How to get the next i that meets the condition under the if not condition in Python？-CodePudding

My problem is a bit difficult to describe, I want to use an example to show my problem.

For example, each number corresponds to a corresponding value,

a=[0,3,5]
b=[0,1,2,3,4,5,6,7]
for i in range(b):
  if i not in a:
    if ***value of i < value of next i that meet the condition***
      a.append(i)
    else:
      a.append(***next i that meet the condition***)

I want to write code like this but the problem is that I don't know how to express the next i that meet the condition. Simply use i 1 is definitely wrong. Can somebody help me? Thank you very much, guys!!

CodePudding user response：

Adjusting the range for your loop makes it easier to access the current and next element of b:

a = [0, 3, 5]
b = [0, 1, 12, 2, 16, 3, 18, 4, 20, 5, 22]

for index in range(1, len(b) - 1):
    current_item = b[index - 1]
    next_item = b[index]

    if current_item in a:
        continue

    if current_item < next_item:
        a.append(current_item)
    else:
        continue

print(a)

Out:

[0, 3, 5, 1, 2, 4]

CodePudding user response：

You didn't mention about additional restrictions so there are no checking if our lists are empty or contain just 1 element.

You can consider using this:

a = [0, 3, 5]
b = [0, 1, 2, 3, 4, 5, 6, 7]

for i in range(len(b) - 1):
    if (b[i] < b[i   1]) and b[i] not in a:
        a.append(b[i])

print(a)

And this is another version if you don't want to use "i 1":

a = [0, 3, 5]
b = [0, 1, 2, 3, 4, 5, 6, 7]
for n, m in zip(b[:-2], b[1:]):
    if n < m and n not in a:
        a.append(n)

print(a)