Sorting a nested list using lambda

Question

Given a nested list representing gradebooks of different courses:

gradebooks = [[['Troy', 92], ['Alice', 95]], [['James', 89], ['Charles', 100], ['Bryn', 59]]]

Using the builtin sorted() function, write code to sort courses by course mean. The expected output is

[[['James', 89], ['Charles', 100], ['Bryn', 59]], [['Troy', 92], ['Alice', 95]]]

Using the builtin sorted() function, write code to sort students of each course by score in descending order. The expected output is

[[['Alice', 95], ['Troy', 92]], [['Charles', 100], ['James', 89], ['Bryn', 59]]]

I'm having trouble with these questions, I tried to answer the first question using codes below but it doesnt work.

sort1 = sorted(gradebooks, key=lambda x : sum(gradebooks[x]) / len(gradebooks[x]))

Yi Zhao · Accepted Answer · 2020-04-18 08:05:03Z

1

First, the reason your code does not work is:

sort1 = sorted(gradebooks, key=lambda x : sum(gradebooks[x]) / len(gradebooks[x]))

The parameter key of sorted function is a function that takes a list element as arguments. So x would be [['Troy', 92], ['Alice', 95]] or [['Charles', 100], ['James', 89], ['Bryn', 59]]. Therefore, gradebooks[x] will cause error:

TypeError: list indices must be integers or slices, not list

For the first problem, your lambda function should be:

sort1 = sorted(gradebooks, key=lambda x: sum([grade for _, grade in x]) / len(x))

For the second problem, my thoughts are: overall it is a list comprehension, in the list comprehension, use sorted function to process each element.

sort2 = [sorted(course, key=lambda x: x[1], reverse=True) for course in gradebooks]

answered Apr 18, 2020 at 8:05

Yi Zhao

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Shubham Sharma · Accepted Answer · 2020-04-18 07:55:32Z

0

Use:

# sort courses by course mean
sort1 = sorted(gradebooks, key=lambda s: sum(dict(s).values()) / len(s))

# sort students of each course by score in descending order. 
sort2 = list(map(lambda s: sorted(s, reverse=True, key=lambda y: y[1]),  gradebooks))

print(sort1)
print(sort2)

This prints:

[[['James', 89], ['Charles', 100], ['Bryn', 59]], [['Troy', 92], ['Alice', 95]]]
[[['Alice', 95], ['Troy', 92]], [['Charles', 100], ['James', 89], ['Bryn', 59]]]

answered Apr 18, 2020 at 7:55

Shubham Sharma

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