使用 For 循环编辑列表中的后续值

Question

我有以下名为“test_list”的列表：

test_list = [1.0, -1.0, -1.0, -1.0, -1.0, 1.0, 1.0, 1.0, -1.0, -1.0, 1.0, 
 1.0, -1.0, 1.0, -1.0, 1.0, 1.0, 1.0, 1.0, -1.0, 1.0, -1.0, -1.0, -1.0,
-1.0, 1.0, -1.0, -1.0, -1.0, -1.0, 1.0, 1.0, 1.0, 1.0, 1.0, -1.0, -1.0, 1.0, 
-1.0, -1.0, -1.0, -1.0, 1.0, -1.0, -1.0, -1.0, -1.0, -1.0, -1.0, 1.0, -1.0, 1.0, 
-1.0, -1.0, -1.0, -1.0, 1.0]

我想遍历列表，并且：

1. 如果列表元素 = 1；
2. 我想将列表中接下来的 5 个元素的值设置为 0；考虑等待期。

为此，我创建了以下辅助函数：

def adjust_for_wait_period(series_list, wait_period=5):
    
    for i in series_list:
        if i == 1:
            index = series_list.index(i)
            wait_period_list = list(range(1, (wait_period + 1)))
            for wait in wait_period_list:
                try:
                    series_list[index + wait] = 0
                except Exception as e:
                    pass
                
    return series_list

当我执行函数时，它只遍历列表的第一个元素并输出这个列表：

[1.0, 0, 0, 0, 0, 0, 1.0, 1.0, -1.0, -1.0, 1.0, 1.0, -1.0, 
 1.0, -1.0, 1.0, 1.0, 1.0, 1.0, -1.0, 1.0, -1.0, -1.0, -1.0, 
-1.0, 1.0, -1.0, -1.0, -1.0, -1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 
-1.0, -1.0, 1.0, -1.0, -1.0, -1.0, -1.0, 1.0, -1.0, -1.0, -1.0, 
-1.0, -1.0, -1.0, 1.0, -1.0, 1.0, -1.0, -1.0, -1.0, -1.0, 1.0]

而不是这个列表：

[1.0, 0, 0, 0, 0, 0, 1.0, 0, 0, 0, 0, 0, -1.0, 
 1.0, 0, 0, 0, 0, 0, -1.0, 1.0, 0, 0, 0, 
 0, 0, -1.0, -1.0, -1.0, -1.0, 1.0, 0, 0, 0, 0, 
 0, -1.0, 0, 0, 0, 0, 0, 1.0, 0, 0, 0, 
 0, 0, -1.0, 1.0, 0, 0, 0, 0, 0, -1.0, 1.0]

这意味着我的函数只遍历第一组值，而不遍历列表的其余部分。

我哪里出错了？

Answer 1

您的代码的错误确实是提到的index = series_list.index(i)@Adam.Er8。这是一种使用计数器查看剩余等待时间的方法。请注意，此方法会修改给定的列表，因此它会覆盖值。如果您需要之前和之后的值，则必须对其进行一些修改。

def adjust_for_wait_period(series_list, waiting_period=5):
    waiting_period_left = 0
    for i in range(len(series_list)):
        if waiting_period_left > 0:
            series_list[i] = 0.0
            waiting_period_left -= 1
        else:
            if series_list[i] == 1.0:   
                waiting_period_left = waiting_period

test_list = [1.0, -1.0, -1.0, -1.0, -1.0, 1.0, 1.0, 1.0, -1.0, -1.0, 1.0, 
 1.0, -1.0, 1.0, -1.0, 1.0, 1.0, 1.0, 1.0, -1.0, 1.0, -1.0, -1.0, -1.0,
-1.0, 1.0, -1.0, -1.0, -1.0, -1.0, 1.0, 1.0, 1.0, 1.0, 1.0, -1.0, -1.0, 1.0, 
-1.0, -1.0, -1.0, -1.0, 1.0, -1.0, -1.0, -1.0, -1.0, -1.0, -1.0, 1.0, -1.0, 1.0, 
-1.0, -1.0, -1.0, -1.0, 1.0]

adjust_for_wait_period(test_list)

print(test_list)

