CODEWITHSUNDEEP
pythondictionary-comprehension
Otto Rodríguez
Otto Rodríguez

Reputation: 11

Simplified Dictionary using comprehensions

everyone i'd like to know a way to simplify the following code with a dictcomprehension:

import random

vowel = "aeiou"
consonants = "bcdfghjklmnpqrstvwxyz"
hand = {}

for i in range(numVowels):
        x = VOWELS[random.randrange(0,len(VOWELS))]
        hand[x] = hand.get(x, 0) + 1
        
for i in range(numVowels, n):    
        x = CONSONANTS[random.randrange(0,len(CONSONANTS))]
        hand[x] = hand.get(x, 0) + 1

Upvotes: 0

Views: 61

Answers (1)

TheNightHawk
TheNightHawk

Reputation: 515

No. As far as I know, list and dict comprehensions don't work when you want to get the value of elements in the middle of the comprehension. This can cause unexpected outputs.

The for loop changes the contents of the dict with each iteration, while a dict comprehension first creates the entire dict, and then assigns it to the var.

Here's an example with a list comprehension:

# for loop
>>> l = [1] * 20
>>> for i in range(1, 20):
...     l[i] = l[i - 1] + l[i]
...
>>> l
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]

# list comprehension
>>> l = [1] * 20
>>> l = [ l[i - 1] + l[1] for i in range(1, 20) ]
>>> l
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

And a dict comprehension

# for loop
>>> s = "abcdefghij"
>>> d = {}
>>> for i in range(10):
...     d[s[i]] = d.get(s[i - 1], 0) + 1
...
>>> d
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10}


# dict comprehension
>>> s = "abcdefghij"
>>> d = {}
>>> d = {s[i]: d.get(s[i - 1], 0) + 1 for i in range(10)}
>>> d
{'a': 1, 'b': 1, 'c': 1, 'd': 1, 'e': 1, 'f': 1, 'g': 1, 'h': 1, 'i': 1, 'j': 1}

Upvotes: 1

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