在Python的重复列表中跟踪值更改 [英] Track value changes in a repetitive list in Python

问题描述

我有一个包含重复值的列表，如下所示:

I have a list with repeating values as shown below:

x = [1, 1, 1, 2, 2, 2, 1, 1, 1]

此列表是从匹配正则表达式的模式(此处未显示)生成的.该列表保证具有重复值(很多重复，很多-数百个，如果不是数千个)，并且永远不会随机排列，因为正则表达式每次都匹配.

This list is generated from a pattern matching regular expression (not shown here). The list is guaranteed to have repeating values (many, many repeats - hundreds, if not thousands), and is never randomly arranged because that's what the regex is matching each time.

我想要的是 跟踪条目从前一个值开始更改的列表索引 .因此对于上面的列表x，我想获取一个更改跟踪列表[3, 6]，该列表指示x[3]和x[6]与列表中的先前条目不同.

What I want is to track the list indices at which the entries change from the previous value. So for the above list x, I want to obtain a change-tracking list [3, 6] indicating that x[3] and x[6] are different from their previous entries in the list.

我设法做到了，但是我想知道是否有更清洁的方法.这是我的代码:

I managed to do this, but I was wondering if there was a cleaner way. Here's my code:

x = [1, 1, 1, 2, 2, 2, 1, 1, 1]

flag = []
for index, item in enumerate(x):
    if index != 0:
        if x[index] != x[index-1]:
            flag.append(index)

print flag

输出:[3, 6]

问题:是否有一种更简洁的方法可以用更少的代码行来完成我想做的事情?

Question: Is there a cleaner way to do what I want, in fewer lines of code?

推荐答案

可以使用列表理解功能和range函数

It can be done using a list comprehension, with a range function

>>> x = [1, 1, 1, 2, 2, 2, 3, 3, 3]
>>> [i for i in range(1,len(x)) if x[i]!=x[i-1] ]
[3, 6]
>>> x = [1, 1, 1, 2, 2, 2, 1, 1, 1]
>>> [i for i in range(1,len(x)) if x[i]!=x[i-1] ]
[3, 6]

这篇关于在Python的重复列表中跟踪值更改的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！