Answer 2

def adjust_for_wait_period(series_list, wait_period=5):
    for idx, item in enumerate(series_list):
        if item == 1:
            for offset in range(wait_period):
                try:
                    series_list[idx + offset + 1] = 0
                except Exception as e:
                    pass
                
    return series_list

test_list = [1.0, -1.0, -1.0, -1.0, -1.0, 1.0, 1.0, 1.0, -1.0, -1.0, 1.0, 
 1.0, -1.0, 1.0, -1.0, 1.0, 1.0, 1.0, 1.0, -1.0, 1.0, -1.0, -1.0, -1.0,
-1.0, 1.0, -1.0, -1.0, -1.0, -1.0, 1.0, 1.0, 1.0, 1.0, 1.0, -1.0, -1.0, 1.0, 
-1.0, -1.0, -1.0, -1.0, 1.0, -1.0, -1.0, -1.0, -1.0, -1.0, -1.0, 1.0, -1.0, 1.0, 
-1.0, -1.0, -1.0, -1.0, 1.0]

print(adjust_for_wait_period(test_list))

Answer 3

正如评论中提到的，您总是使用 index() 找到相同的 first -1 值 ...您可以使用以下方法更改：

test_list = [1.0, -1.0, -1.0, -1.0, -1.0, 1.0, 1.0, 1.0, -1.0, -1.0, 1.0, 1.0, -1.0,
            1.0, -1.0, 1.0, 1.0, 1.0, 1.0, -1.0, 1.0, -1.0, -1.0, -1.0, -1.0, 1.0, -1.0,
            -1.0, -1.0, -1.0, 1.0, 1.0, 1.0, 1.0, 1.0, -1.0, -1.0, 1.0, -1.0, -1.0, -1.0,
            -1.0, 1.0, -1.0, -1.0, -1.0, -1.0, -1.0, -1.0, 1.0, -1.0, 1.0, -1.0, -1.0,
            -1.0, -1.0, 1.0]

def adjust_for_wait_period(series_list, wait_period=5):
  # changing the iterating list is fine because we do not lengthen/shorten it.
  # if you change the list-length by removing/adding elements you shoud enumerate
  # over a copy of it!
  for idx, value in enumerate(series_list):
    if value == 1.0:  # float equality tests are error prone, should be fine here
        # slice assignment: replaces exact amount of items needed even if < wait_period
        l[idx+1:idx+1+wait_period] = [0 for _ in range(len(l[idx+1:idx+1+wait_period]))] 
  return l
    
print(test_list, len(test_list))
tl = adjust_for_wait_period(test_list)
print(tl, len(tl))

输出：

# input
[1.0, -1.0, -1.0, -1.0, -1.0, 1.0, 1.0, 1.0, -1.0, -1.0, 1.0, 1.0, -1.0, 1.0, -1.0, 1.0, 
 1.0, 1.0, 1.0, -1.0, 1.0, -1.0, -1.0, -1.0, -1.0, 1.0, -1.0, -1.0, -1.0, -1.0, 1.0, 
 1.0, 1.0, 1.0, 1.0, -1.0, -1.0, 1.0, -1.0, -1.0, -1.0, -1.0, 1.0, -1.0, -1.0, -1.0, 
 -1.0, -1.0, -1.0, 1.0, -1.0, 1.0, -1.0, -1.0, -1.0, -1.0, 1.0] 57

# output
[1.0, 0, 0, 0, 0, 0, 1.0, 0, 0, 0, 0, 0, -1.0, 1.0, 0, 0, 0, 0, 0, -1.0, 1.0, 0, 0, 
 0, 0, 0, -1.0, -1.0, -1.0, -1.0, 1.0, 0, 0, 0, 0, 0, -1.0, 1.0, 0, 0, 0, 0, 0, -1.0, 
-1.0, -1.0, -1.0, -1.0, -1.0, 1.0, 0, 0, 0, 0, 0, -1.0, 1.0] 57

此解决方案在列表中使用slice-assignment，这应该会导致替换成本更低，然后逐个进行替换 - 但更改的值总量仍然相同。