sorted python lambda
lst = [('candy','30','100'), ('apple','10','200'), ('baby','20','300')]
lst.sort(key=lambda x:x[1])
print(lst)sorted python lambda
lst = [('candy','30','100'), ('apple','10','200'), ('baby','20','300')]
lst.sort(key=lambda x:x[1])
print(lst)python sort based off lambda
a = sorted(a, key=lambda x: x.modified, reverse=True)sorted list python
sorted(iterable, key=None, reverse=False)
type(sorted(iterable, key=None, reverse=False)) = listpython lambda key sort
>>> student_tuples = [
... ('john', 'A', 15),
... ('jane', 'B', 12),
... ('dave', 'B', 10),
... ]
>>> sorted(student_tuples, key=lambda student: student[2]) # sort by age
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]python sort
nums = [4,8,5,2,1]
#1 sorted() (Returns sorted list)
sorted_nums = sorted(nums)
print(sorted_nums)#[1,2,4,5,8]
print(nums)#[4,8,5,2,1]
#2 .sort() (Changes original list)
nums.sort()
print(nums)#[1,2,4,5,8]python sorted lambda
a = ["tim", "bob", "anna", "steve", "john"]
# sorts the list by the first letter of each name
b = sorted(a, key=lambda x : x[0])
# x = each of the values in the list a
# sorts the list by length FIRST, then alphabetical order SECOND
c = sorted(a, key=lambda x : (len(x), x))Copyright © 2021 Codeinu
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