Create a dictionary with comprehension

Question

Can I use list comprehension syntax to create a dictionary?

For example, by iterating over pairs of keys and values:

d = {... for k, v in zip(keys, values)}

Answer 1

Use a dict comprehension (Python 2.7 and later):

{key: value for key, value in zip(keys, values)}

---

Alternatively, use the dict constructor (for str keys only):

pairs = [('a', 1), ('b', 2)]
dict(pairs)                          # → {'a': 1, 'b': 2}
dict((k, v + 10) for k, v in pairs)  # → {'a': 11, 'b': 12}

Given separate lists of keys and values, use the dict constructor with zip:

keys = ['a', 'b']
values = [1, 2]
dict(zip(keys, values))              # → {'a': 1, 'b': 2}

Answer 2

In Python 3 and Python 2.7+, dictionary comprehensions look like the below:

d = {k:v for k, v in iterable}

For Python 2.6 or earlier, see fortran's answer.

Answer 3

In fact, you don't even need to iterate over the iterable if it already comprehends some kind of mapping, the dict constructor doing it graciously for you:

>>> ts = [(1, 2), (3, 4), (5, 6)]
>>> dict(ts)
{1: 2, 3: 4, 5: 6}
>>> gen = ((i, i+1) for i in range(1, 6, 2))
>>> gen
<generator object <genexpr> at 0xb7201c5c>
>>> dict(gen)
{1: 2, 3: 4, 5: 6}

Answer 4

Create a dictionary with list comprehension in Python

I like the Python list comprehension syntax.

Can it be used to create dictionaries too? For example, by iterating over pairs of keys and values:

mydict = {(k,v) for (k,v) in blah blah blah}

You're looking for the phrase "dict comprehension" - it's actually:

mydict = {k: v for k, v in iterable}

Assuming blah blah blah is an iterable of two-tuples - you're so close. Let's create some "blahs" like that:

blahs = [('blah0', 'blah'), ('blah1', 'blah'), ('blah2', 'blah'), ('blah3', 'blah')]

Dict comprehension syntax:

Now the syntax here is the mapping part. What makes this a dict comprehension instead of a set comprehension (which is what your pseudo-code approximates) is the colon, : like below:

mydict = {k: v for k, v in blahs}

And we see that it worked, and should retain insertion order as-of Python 3.7:

>>> mydict
{'blah0': 'blah', 'blah1': 'blah', 'blah2': 'blah', 'blah3': 'blah'}

In Python 2 and up to 3.6, order was not guaranteed:

>>> mydict
{'blah0': 'blah', 'blah1': 'blah', 'blah3': 'blah', 'blah2': 'blah'}

Adding a Filter:

All comprehensions feature a mapping component and a filtering component that you can provide with arbitrary expressions.

So you can add a filter part to the end:

>>> mydict = {k: v for k, v in blahs if not int(k[-1]) % 2}
>>> mydict
{'blah0': 'blah', 'blah2': 'blah'}

Here we are just testing for if the last character is divisible by 2 to filter out data before mapping the keys and values.

Answer 5

In Python 2.7, it goes like:

>>> list1, list2 = ['a', 'b', 'c'], [1,2,3]
>>> dict( zip( list1, list2))
{'a': 1, 'c': 3, 'b': 2}

Zip them!

Answer 6

Python version >= 2.7, do the below:

d = {i: True for i in [1,2,3]}

Python version < 2.7_{^{(RIP, 3 July 2010 - 31 December 2019)}}, do the below:

d = dict((i,True) for i in [1,2,3])

Answer 7

To add onto @fortran's answer, if you want to iterate over a list of keys key_list as well as a list of values value_list:

d = dict((key, value) for (key, value) in zip(key_list, value_list))

or

d = {key: value for (key, value) in zip(key_list, value_list)}

Answer 8

Just to throw in another example. Imagine you have the following list:

nums = [4,2,2,1,3]

and you want to turn it into a dict where the key is the index and value is the element in the list. You can do so with the following line of code:

{index:nums[index] for index in range(0,len(nums))}

Answer 9

Here is another example of dictionary creation using dict comprehension:

What i am tring to do here is to create a alphabet dictionary where each pair; is the english letter and its corresponding position in english alphabet

>>> import string
>>> dict1 = {value: (int(key) + 1) for key, value in 
enumerate(list(string.ascii_lowercase))}
>>> dict1
{'a': 1, 'c': 3, 'b': 2, 'e': 5, 'd': 4, 'g': 7, 'f': 6, 'i': 9, 'h': 8, 
'k': 11, 'j': 10, 'm': 13, 'l': 12, 'o': 15, 'n': 14, 'q': 17, 'p': 16, 's': 
19, 'r': 18, 'u': 21, 't': 20, 'w': 23, 'v': 22, 'y': 25, 'x': 24, 'z': 26}
>>> 

