Create a dictionary with comprehension

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))}

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

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

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}

>>> {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}

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}

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.

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}

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}