add('id', 123)

…is that your ‘add’ method (…and any method you want to be a member of a class) needs to have an explicit ‘self’ declared as its first argument, like:

def add(self, 'id', 23):

To implement the operator overloading to access items by key, look in the docs for the magic methods __getitem__ and __setitem__.

Note that because Python uses Duck Typing, there may actually be no reason to derive your custom dict class from the language’s dict class — without knowing more about what you’re trying to do (e.g, if you need to pass an instance of this class into some code someplace that will break unless isinstance(MyDict(), dict) == True), you may be better off just implementing the API that makes your class sufficiently dict-like and stopping there.

Here is an alternative solution:

class AttrDict(dict):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        self.__dict__ = self

a = AttrDict()
a.a = 1
a.b = 2

I really don’t see the right answer to this anywhere

class MyClass(dict):
    
    def __init__(self, a_property):
        self[a_property] = a_property

All you are really having to do is define your own __init__ – that really is all that there is too it.

Another example (little more complex):

class MyClass(dict):

    def __init__(self, planet):
        self[planet] = planet
        info = self.do_something_that_returns_a_dict()
        if info:
            for k, v in info.items():
                self[k] = v

    def do_something_that_returns_a_dict(self):
        return {"mercury": "venus", "mars": "jupiter"}

This last example is handy when you want to embed some kind of logic.

Anyway… in short class GiveYourClassAName(dict) is enough to make your class act like a dict. Any dict operation you do on self will be just like a regular dict.

This is my best solution. I used this many times.

class DictLikeClass:
    ...
    def __getitem__(self, key):
        return getattr(self, key)

    def __setitem__(self, key, value):
        setattr(self, key, value)
    ...

You can use like:

>>> d = DictLikeClass()
>>> d["key"] = "value"
>>> print(d["key"])

Don’t inherit from Python built-in dict, ever! for example update method woldn’t use __setitem__, they do a lot for optimization. Use UserDict.

from collections import UserDict

class MyDict(UserDict):
    def __delitem__(self, key):
        pass
    def __setitem__(self, key, value):
        pass

What is the best way to remove an item from a dictionary by value, i.e. when the item’s key is unknown? Here’s a simple approach:

for key, item in some_dict.items():
    if item is item_to_remove:
        del some_dict[key]

Are there better ways? Is there anything wrong with mutating (deleting items) from the dictionary while iterating it?

Be aware that you’re currently testing for object identity (is only returns True if both operands are represented by the same object in memory – this is not always the case with two object that compare equal with ==). If you are doing this on purpose, then you could rewrite your code as

some_dict = {key: value for key, value in some_dict.items() 
             if value is not value_to_remove}

But this may not do what you want:

>>> some_dict = {1: "Hello", 2: "Goodbye", 3: "You say yes", 4: "I say no"}
>>> value_to_remove = "You say yes"
>>> some_dict = {key: value for key, value in some_dict.items() if value is not value_to_remove}
>>> some_dict
{1: 'Hello', 2: 'Goodbye', 3: 'You say yes', 4: 'I say no'}
>>> some_dict = {key: value for key, value in some_dict.items() if value != value_to_remove}
>>> some_dict
{1: 'Hello', 2: 'Goodbye', 4: 'I say no'}

So you probably want != instead of is not.

The dict.pop(key[, default]) method allows you to remove items when you know the key. It returns the value at the key if it removes the item otherwise it returns what is passed as default. See the docs.’

Example:

>>> dic = {'a':1, 'b':2}
>>> dic
{'a': 1, 'b': 2}
>>> dic.pop('c', 0)
0
>>> dic.pop('a', 0)
1
>>> dic
{'b': 2}

a = {'name': 'your_name','class': 4}
if 'name' in a: del a['name']

A simple comparison between del and pop():

import timeit
code = """
results = {'A': 1, 'B': 2, 'C': 3}
del results['A']
del results['B']
"""
print timeit.timeit(code, number=100000)
code = """
results = {'A': 1, 'B': 2, 'C': 3}
results.pop('A')
results.pop('B')
"""
print timeit.timeit(code, number=100000)

result:

0.0329667857143
0.0451040902256

So, del is faster than pop().

items() returns a list, and it is that list you are iterating, so mutating the dict in the loop doesn’t matter here. If you were using iteritems() instead, mutating the dict in the loop would be problematic, and likewise for viewitems() in Python 2.7.

I can’t think of a better way to remove items from a dict by value.

I’d build a list of keys that need removing, then remove them. It’s simple, efficient and avoids any problem about simultaneously iterating over and mutating the dict.

keys_to_remove = [key for key, value in some_dict.iteritems()
                  if value == value_to_remove]
for key in keys_to_remove:
    del some_dict[key]

c is the new dictionary, and a is your original dictionary, {‘z’,’w’} are the keys you want to remove from a

c = {key:a[key] for key in a.keys() - {'z', 'w'}}

Also check: https://www.safaribooksonline.com/library/view/python-cookbook-3rd/9781449357337/ch01.html

y={'username':'admin','machine':['a','b','c']}
if 'c' in y['machine'] : del y['machine'][y['machine'].index('c')]

There is nothing wrong with deleting items from the dictionary while iterating, as you’ve proposed. Be careful about multiple threads using the same dictionary at the same time, which may result in a KeyError or other problems.

Of course, see the docs at http://docs.python.org/library/stdtypes.html#typesmapping

This is how I would do it.

for key in some_dict.keys():
    if some_dict[key] == item_to_remove:
        some_dict.pop(key)
        break

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())