Notice the use of enumerate here to get a list of alphabets and their indexes in the list and swapping the alphabets and indices to generate the key value pair for dictionary

Hope it gives a good idea of dictionary comp to you and encourages you to use it more often to make your code compact

Answer 10

This code will create dictionary using list comprehension for multiple lists with different values that can be used for pd.DataFrame()

#Multiple lists 
model=['A', 'B', 'C', 'D']
launched=[1983,1984,1984,1984]
discontinued=[1986, 1985, 1984, 1986]

#Dictionary with list comprehension
keys=['model','launched','discontinued']
vals=[model, launched,discontinued]
data = {key:vals[n] for n, key in enumerate(keys)}

#Convert dict to dataframe
df=pd.DataFrame(data)
display(df)

enumerate will pass n to vals to match each key with its list

Answer 11

Try this,

def get_dic_from_two_lists(keys, values):
    return { keys[i] : values[i] for i in range(len(keys)) }

Assume we have two lists country and capital

country = ['India', 'Pakistan', 'China']
capital = ['New Delhi', 'Islamabad', 'Beijing']

Then create dictionary from the two lists:

print get_dic_from_two_lists(country, capital)

The output is like this,

{'Pakistan': 'Islamabad', 'China': 'Beijing', 'India': 'New Delhi'}

Answer 12

Adding to @Ekhtiar answer, if you want to make look up dict from list, you can use this:

names = ['a', 'b', 'd', 'f', 'c']
names_to_id = {v:k for k, v in enumerate(names)}
# {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'f': 4}

Or in rare case that you want to filter duplicate, use set first (best in list of number):

names = ['a', 'b', 'd', 'f', 'd', 'c']
sorted_list = list(set(names))
sorted_list.sort()
names_to_id = {v:k for k, v in enumerate(sorted_list)}
# {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'f': 4}

names = [1,2,5,5,6,2,1]
names_to_id = {v:k for k, v in enumerate(set(names))}
# {1: 0, 2: 1, 5: 2, 6: 3}

Answer 13

>>> {k: v**3 for (k, v) in zip(string.ascii_lowercase, range(26))}

Python supports dict comprehensions, which allow you to express the creation of dictionaries at runtime using a similarly concise syntax.

A dictionary comprehension takes the form {key: value for (key, value) in iterable}. This syntax was introduced in Python 3 and backported as far as Python 2.7, so you should be able to use it regardless of which version of Python you have installed.

A canonical example is taking two lists and creating a dictionary where the item at each position in the first list becomes a key and the item at the corresponding position in the second list becomes the value.

The zip function used inside this comprehension returns an iterator of tuples, where each element in the tuple is taken from the same position in each of the input iterables. In the example above, the returned iterator contains the tuples (“a”, 1), (“b”, 2), etc.

Output:

{'i': 512, 'e': 64, 'o': 2744, 'h': 343, 'l': 1331, 's': 5832, 'b': 1, 'w': 10648, 'c': 8, 'x': 12167, 'y': 13824, 't': 6859, 'p': 3375, 'd': 27, 'j': 729, 'a': 0, 'z': 15625, 'f': 125, 'q': 4096, 'u': 8000, 'n': 2197, 'm': 1728, 'r': 4913, 'k': 1000, 'g': 216, 'v': 9261}

Answer 14

Yes, it's possible. In python, Comprehension can be used in List, Set, Dictionary, etc. You can write it this way

mydict = {k:v for (k,v) in blah}

Another detailed example of Dictionary Comprehension with the Conditional Statement and Loop:

parents = [father, mother]
            
parents = {parent:1 - P["mutation"] if parent in two_genes else 0.5 if parent in one_gene else P["mutation"] for parent in parents}

Answer 15

You can create a new dict for each pair and merge it with the previous dict:

reduce(lambda p, q: {**p, **{q[0]: q[1]}}, bla bla bla, {})

Obviously this approaches requires reduce from functools.

Answer 16

Assuming blah blah blah is a two-tuples list:

Let's see two methods:

# method 1
>>> lst = [('a', 2), ('b', 4), ('c', 6)]
>>> dict(lst)
{'a': 2, 'b': 4, 'c': 6}

# method 2
>>> lst = [('a', 2), ('b', 4), ('c', 6)]
>>> d = {k:v for k, v in lst}
>>> d
{'a': 2, 'b': 4, 'c': 6}

Answer 17

this approach uses iteration over the given date using a for loop.

Syntax: {key: value for (key, value) in data}

Eg:

# create a list comprehension with country and code:
    Country_code = [('China', 86), ('USA', 1),
            ('Ghana', 233), ('Uk', 44)]

# use iterable method to show results
{key: value for (key, value) in Country_code}