仅当较早的数字小于当前数字时，如何在 python 列表中找到数字的差异

Question

I have a list

my_list = [6, 9, 10, 0, 5]

how to write a code to get a list of all the differences of a number from it's earlier number only if it's greater than the earlier number

my_list[0] has no earlier number

my_list[1] = 9 which is greater than my_list[0] we can find the difference 9-6 = 3 . now answer_list=[3] .

my_list[2] = 10 which is greater than my_list[0] , my_list[1] answer_list = [3, 1, 4]

my_list[3] = 0 no previous number is less than 2 . Do nothing

my_list[4] = 5 . my_list[3] is less than my_list[4] . answer_list = [3, 1, 4, 5]

should return [3, 1, 4, 5]

my solution is

def ans(my_list):
    new_list = []
    for x in my_list:
        for i in range(my_list.index(x)):
            if x >= my_list[i]:
                diff = x - my_list[i]
                new_list.append(diff)
    return new_list

is there a better way to this, the nested loops is bit over kill and time consuming

Answer 1

You need two loops, but you can drop the need to use index if you use enumerate .

my_list = [6, 9, 10, 0, 5]

result = []
for index, value in enumerate(my_list[1:], start=1):
    for previous_value in my_list[:index]:
        difference = value - previous_value
        if difference > 0:
            result.append(difference)

print(result)

Bonus: your algorithm has problems if you have duplicate numbers in the orginal list, like [6, 9, 10, 9, 10, 0, 5] . This approach takes care of it. Your result: [3, 4, 1, 3, 4, 1, 5] , correct result: [3, 4, 1, 3, 4, 1, 1, 5]

---

Less nested, but less readable:

result = []
for index, value in enumerate(my_list[1:], start=1):
    result.extend(filter(lambda x: x > 0, (value - previous_value for previous_value in my_list[:index])))
print(result)