Dict comprehension Python

user_17696360 0
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I want to create a dict with dict comprehension where self.students[i] are some names and j should be 0,2,4,6,8.

Right now x returns

{'Patricia': 6, 'Roger': 6, 'Samantha': 6, 'Xander': 6}

What I want it to return is

{'Patricia': 0, 'Roger': 2, 'Samantha': 4, 'Xander': 6}`

Here is the code

x = {self.students[i] : j for i in range(0,len(self.students)) for j in range(0,len(self.students)+3, 2)}
dictionary python
6 Answers

Use enumerate:

x = {student : i * 2 for i, student in enumerate(self.students)}
user_54598390

You don't need to use range when looping over a list, you can just loop over the list itself. Also, you don't need to use a 2nd loop for j, you can just calculate your number based off of the index in the list.

To get the indexes as you loop, you can use enumerate.

x = {student:i*2 for i,student in enumerate(self.students)}
user_2064030

Your current code is looping through every student every time.

Hence on the last loop, when j = 6, all of the values are changed to 6.

user_133696930

This works

students = ['Ivan', 'Masha', 'Sasha']
students_dict = {i: k*2 for k, i in enumerate(students)}
print(students_dict)

What you were doing was looping twice. This is used for making 2 dimensional lists. To understand your old code better, see here

user_128711470

zip the range with the list to get pairs of values, instead of using nested for loops (which are overwriting each value in turn across the entire dict, leaving you with only the last value of j):

>>> students = ['Patricia', 'Roger', 'Samantha', 'Xander']
>>> {s: j for s, j in zip(students, range(0, len(students) * 2, 2))}
{'Patricia': 0, 'Roger': 2, 'Samantha': 4, 'Xander': 6}
user_37997590

You could use zip to build the dictionary from its keys combined with an even number counter (from itertools):

x = dict(zip(self.students,itertools.count(0,2)))
user_52375600
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